Matrices and Determinants

Applying, ${C_1} \to {C_1} + {C_2} + {C_3}\,\,\,$ we get

$f\left( x \right) = \left| {\matrix{ {1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x} & {\left( {1 + {b^2}} \right)x} & {\left( {1 + {c^2}} \right)x} \cr {1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x} & {1 + {b^2}x} & {\left( {1 + {c^2}x} \right)} \cr {1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x} & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \cr } } \right|$

$ = \left| {\matrix{ 1 & {\left( {1 + {b^2}} \right)x} & {\left( {1 + {c^2}} \right)x} \cr 1 & {1 + {b^2}x} & {\left( {1 + {c^2}x} \right)} \cr 1 & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \cr } } \right|$

$\left[ \, \right.$ As given that ${a^2} + {b^2} + {c^2} = - 2$ $\left. {} \right]$

$\therefore$ ${a^2} + {b^2} + {c^2} + 2 = 0$

Applying ${R_1} \to {R_1} - {R_2},\,\,\,{R_2} \to {R_2} - {R_3}$

$\therefore$ $f\left( x \right) = \left| {\matrix{ 0 & {x - 1} & 0 \cr 0 & {1 - x} & {x - 1} \cr 1 & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \cr } } \right|$

$f\left( x \right) = {\left( {x - 1} \right)^2}$

Hence degree $=2.$

Given that $10B$ $\,\,\, = \left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right]$

$ \Rightarrow B = {1 \over {10}}\left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right]$

Also since, $B = {A^{ - 1}} \Rightarrow AB = I$

$ \Rightarrow {1 \over {10}}\left[ {\matrix{ 1 & { - 1} & 1 \cr 2 & 1 & { - 3} \cr 1 & 1 & 1 \cr } } \right]\left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$

$ \Rightarrow {1 \over {10}}\left[ {\matrix{ {10} & 0 & {5 - 2} \cr 0 & {10} & { - 5 + \alpha } \cr 0 & 0 & {5 + \alpha } \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$

$ \Rightarrow {{5 - \alpha } \over {10}} = 0$

$ \Rightarrow \alpha = 5$

$\left| {\matrix{ {\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr {\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr {\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr } } \right|$

$ = \left| {\matrix{ {\log {a_1}r{}^{n - 1}} & {\log {a_1}r{}^n} & {\log {a_1}r{}^{n + 1}} \cr {\log {a_1}r{}^{n + 2}} & {\log {a_1}r{}^{n + 3}} & {\log {a_1}r{}^{n + 4}} \cr {\log {a_1}r{}^{n + 5}} & {\log {a_1}r{}^{n + 6}} & {\log {a_1}r{}^{n + 7}} \cr } } \right|$

$ = \left| {\matrix{ {\log {a_1} + \left( {n - 1} \right)\log r} & {\log {a_1} + n\log r} & {\log {a_1} + \left( {n + 1} \right)\log r} \cr {\log {a_1} + \left( {n + 2} \right)\log r} & {\log {a_1} + \left( {n + 3} \right)\log r} & {\log {a_1} + \left( {n + 4} \right)\log r} \cr {\log {a_1} + \left( {n + 5} \right)\log r} & {\log {a_1} + \left( {n + 6} \right)\log r} & {\log {a_1} + \left( {n + 7} \right)\log r} \cr } } \right|$

$ = 0\left[ \, \right.$ Apply $\,\,\,\,{c_2} \to {c_2} - {1 \over 2}{c_1} - {1 \over 2}{c_3}\,\left. \, \right]$

$A = \left[ {\matrix{ 0 & 0 & { - 1} \cr 0 & { - 1} & 0 \cr { - 1} & 0 & 0 \cr } } \right]$

clearly $\,\,\,A \ne 0.\,$ Also $\,\,\left| A \right| = - 1 \ne 0$

$\therefore$ ${A^{ - 1}}\,\,$ exists, further

$\left( { - 1} \right)I = \left[ {\matrix{ { - 1} & 0 & 0 \cr 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr } } \right] \ne A$

Also ${A^2} = \left[ {\matrix{ 0 & 0 & { - 1} \cr 0 & { - 1} & 0 \cr { - 1} & 0 & 0 \cr } } \right]\left[ {\matrix{ 0 & 0 & { - 1} \cr 0 & { - 1} & 0 \cr { - 1} & 0 & 0 \cr } } \right]$

$ = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] = I$

$\Delta = \left| {\matrix{ 1 & {{\omega ^n}} & {{\omega ^{2n}}} \cr {{\omega ^n}} & {{\omega ^{2n}}} & 1 \cr {{\omega ^{2n}}} & 1 & {{\omega ^n}} \cr } } \right|$

$ = 1\left( {{\omega ^{3n}} - 1} \right) - {\omega ^n}\left( {{\omega ^{2n}} - {\omega ^{2n}}} \right) + $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^{2n}}\left( {{\omega ^n} - {\omega ^{4n}}} \right)$

$ = {\omega ^{3n}} - 1 - 0 + {\omega ^{3n}} - {\omega ^{6n}}$

$ = 1 - 1 + 1 - 1 = 0$ $\left[ {} \right.$ as $\,\,\,\,\,$ ${\omega ^{3n}} = 1$ $\left. {} \right]$

For homogeneous system of equations to have non zero solution, $\Delta = 0$

$\left| {\matrix{ 1 & {2a} & a \cr 1 & {3b} & b \cr 1 & {4c} & c \cr } } \right| = 0\,{C_2} \to {C_2} - 2{C_3}$

$\left| {\matrix{ 1 & 0 & a \cr 1 & b & b \cr 1 & {2c} & c \cr } } \right| = 0\,\,{R_3} \to {R_3} - {R_2},{R_2} \to {R_2} - {R_1}$

$\left| {\matrix{ 1 & 0 & a \cr 0 & b & {b - a} \cr 0 & {2c - b} & {c - b} \cr } } \right| = 0$

$b\left( {c - b} \right) - \left( {b - a} \right)\left( {2c - b} \right) = 0$

On simplification, ${2 \over b} = {1 \over a} + {1 \over c}$

$\therefore$ $a,b,c$ are in Harmonic Progression.

${A^2} = \left[ {\matrix{ \alpha & \beta \cr \beta & \alpha \cr } } \right] = \left[ {\matrix{ a & b \cr b & a \cr } } \right]\left[ {\matrix{ a & b \cr b & a \cr } } \right]$

$ = \left[ {\matrix{ {{a^2} + {b^2}} & {2ab} \cr {2ab} & {{a^2} + {b^2}} \cr } } \right]$

$\alpha = {a^2} + {b^2};\,\,\beta = 2ab$

We have $\left| {\matrix{ a & b & {ax + b} \cr b & c & {bx + c} \cr {ax + b} & {bx + c} & 0 \cr } } \right|$

By $\,\,\,{R_3} \to {R_3} - \left( {x{R_1} + {R_2}} \right);$

$ = \left| {\matrix{ a & b & {ax + b} \cr b & c & {bx + c} \cr 0 & 0 & { - \left( {a{x^2} + 2bx + C} \right)} \cr } } \right|$

$ = \left( {a{x^2} + 2bx + c} \right)\left( {{b^2} - ac} \right)$

$ = \left( + \right)\left( - \right) = - ve.$

Let us consider the matrices given:

$ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $

$ P = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} $

$ Q = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix} $

$ R = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $

where $ x, y, z, a, b, c, d $ are non-zero real numbers. We aim to verify properties of matrix $ Q $ given the relation $ Q R = R P $.

Steps and Calculations

Matrix Equation Analysis:

Given $ Q R = R P $, we equate both sides:

$ \begin{pmatrix} a x + c y & b x + d y \\ a z + 4c & b z + 4d \end{pmatrix} = \begin{pmatrix} 2a & 3b \\ 2c & 3d \end{pmatrix} $

From the matrix equality, we derive the following key equations:

$ a x + c y = 2a $

$ b x + d y = 3b $

$ a z + 4c = 2c $

$ b z + 4d = 3d $

Determinant Condition:

Comparing the equations:

Set $ a z = -2c $ and $ b z = -d $, resulting in $ \frac{a}{b} = \frac{2c}{d} $.

The requirement $ |R| \neq 0 $ implies $ |P| = |Q| $. So, equating determinants gives us:

$ |P| = |Q| \Rightarrow 6 = 4x - yz \quad \text{(Equation 1)} $

Condition for Non-Trivial Solutions:

From:

$ a(x - z) + c y = 0 $

$ a z + 2c = 0 $

implies a non-trivial solution matrix must have:

$ \left|\begin{array}{cc} x-2 & y \\ z & 2 \end{array}\right| = 0 \Rightarrow 2x - yz = 4 \quad \text{(Equation 2)} $

Solve for $ x $, $ yz $:

Solving Equations 1 and 2:

$ 4x - yz = 6 $

$ 2x - yz = 4 $

Subtract the second from the first:

$ (4x - yz) - (2x - yz) = 6 - 4 \Rightarrow 2x = 2 \Rightarrow x = 1 $

Plugging $ x = 1 $ back, we find:

$ 4(1) - yz = 6 \Rightarrow 4 - yz = 6 \Rightarrow yz = -2 $

Verify Determinants:

For $ Q - 2I $:

$ |Q - 2I| = \left|\begin{array}{cc} x-2 & y \\ z & 2 \end{array}\right| = 2x - yz = 0 $

For $ Q - 6I $:

$ |Q - 6I| = \left|\begin{array}{cc} x-6 & y \\ z & -2 \end{array}\right| = -2x + 12 - yz = 12 $

For $ Q - 3I $:

$ |Q - 3I| = \left|\begin{array}{cc} x-3 & y \\ z & 1 \end{array}\right| = x - 3 - yz = 0 $

Therefore, using the derived values, the statements for the determinants align correctly. The findings are:

$ |Q - 2I| = 0 $

$ |Q - 6I| = 12 $

$ yz = -2 $

These calculations validate the conditions for matrix $ Q $ concerning the predefined matrix properties.

$\begin{aligned} \text{(A)}\quad & \mathrm{ax}^2+2 \mathrm{bxy}+\mathrm{cy}^2>0 \forall(\mathrm{x}, \mathrm{y}) \in \mathbb{R}^2-\{(0,0)\} \\ & \Rightarrow \mathrm{ax}+2 \mathrm{bxy}+\mathrm{cy}^2 \text { must represent pair of imaginary lines and } \mathrm{a}, \mathrm{c}>0 . \\ & \Rightarrow \mathrm{b}^2<\mathrm{ac} \end{aligned}$

$\mathrm{(B)}\quad\mathrm{b}^2<3 \times \frac{1}{12} \Rightarrow|2 \mathrm{~b}|<1$

$\mathrm{(C)}\quad$ since $\mathrm{b}^2 \neq \mathrm{ac}$

$\Rightarrow \mathrm{ax}+\mathrm{by}=1 \text { and } \mathrm{bx}+\mathrm{cy}=-1$

are not parallel lines.

$\mathrm{(D)}\quad\mathrm{ac}+\mathrm{a}+\mathrm{c}>\mathrm{b}^2 \Rightarrow$ lines are not parallel.

$\Rightarrow$ Options (B), (C), (D) are Correct.

$M=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]$

$|M|=-1+1=0 \Rightarrow M$ is singular so non-invertible $ \Rightarrow $ [A] is wrong.

(B) $M\left[\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right]=\left[\begin{array}{l}-a_1 \\ -a_2 \\ -a_3\end{array}\right] \Rightarrow\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]\left[\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right]=\left[\begin{array}{l}-a_1 \\ -a_2 \\ -a_3\end{array}\right]$

$ \left.\begin{array}{l} a_1+a_2+a_3=-a_1 \\ a_1+a_3=-a_2 \\ a_2=-a_3 \end{array}\right\} \Rightarrow a_1=0 \text { and } a_2+a_3=0 \text { infinite solutions exists [B] is correct. } $

Option (D) :

$ \begin{aligned} & M-2 I=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]-2\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -2 & 1 \\ 0 & 1 & -2 \end{array}\right] \\\\ & |M-2 I|=0 \Rightarrow[D] \text { is wrong } \end{aligned} $

Option (C) :

$ \begin{aligned} & \mathrm{MX}=0 \Rightarrow\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \\\\ & x+y+z=0 \\\\ & x+z=0 \\\\ & y=0 \\\\ & \therefore \text { Infinite solution } \end{aligned} $

$[\mathrm{C}]$ is correct

For option (a)

$PEP = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]\left[ {\matrix{ 1 & 2 & 3 \cr 2 & 3 & 4 \cr 8 & {13} & {18} \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]$

$ = \left[ {\matrix{ 1 & 2 & 3 \cr 8 & {13} & {18} \cr 2 & 3 & 4 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right] = \left[ {\matrix{ 1 & 3 & 2 \cr 8 & {18} & {13} \cr 2 & 4 & 3 \cr } } \right] = F$

and ${P^2} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$

Hence, option (a) is correct.

For option (b)

$\left| {EQ + PF{Q^{ - 1}}} \right| = \left| {EQ} \right| + \left| {PF{Q^{ - 1}}} \right|$ .... (i)

$\because$ | E | = 0 and | F | = 0 and | Q | $\ne$ 0

$\therefore$ $\left| {EQ} \right| = \left| E \right|\left| Q \right| = 0$

and $\left| {PF{Q^{ - 1}}} \right| = {{\left| P \right|\left| F \right|} \over {\left| Q \right|}} = 0$

Let $R = EQ + PF{Q^{ - 1}}$ .... (ii)

$ \Rightarrow RQ = E{Q^2} + PF = E{Q^2} + {P^2}EP = E{Q^2} + EP$ [$\because$ P² = I]

$ = E({Q^2} + P)$

$ \Rightarrow \left| {RQ} \right| = \left| {E({Q^2} + P)} \right|$

$ \Rightarrow \left| R \right|\left| Q \right| = \left| E \right|\left| {{Q^2} + P} \right| = 0$ [$\because$ | E | = O]

$ \Rightarrow \left| R \right| = 0$ (as $\left| Q \right| \ne 0$) ..... (iii)

From Eqs. (ii) and (iii), we get Eq. (i) is true.

Hence, option (b) is correct.

For option (c)

$\left| {{{(EF)}^3}} \right| > {\left| {EF} \right|^2}$

i.e. 0 > 0 which is false.

For option (d)

$\because$ ${P^2} = I \Rightarrow {P^{ - 1}} = P$

$\therefore$ ${P^{ - 1}}FP = PFP = PPEPP = E$

So, $E + {P^{ - 1}}FP = E + E = 2E$

$ \Rightarrow Tr(E + {P^{ - 1}}FP) = Tr(2E) = 2Tr(E)$ ..... (iv)

and ${P^{ - 1}}EP + F$

$ \Rightarrow PEP + F = 2PEP$ [$\because$ F = PEP]

$\therefore$ $Tr(2PEP) = 2Tr(PEP) = 2Tr(F) = 2Tr(E)$ ..... (v)

From Eqs. (iv) and (v) option (d) is also correct.

$\because$ I $-$ EF = G^$-$1 $\Rightarrow$ G $-$ GEF = I ..... (i)

and G $-$ EFG = I ..... (ii)

Clearly, GEF = EFG $\to$ option (c) is correct.

Also, (I $-$ FE) (I + FGE)

= I $-$ FE + FGE $-$ FEFGE

= I $-$ FE + FGE $-$ F(G $-$ I) E

= I $-$ FE + FGE $-$ FGE + FE

= I $\to$ option (b) is correct but option (d) is incorrect.

$\because$ (I $-$ FE) (I $-$ FGE) = I $-$ FE $-$ FGE + F(G $-$ I) E

= I $-$ 2FE

Now, (I $-$ FE) ($-$ FGE) = $-$ FE

$\Rightarrow$ | I $-$ FE | | FGE | = | FE |

$\to$ option (a) is correct.

It is given that matrix M be a 3 $ \times $ 3 invertible matrix, such that

M^$-$1 = adj(adj M) $ \Rightarrow $ M^$-$1 = |M| M

($ \because $ for a matrix A of order 'n' adj(adjA) = |A|^n$-$2 A}

$ \Rightarrow $ M^$-$1 M = |M|M² $ \Rightarrow $ M²|M| = I .....(i)

$ \because $ det(M² |M|) = det(I) = 1

$ \Rightarrow $ |M|³|M|² = 1 $ \Rightarrow $ |M| = 1 .....(ii)

from Eqs. (i) and (ii), we get

M² = I

As, adj M = |M|M^$-$1 = M

$ \Rightarrow $ (adj M)² = M² (adj M)² = I

It is given, that matrices

$P = \left[ {\matrix{ 1 & 1 & 1 \cr 0 & 2 & 2 \cr 0 & 0 & 3 \cr } } \right]$, $Q = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 6 \cr } } \right]$

$ \therefore $ ${P^{ - 1}} = {{adj(P)} \over {|P|}}$

as |P| = 6 and adj P = ${\left[ {\matrix{ 6 & 0 & 0 \cr { - 3} & 3 & 0 \cr 0 & { - 2} & 2 \cr } } \right]^T}$

$ \Rightarrow $ ${P^{ - 1}} = {1 \over 6}\left[ {\matrix{ 6 & { - 3} & 0 \cr 0 & 3 & { - 2} \cr 0 & 0 & 2 \cr } } \right]$

$ \therefore $ |R| = |PQP^$-$1| [$ \because $ R = PQP^$-$1 (given)]

$ \Rightarrow $ |R| = |P||Q||P^$-$1| = |Q| [$ \because $ |P| |P^$-$1| = |I| = 1]

$ = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 6 \cr } } \right] = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 5 \cr } } \right] + \left[ {\matrix{ 2 & x & 0 \cr 0 & 4 & 0 \cr x & x & 1 \cr } } \right]$

$ = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 5 \cr } } \right] + 2(4 - 0) - x(0 - 0) + 0(0 - 4x)$

$ = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 5 \cr } } \right] + 8$, for all x $ \in $ R

$ \because $ $PQ = \left[ {\matrix{ 1 & 1 & 1 \cr 0 & 2 & 2 \cr 0 & 0 & 3 \cr } } \right]\left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 6 \cr } } \right] = \left[ {\matrix{ {2 + x} & {4 + 2x} & {x + 6} \cr {2x} & {2x + 8} & {12} \cr {3x} & {3x} & {18} \cr } } \right]$

and $QP = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 6 \cr } } \right]\left[ {\matrix{ 1 & 1 & 1 \cr 0 & 2 & 2 \cr 0 & 0 & 3 \cr } } \right] = \left[ {\matrix{ 2 & {2 + 2x} & {2 + 5x} \cr 0 & 8 & 8 \cr x & {3x} & {3x + 18} \cr } } \right]$

There is no common value of 'x', for which each corresponding element of matrices PQ and QP is equal.

For x = 0, $Q = \left[ {\matrix{ 2 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 6 \cr } } \right]$

then, if $R\left[ {\matrix{ 1 \cr a \cr b \cr } } \right] = 6\left[ {\matrix{ 1 \cr a \cr b \cr } } \right]$

$ \Rightarrow $ $PQ{P^{ - 1}}\left[ {\matrix{ 1 \cr a \cr b \cr } } \right] = 6\left[ {\matrix{ 1 \cr a \cr b \cr } } \right]$ [$ \because $ R = PQP^$-$1]

$ \Rightarrow {1 \over 6}\left[ {\matrix{ 1 & 1 & 1 \cr 0 & 2 & 2 \cr 0 & 0 & 3 \cr } } \right]\left[ {\matrix{ 2 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 6 \cr } } \right]\left[ {\matrix{ 6 & { - 3} & 0 \cr 0 & 3 & { - 2} \cr 0 & 0 & 2 \cr } } \right]\left[ {\matrix{ 1 \cr a \cr b \cr } } \right] = 6\left[ {\matrix{ 1 \cr a \cr b \cr } } \right]$

$ \Rightarrow {1 \over 6}\left[ {\matrix{ 2 & 4 & 6 \cr 0 & 8 & {12} \cr 0 & 0 & {18} \cr } } \right]\left[ {\matrix{ 6 & { - 3} & 0 \cr 0 & 3 & { - 2} \cr 0 & 0 & 2 \cr } } \right]\left[ {\matrix{ 1 \cr a \cr b \cr } } \right] = 6\left[ {\matrix{ 1 \cr a \cr b \cr } } \right]$

$ \Rightarrow \left[ {\matrix{ {12} & 6 & 4 \cr 0 & {24} & 8 \cr 0 & 0 & {36} \cr } } \right]\left[ {\matrix{ 1 \cr a \cr b \cr } } \right] = 36\left[ {\matrix{ 1 \cr a \cr b \cr } } \right]$

$ \Rightarrow \left[ {\matrix{ {12} & + & {6a} & + & {4b} \cr 0 & + & {24a} & + & {8b} \cr 0 & + & 0 & + & {36b} \cr } } \right] = \left[ {\matrix{ {36} \cr {36a} \cr {36b} \cr } } \right]$

$ \Rightarrow 6a + 4b = 24$ and $12a = 8b$

$ \Rightarrow 3a + 2b = 12$ and $3a = 2b$

$ \Rightarrow a = 2$ and $b = 3$

So, a + b = 5.

Now, $R\left[ {\matrix{ \alpha \cr \beta \cr \gamma \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$

and $\alpha \widehat i + \beta \widehat j + \gamma \widehat k$ is a unit vector,

so det (R) = 0

$ \Rightarrow $ det (Q) = 0 [$ \because $ R = PQP^$-$1 So, |R| = |Q|]

$ \Rightarrow \left| {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 6 \cr } } \right| = 0$

$ \Rightarrow 2(24 - 0) - x(0 - 0) + x(0 - 4x) = 0 \Rightarrow 48 - 4{x^2} = 0$

$ \Rightarrow {x^2} = 12 \Rightarrow x = \pm 2\sqrt 3 $

So, for x = 1, there does not exist a unit vector $\alpha \widehat i + \beta \widehat j + \gamma \widehat k$, for which $R\left[ {\matrix{ \alpha \cr \beta \cr \gamma \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$

Hence, options (b) and (d) are correct.

${P_1} = I = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right],\,{P_2} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right],\,{P_3} = \left[ {\matrix{ 0 & 1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 \cr } } \right],\,{P_4} = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right],\,{P_5} = \left[ {\matrix{ 0 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr } } \right],\,{P_6} = \left[ {\matrix{ 0 & 0 & 1 \cr 0 & 1 & 0 \cr 1 & 0 & 0 \cr } } \right]$ and $X = \sum\limits_{k = 1}^6 {{P_k}} \left[ {\matrix{ 2 & 1 & 3 \cr 1 & 0 & 2 \cr 3 & 2 & 1 \cr } } \right]P_k^T$

$ \because $ $P_1^T = {P_1},\,P_2^T = {P_2},\,P_3^T = {P_3},\,P_4^T = {P_5},\,P_5^T = {P_4}$ and $P_6^T = {P_6}$

and Let $Q = \left[ {\matrix{ 2 & 1 & 3 \cr 1 & 0 & 2 \cr 3 & 2 & 1 \cr } } \right]$

and $ \because $ Q^T = Q

Now,

$X = ({P_1}QP_1^T) + ({P_2}QP_2^T) + ({P_3}QP_3^T) + ({P_4}QP_4^T) + ({P_5}QP_5^T) + ({P_6}QP_6^T)$

So, ${X^T} = {({P_1}QP_1^T)^T} + {({P_2}QP_2^T)^T} + {({P_3}QP_3^T)^T} + {({P_4}QP_4^T)^T} + {({P_5}QP_5^T)^T} + {({P_6}QP_6^T)^T}$

= ${P_1}QP_1^T$ + ${P_2}QP_2^T$ + ${P_3}QP_3^T$ + ${P_4}QP_4^T$ + ${P_5}QP_5^T$ + ${P_6}QP_6^T$

[$ \because $ (ABC)^T = C^TB^TA^T and (A^T)^T = A and Q^T = Q]

$ \Rightarrow $ X^T = X

$ \Rightarrow $ X is a symmetric matrix.

The sum of diagonal entries of X = Tr(X)

$ = \sum\limits_{i = 1}^6 {{T_r}({P_i}QP_i^T) = \sum\limits_{i = 1}^6 {{T_r}(QP_i^T{P_i})} } $ [$ \because $ T_r(ABC) = T_r(BCA)]

$ = \sum\limits_{i = 1}^6 {{T_r}(QI)} $

[$ \because $ P_i's are orthogonal matrices]

$ = \sum\limits_{i = 1}^6 {{T_r}(Q) = 6{T_r}(Q) = 6 \times 3 = 18} $

Now, Let $R = \left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$, then

$XR = \sum\limits_{K = 1}^6 {({P_K}} QP_K^T)R = \sum\limits_{K = 1}^6 {({P_K}QP_K^TR)} $

$ = \sum\limits_{K = 1}^6 {({P_K}QR)} $ [$ \because $ $P_K^T$R = R]

$ = \sum\limits_{K = 1}^6 {{P_K}\left[ {\matrix{ 6 \cr 3 \cr 6 \cr } } \right]} $

$ = \sum\limits_{K = 1}^6 {{P_K}\left[ {\matrix{ 6 \cr 3 \cr 6 \cr } } \right]} = \left[ {\matrix{ 2 & 2 & 2 \cr 2 & 2 & 2 \cr 2 & 2 & 2 \cr } } \right]\left[ {\matrix{ 6 \cr 3 \cr 6 \cr } } \right]$

$ \Rightarrow XR = \left[ {\matrix{ {30} \cr {30} \cr {30} \cr } } \right] \Rightarrow XR = 30R$

$ \Rightarrow X\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right] = 30\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$

$ \Rightarrow $ (X $-$ 30I) R = 0 $ \Rightarrow $ |X $-$ 30I| = 0

So, (X $-$ 30I) is not invertible and value of $\alpha $ = 30.

Hence, options (a), (b) and (d) are correct.

Given square matrix

$M = \left[ {\matrix{ 0 & 1 & a \cr 1 & 2 & 3 \cr 3 & b & 1 \cr } } \right]$

and adj $(M) = \left[ {\matrix{ { - 1} & 1 & { - 1} \cr 8 & { - 6} & 2 \cr { - 5} & 3 & { - 1} \cr } } \right]$

$ \because $ |adj (M)| = |M|² = $\left[ {\matrix{ { - 1} & 1 & { - 1} \cr 8 & { - 6} & 2 \cr { - 5} & 3 & { - 1} \cr } } \right]$

$ \Rightarrow $ $|M{|^2} = - 1(6 - 6) - 1( - 8 + 10) - 1(24 - 30) = - 2 + 6 = 4$

$ \Rightarrow $ $|M| = \pm 2$

$ \therefore $ det (adj M²) = $|{M^2}{|^2} = |M{|^4} = 16$

As we know A(adj A) = |A| I

$ \Rightarrow $ M = |M| (adj M)^-1 ....(i)

$ \because $ (adj M)^-1 = ${1 \over {|adj\,M|}}{\left[ {\matrix{ 0 & { - 2} & { - 6} \cr { - 2} & { - 4} & { - 2} \cr { - 4} & { - 6} & { - 2} \cr } } \right]^T}$

= ${1 \over 4}\left[ {\matrix{ 0 & { - 2} & { - 4} \cr { - 2} & { - 4} & { - 6} \cr { - 6} & { - 2} & { - 2} \cr } } \right]$

So$\left[ {\matrix{ 0 & 1 & a \cr 1 & 2 & 3 \cr 3 & b & 1 \cr } } \right] = {{|M|} \over 4}\left[ {\matrix{ 0 & { - 2} & { - 4} \cr { - 2} & { - 4} & { - 6} \cr { - 6} & { - 2} & { - 2} \cr } } \right]$

$ \Rightarrow $ M = $\left[ {\matrix{ 0 & 1 & 2 \cr 1 & 2 & 3 \cr 3 & 1 & 1 \cr } } \right]$

Now, If M$\left[ {\matrix{ \alpha \cr \beta \cr \gamma \cr } } \right] = \left[ {\matrix{ 1 \cr 2 \cr 3 \cr } } \right]$

$ \Rightarrow $ $\left[ {\matrix{ 0 & 1 & 2 \cr 1 & 2 & 3 \cr 3 & 1 & 1 \cr } } \right]\left[ {\matrix{ \alpha \cr \beta \cr \gamma \cr } } \right] = \left[ {\matrix{ 1 \cr 2 \cr 3 \cr } } \right]$

$ \Rightarrow $ $\beta $ + 2$\gamma $ = 1, $\alpha $ + 2$\beta $ + 3$\gamma $ = 2 and 3$\alpha $ + $\beta $ + $\gamma $ = 3

$ \Rightarrow $ $\alpha $ = 1, $\beta $ = -1 and $\gamma $ = 1

$ \therefore $ $\alpha $ - $\beta $ + $\gamma $ = 3

And ${(adj\,M)^{ - 1}} + adj\,({M^{ - 1}}) = 2{(adj\,M)^{ - 1}}$

[$ \because $ adj (M^-1) = (adj M)^-1]

= $2\left( { - {M \over 2}} \right) = - M$

[$ \because $ (adj M)-¹ = ${M \over {|M|}}$ from Eq. (i) ]

and $ \because $ a = 2 and b = 1, so a + b = 3

Hence, options (b), (c) and (d) are correct.

We have,

$\eqalign{ & - x + 2y + 5z = {b_1} \cr & 2x - 4y + 3z = {b_2} \cr & x - 2y + 2z = {b_3} \cr} $

has at least one solution.

$ \therefore $ $D = \left| {\matrix{ { - 1} & 2 & 5 \cr 2 & { - 4} & 3 \cr 1 & { - 2} & 2 \cr } } \right|$

and ${D_1} = {D_2} = {D_3} = 0$

$ \Rightarrow {D_1} = \left| {\matrix{ {{b_1}} & 2 & 5 \cr {{b_2}} & { - 4} & 3 \cr {{b_3}} & { - 2} & 2 \cr } } \right|$

$ = - 2{b_1} - 14{b_2} + 26{b_3} = 0$

$ \Rightarrow {b_1} + 7{b_2} = 13{b_3}$ ....(i)

(a) $ \Rightarrow D = \left| {\matrix{ 1 & 2 & 3 \cr 0 & 4 & 5 \cr 1 & 2 & 6 \cr } } \right| = 1(24 - 10) + 1(10 - 12)$

$ = 14 - 2 = 12 \ne 0$

Here, D $ \ne $ 0 $ \Rightarrow $ unique solution for any b₁, b₂, b₃.

(b) $D = \left| {\matrix{ 1 & 1 & 3 \cr 5 & 2 & 6 \cr { - 2} & { - 1} & { - 3} \cr } } \right|$

$ = 1( - 6 + 6) - 1( - 15 + 12) + 3( - 5 + 4) = 0$

For at least one solution

${D_1} = {D_2} = {D_3} = 0$

Now, ${D_1} = \left| {\matrix{ {{b_1}} & 1 & 3 \cr {{b_2}} & 2 & 6 \cr {{b_3}} & { - 1} & { - 3} \cr } } \right|$

$ = {b_1}( - 6 + 6) - {b_2}( - 3 + 3) + {b_3}(6 - 6) = 0$

${D_2} = \left| {\matrix{ 1 & {{b_1}} & 3 \cr 5 & {{b_2}} & 6 \cr { - 2} & {{b_3}} & { - 3} \cr } } \right|$

$ = {b_1}( - 15 + 12) + {b_2}( - 3 + 6) - {b_3}(6 - 15)$

$ = 3{b_1} + 3{b_2} + 9{b_3} = 0 \Rightarrow {b_1} + {b_2} + 3{b_3} = 0$

not satisfies the Eq. (i)

$ \therefore $ It has no solution.

(c) $D = \left| {\matrix{ { - 1} & 2 & { - 5} \cr 2 & { - 4} & {10} \cr 1 & { - 2} & 5 \cr } } \right|$

= $ - 1( - 20 + 20) - 2(10 - 10) - 5( - 4 + 4) = 0$

Here, b₂ = $-$2b₁ and b₃ = $-$b₁ satisfies the Eq. (i)

Planes are parallel.

(d) $D = \left| {\matrix{ 1 & 2 & 5 \cr 2 & 0 & 3 \cr 1 & 4 & { - 5} \cr } } \right|$

$ = 1(0 - 12) - 2( - 10 - 3) + 5(8 - 0) = 54$

$D \ne 0$

$ \therefore $ It has unique solutino for any b₁, b₂, b₃.

For a matrix to be square of matrix with real entries, its determinant should be positive.

Option (A) : $\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & { - 1} \cr } } \right]$, determinant is NOT possible:

1($-$1) $-$ 0(0) + 0(0) = $-$1

Option (B) : $\left[ {\matrix{ 1 & 0 & 0 \cr 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr } } \right]$, determinant is possible:

1(1) $-$ 0(0) + 0(0) = +1

Option (C) : $\left[ {\matrix{ { - 1} & 0 & 0 \cr 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr } } \right]$, determinant is NOT possible :

$-$1(1) $-$ 0(0) + 0(0) = $-$1

Option (D) : $\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$, determinant is possible:

(1) $-$ 0(0) + 0(0) = +1

Thus, options (A) and (C) are NOT the square of a 3 $\times$ 3 matrix with real entries.

The given system of linear equation is

ax + 2y = $\lambda$

3x $-$ 2y = $\mu$

By Cramer's rule, we have

$\Delta = \left| {\matrix{ a & 2 \cr 3 & { - 2} \cr } } \right| = - 2a - 6 = - 2(a + 3)$

${\Delta _1} = \left| {\matrix{ \lambda & 2 \cr \mu & { - 2} \cr } } \right| = - 2\lambda - 2\mu = - 2(\lambda + \mu )$

${\Delta _2} = \left| {\matrix{ a & \lambda \cr 3 & \mu \cr } } \right| = (a\mu - 3\lambda )$

For unique solution: $\Delta$ $\ne$ 0 $\Rightarrow$ a + 3 $\ne$ 0 $\Rightarrow$ a $\ne$ $-$3, where $\lambda$ and $\mu$ can take any values.

Hence, option (B) is correct.

For infinite solution: $\Delta$ = 0, $\Delta$₁ and $\Delta$₂ are zero.

That is, a = $-$3 and $\lambda$ + $\mu$ = 0 or a$\mu$ $-$ 3$\lambda$ = 0.

Hence, options (C) are correct.

For no solution: $\Delta$ = 0 and either $\Delta$₁ or $\Delta$₂ is non-zero.

That is, a = $-$3 and $\lambda$ + $\mu$ $\ne$ 0.

Hence, option (D) is correct.

$P = \left[ {\matrix{ 3 & { - 1} & { - 2} \cr 2 & 0 & \alpha \cr 3 & { - 5} & 0 \cr } } \right]$

Now, $\left| P \right| = 3(5\alpha ) + 1( - 3\alpha ) - 2( - 10)$

$ = 12\alpha + 20$ ....... (i)

$\therefore$ $(P) = {\left[ {\matrix{ {5\alpha } & {2\alpha } & { - 10} \cr { - 10} & 6 & {12} \cr { - \alpha } & { - (3\alpha + 4)} & 2 \cr } } \right]^T}$

$ = \left[ {\matrix{ {5\alpha } & { - 10} & { - \alpha } \cr {2\alpha } & 6 & { - 3\alpha - 4} \cr { - 10} & {12} & 2 \cr } } \right]$ ...... (ii)

As, $PQ = kI$

$ \Rightarrow \left| P \right|\left| Q \right| = \left| {kI} \right|$

$ \Rightarrow \left| P \right|\left| Q \right| = {k^3}$

$ \Rightarrow \left| P \right|\left( {{{{k^2}} \over 2}} \right) = {k^3}$ [given, $\left| Q \right| = {{{k^2}} \over 2}$]

$ \Rightarrow \left| P \right| = 2k$ ......(iii)

$\because$ $PQ = ki$

$\therefore$ $Q = k{p^{ - 1}}I$

$ = k\,.\,{{adj\,P} \over {\left| P \right|}} = {{k(adj\,P)} \over {2k}}$ [from Eq. (iii)]

$ = {{adj\,P} \over 2}$

$ = {1 \over 2}\left[ {\matrix{ {5\alpha } & { - 10} & { - \alpha } \cr {2\alpha } & 6 & { - 3\alpha - 4} \cr { - 10} & {12} & 2 \cr } } \right]$

$\therefore$ ${q_{23}} = {{ - 3\alpha - 4} \over 2}$ [given, ${q_{23}} = - {k \over 8}$]

$ \Rightarrow - {{(3\alpha + 4)} \over 2} = - {k \over 8}$

$ \Rightarrow (3\alpha + 4) \times 4 = k$

$ \Rightarrow 12\alpha + 16 = k$ ...... (iv)

From Eq. (iii), $\left| P \right| = 2k$

$ \Rightarrow 12\alpha + 20 = 2k$ [from Eq. (i)] ...... (v)

On solving Eqs. (iv) and (v), we get

$\alpha$ = $-$1 and k = 4 ....... (vi)

$\therefore$ $4\alpha - k + 8 = - 4 - 4 + 8 = 0$

$\therefore$ Option (b) is correct.

Now, $\left| {P\,adj(Q)} \right| = \left| P \right|\left| {adj\,Q} \right|$

$ = 2k{\left( {{{{k^2}} \over 2}} \right)^2} = {{{k^5}} \over 2} = {{{2^{10}}} \over 2} = {2^9}$

$\therefore$ Option (c) is correct.

Given,

${X^T} = - X,{Y^T} = - Y,{Z^T} = Z$

(a) Let $P = {Y^3}{Z^4} - {Z^4}{Y^3}$

Then, ${P^T} = {({Y^3}{Z^4})^T} - {({Z^4}{Y^3})^T}$

$ = {({Z^T})^4}{({Y^T})^3} - {({Y^T})^3}{({Z^T})^4}$

$ = - {Z^4}{Y^3} + {Y^3}{Z^4} = P$

$\therefore$ P is symmetric matrix.

(b) Let $P = {X^{44}} + {Y^{44}}$

Then, ${P^T} = {({X^T})^{44}} + {({Y^T})^{44}}$

$ = {X^{44}} + {Y^{44}} = P$

$\therefore$ P is symmetric matrix.

(c) Let $P = {X^4}{Z^3} - {Z^3}{X^4}$

Then, ${P^T} = {({X^4}{Z^3})^T} - {({Z^3}{X^4})^T}$

$ = {({Z^T})^3}{({X^T})^4} - {({X^T})^4}{({Z^T})^3}$

$ = {Z^3}{X^4} - {X^4}{Z^3}$

$ = - P$

$\therefore$ P is skew-symmetric matrix.

(d) Let $P = {X^{23}} + {Y^{23}}$

Then, ${P^T} = {({X^T})^{23}} + {({Y^T})^{23}}$

$ = - {X^{23}} - {Y^{23}}$

$ = - P$

$\therefore$ P is skew-symmetric matrix.

$\left| {\matrix{ {{{(1 - \alpha )}^2}} & {{{(1 + 2\alpha )}^2}} & {{{(1 + 3\alpha )}^2}} \cr {{{(2 + \alpha )}^2}} & {{{(2 + 2\alpha )}^2}} & {{{(2 + 3\alpha )}^2}} \cr {{{(3 + \alpha )}^2}} & {{{(3 + 2\alpha )}^2}} & {{{(3 + 3\alpha )}^2}} \cr } } \right| = - 648\alpha $

Applying ${R_3} \to {R_3} - {R_2},{R_2} \to {R_2},{R_1}$, we get

$\left| {\matrix{ {{{(1 - \alpha )}^2}} & {{{(1 + 2\alpha )}^2}} & {{{(1 + 3\alpha )}^2}} \cr {3 + 2\alpha } & {3 + 4\alpha } & {3 + 6\alpha } \cr {5 + 2\alpha } & {5 + 4\alpha } & {5 + 6\alpha } \cr } } \right| = - 648\alpha $

Applying ${R_3} \to {R_3} - {R_2}$, we get

$\left| {\matrix{ {{{(1 - \alpha )}^2}} & {{{(1 + 2\alpha )}^2}} & {{{(1 + 3\alpha )}^2}} \cr {3 + 2\alpha } & {3 + 4\alpha } & {3 + 6\alpha } \cr 2 & 2 & 2 \cr } } \right| = - 648\alpha $

Applying ${C_3} \to {C_3} - {C_2},{C_2} \to {C_2} - {C_1}$, we get

$\left| {\matrix{ {{{(1 - \alpha )}^2}} & {{{(1 + 2\alpha )}^2}} & {{{(1 + 3\alpha )}^2}} \cr {3 + 2\alpha } & {2\alpha } & {2\alpha } \cr 2 & 0 & 0 \cr } } \right| = - 648\alpha $

Expanding, $2{\alpha ^2}(3\alpha + 2) - 2{\alpha ^2}(5\alpha + 2) = - 324\alpha $

$ \Rightarrow - 4{\alpha ^3} = - 324\alpha \Rightarrow \alpha ({\alpha ^2} - 81) = 0$

$\therefore$ $\alpha = 0, - 9,9$

Note : A square matrix M is invertible if det(M) or | M | $\ne$ 0.

Let $M = \left[ {\matrix{ a & b \cr b & c \cr } } \right]$

(a) Given that $\left[ {\matrix{ a \cr b \cr } } \right] = \left[ {\matrix{ b \cr c \cr } } \right]$

$ \Rightarrow a = b = c = \alpha $ (let)

$ \Rightarrow M = \left[ {\matrix{ \alpha & \alpha \cr \alpha & \alpha \cr } } \right]$

$\Rightarrow$ | M | = 0

$\Rightarrow$ M is non-invertible.

(b) Given that [bc] = [ab]

$ \Rightarrow a = b = c = \alpha $ (let)

Again | M | = 0

$\Rightarrow$ M is non-invertible.

(c) As given $M = \left[ {\matrix{ a & 0 \cr 0 & c \cr } } \right]$

$ \Rightarrow \left| M \right| = ac \ne 0$ ($\because$ a and c are non-zero)

$\Rightarrow$ M is invertible.

(d) $M = \left[ {\matrix{ a & b \cr b & c \cr } } \right] \Rightarrow \left| M \right| = ac - {b^2} \ne 0$

$\because$ ac is not equal to square of an integer.

$\therefore$ M is invertible.

Given MN = NM. Therefore, a² $-$ b² = (a + b) (a $-$ b) of algebra of numbers is applicable.

Now, M² = N⁴ $\Rightarrow$ M² $-$ N⁴ = 0 (Null matrix)

$\Rightarrow$ (M + N²) (M $-$ N²) = 0

Since M $\ne$ N² (given), the possibilities are

(M + N²) = 0 and M $-$ N² $\ne$ 0 (1)

or (M + N²) $\ne$ 0 and M + N² $\ne$ 0 (2)

Now, we know if A and B are non-null square matrix nd AB = 0, then A and B both are singular, that is, | A | = 0 and | B | = 0 and AB = 0.

Note : For example, let A be non-singular. Therefore, B = In(2) = A^$-$1 AB = 0 (since AB = 0 is assumed).

Hence, B is singular, which is a contradiction and A has to be singular. Similarly, B also has to be singular.

Therefore, from Eqs. (1) and (2), we conclude the only possibility is | M + N² | = 0.

Now checking options :

(A) | M² + MN² | = | M || M + N² | = 0

Hence, option (A) is correct.

(B) (M² + MN²) U = 0

Since, M² + MN² is singular. Therefore, U has infinitely many possible values (non-trivial solutions). Hence, option (B) is true.

(C) False since | M² + MN² | = 0.

(D) Since | M² + MN² | = 0, U is not a necessarily a zero matrix.

The given matrix $ P = [p_{ij}] $ is an $ n \times n $ matrix where $ p_{ij} = \omega^{i + j} $ and $ \omega $ is a complex cube root of unity with $\omega \neq 1$. We need to determine when $ P^2 \neq 0 $.

For $ n = 1 $:

$ P = [p_{ij}]_{1 \times 1} = [\omega^2] $

$ \Rightarrow P^2 = [\omega^4] \neq 0 $

For $ n = 2 $:

$ P = [p_{ij}]_{2 \times 2} = \begin{bmatrix} \omega^2 & \omega^3 \\ \omega^3 & \omega^4 \end{bmatrix} = \begin{bmatrix} \omega^2 & 1 \\ 1 & \omega \end{bmatrix} $

$ P^2 = \begin{bmatrix} \omega^2 & 1 \\ 1 & \omega \end{bmatrix} \begin{bmatrix} \omega^2 & 1 \\ 1 & \omega \end{bmatrix} = \begin{bmatrix} \omega^4 + 1 & \omega^2 + \omega \\ \omega^2 + \omega & 1 + \omega^2 \end{bmatrix} \neq 0 $

For $ n = 3 $:

$ P = [p_{ij}]_{3 \times 3} = \begin{bmatrix} \omega^2 & \omega^3 & \omega^4 \\ 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \end{bmatrix} = \begin{bmatrix} \omega^2 & 1 & \omega \\ 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \end{bmatrix} $

$ P^2 = \begin{bmatrix} \omega^2 & 1 & \omega \\ 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \end{bmatrix} \begin{bmatrix} \omega^2 & 1 & \omega \\ 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = 0 $

Thus, $ P^2 = 0 $ when $ n $ is a multiple of 3. Therefore, $ P^2 \neq 0 $ when $ n $ is not a multiple of 3.

The possible values of $ n $ where $ P^2 \neq 0 $ are:

$ \Rightarrow n = 55, 58, 56 $

In case of option (A),

${({N^T}MN)^T} = {N^T}{M^T}{({N^T})^T} = {N^T}{M^T}N$

Now, ${N^T}{M^T}N = \left\{ \matrix{ {N^T}MN,\,when\,{M^T} = M \hfill \cr - {N^T}MN,\,when\,{M^T} = - M \hfill \cr} \right.$

$\therefore$ N^TMN is symmetric or skew symmetric according as M is symmetric or skew symmetric.

In case of option (B),

${(MN - NM)^T} = {(MN)^T} - {(NM)^T}$

$ = {N^T}{M^T} - {M^T}{N^T}$

$ = NM - MN$ [$\because$ ${M^T} = M,\,{N^T} = N$]

$ = - (MN - NM)$

$\therefore$ $MN - NM$ is skew symmetric matrix.

In case of option (C),

${(MN)^T} = {N^T}{M^T} = NM$ [$\because$ ${M^T} = M,\,{N^T} = N$]

Now, MN cannot always be equal to NM

$\therefore$ MN is not symmetric matrix.

In case of option (D),

$(Adj\,M)(Adj\,N) = Adj(NM) \ne Adj(MN)$

Therefore, (C) and (D) are the correct options.

Concept Involved If $\left| {{A_{n \times n}}} \right| = \Delta $, then $\left| {adj\,A} \right| = {\Delta ^{n - 1}}$

Here, ${P_{3 \times 3}} = \left[ {\matrix{ 1 & 4 & 4 \cr 2 & 1 & 7 \cr 1 & 1 & 3 \cr } } \right]$

$ \Rightarrow \left| {adj\,P} \right| = {\left| P \right|^2}$

$\therefore$ $\left| {adj\,P} \right| = \left| {\matrix{ 1 & 4 & 4 \cr 2 & 1 & 7 \cr 1 & 1 & 3 \cr } } \right|$

$ = 1(3 - 7) - 4(6 - 7) + 4(2 - 1)$

$ = - 4 + 4 + 4 = 4$

$ \Rightarrow \left| P \right| = \pm \,2$

Given, ${M^T} = - M$, ${N^T} = - N$

and $MN = NM$ ..... (i)

$\therefore$ ${M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T}$

$ = {M^2}{N^2}{N^{ - 1}}{({M^T})^{ - 1}}{({N^{ - 1}})^T}.{M^T}$

$ = {M^2}N(N{M^{ - 1}}){( - M)^{ - 1}}{({N^T})^{ - 1}}( - M)$

$ = {M^2}NI( - {M^{ - 1}}){( - N)^{ - 1}}( - M)$

$ = - {M^2}N{M^{ - 1}}{N^{ - 1}}M$

$ = - M.(MN){M^{ - 1}}{N^{ - 1}}M$

$ = - M(NM){M^{ - 1}}{N^{ - 1}}M$

$ = - MN(N{M^{ - 1}}){N^{ - 1}}M$

$ = - M(N{N^{ - 1}})M = - {M^2}$

Note : This question is wrong, as given. An odd order skew symmetric matrix can't be invertible. Had the matrix be of even order, it could have been correct.

To find the number of $3 \times 3$ invertible matrices $Q$ with integer entries that satisfy both $Q^{-1} = Q^T$ and $PQ = QP$, consider the given matrix $P$:

$ P = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} $

The condition $PQ = QP$ implies:

$ \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} $

This results in:

$ \begin{pmatrix} 2a & 2b & 2c \\ 2d & 2e & 2f \\ 3g & 3h & 3i \end{pmatrix} = \begin{pmatrix} 2a & 2b & 3c \\ 2d & 2e & 3f \\ 2g & 2h & 3i \end{pmatrix} $

From this, we derive that:

$ c = 0, \quad f = 0, \quad h = 0, \quad g = 0 $

Thus, the matrix $Q$ is of the form:

$ Q = \begin{pmatrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & e \end{pmatrix} $

Next, using the condition $Q^T = Q^{-1}$, we have the equation:

$ Q Q^T = I $

This results in:

$ \begin{pmatrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & i \end{pmatrix} \begin{pmatrix} a & c & 0 \\ b & d & 0 \\ 0 & 0 & i \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $

This simplifies to:

$ \begin{pmatrix} a^2 + b^2 & ac + bd & 0 \\ ac + bd & c^2 + d^2 & 0 \\ 0 & 0 & i^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $

From which we conclude:

$ a^2 + b^2 = 1, \quad ac + bd = 0, \quad c^2 + d^2 = 1, \quad i^2 = 1 $

Considering all possible integer solutions for these equations:

Case I:

$(a, b) = (0, 1)$ or $(0, -1)$,

$(c, d) = (1, 0)$ or $(-1, 0)$

Case II:

$(a, b) = (1, 0)$ or $(-1, 0)$,

$(c, d) = (0, 1)$ or $(0, -1)$

For each case, $e$ can be either 1 or -1 since $i^2=1$. Therefore, $e = \pm 1$.

Combining these options, we find a total of 16 possible matrices $Q$.

$\alpha, \beta$ are roots of $x^2+x-1=0$

$\begin{aligned} & \therefore \alpha+\beta=-1 \Rightarrow 1+\alpha+\beta=0 \\ & M=\left[\begin{array}{lll} \mathrm{a}_{11} & \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{21} & \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{31} & \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right] \end{aligned}$

(P) $\quad M=\left[\begin{array}{lll}1 & \alpha & \beta \\ \alpha & \beta & 1 \\ \beta & 1 & \alpha\end{array}\right] \Rightarrow 3 ! \times 2=12$

For one arrangement of row 1 we can arrange other two rows exactly in two ways and row 1 can be arranged in 3! ways

$\therefore 3!\times 2=12 \text { ways }$

(Q) $\quad M=\left[\begin{array}{lll}x & a & b \\ a & y & c \\ b & c & z\end{array}\right] \Rightarrow$ Consider one such arrangement with $a=\alpha, b=\beta, c=1$

a, b, c can be arranged in 3! ways and corresponding entries can be arranged in 1 way.

(R)

$\begin{aligned} & {\left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} a \\ 0 \\ -c \end{array}\right]} \\ & a y+b z=a \\ & -a x+c z=0 \\ & -b x-c y=-c \end{aligned}$

It is observed that $\mathrm{D}=\mathrm{D}_{\mathrm{x}}=\mathrm{D}_{\mathrm{y}}=\mathrm{D}_{\mathrm{z}}=0$

$\therefore$ infinite solution

(S)

$\begin{aligned} & {\left[\begin{array}{lll} 1 & \alpha & \beta \\ \beta & \alpha & 1 \\ \alpha & 1 & \beta \end{array}\right] } \\ & \Rightarrow \alpha \beta-1-\alpha \beta^2+\alpha^2+\beta^2-\alpha^2 \beta=0 \quad(\text { since } \alpha \beta=\alpha+\beta=-1) \end{aligned}$

Given

$\begin{gathered}x+2 y+z=7 \\\\ x+\alpha z=11 \\\\ 2 x-3 y+\beta z=\gamma\end{gathered}$

Using Cramer's rule

$ \begin{aligned} \Delta & =\left|\begin{array}{ccc} 1 & 2 & 1 \\ 1 & 0 & \alpha \\ 2 & -3 & \beta \end{array}\right| \\\\ & =1(3 \alpha)-2(\beta-2 \alpha)+1(-3) \\\\ & =3 \alpha-2 \beta+4 \alpha-3 \\\\ & =7 \alpha-2 \beta-3 \end{aligned} $

$\begin{aligned} \Delta_x & =\left|\begin{array}{ccc}7 & 2 & 1 \\ 11 & 0 & \alpha \\ \gamma & -3 & \beta\end{array}\right| \\\\ & =7(3 \alpha)-2(11 \beta-\gamma \alpha)+1(-33) \\\\ & =21 \alpha-22 \beta+22 \gamma-33\end{aligned}$

$\begin{aligned} \Delta_y & =\left|\begin{array}{ccc}1 & 7 & 1 \\ 1 & 11 & \alpha \\ 2 & \gamma & \beta\end{array}\right| \\\\ & =1(11 \beta-\alpha \gamma)-7(\beta-2 \alpha)+1(\gamma-22) \\\\ & =11 \beta-\alpha \gamma-7 \beta+14 \alpha+\gamma-22 \\\\ & =14 \alpha+4 \beta+\gamma-\alpha \gamma-22\end{aligned}$

$\begin{aligned} \Delta_z & =\left|\begin{array}{ccc}1 & 2 & 7 \\ 1 & 0 & 11 \\ 2 & -3 & \gamma\end{array}\right| \\\\ & =1(33)-2(\gamma-22)+7(-3) \\\\ & =33-2 \gamma+44-21 \\\\ & =-2 \gamma+56\end{aligned}$

For unique solution $\Delta \neq 0$

For infinite solution

$ \Delta=\Delta x=\Delta y=\Delta z=0 $

For no solution $\Delta=0$ and atleast one in $\Delta x, \Delta y, \Delta z$ is non zero.

$ \Delta=0 $

$ \Rightarrow \beta=\frac{1}{2}(7 \alpha-3) $

(P) $ \beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$

then $ \Delta=0, \Delta x=\Delta y=\Delta z=0$

$\therefore$ Infinite solution

(Q) $ \beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$

$\therefore \Delta=0$ and $\Delta_2 \neq 0$

$\Rightarrow$ No solution.

$ \begin{aligned} & \text { (R) } \beta \neq \frac{1}{2}(7 \alpha-3), \alpha=1, \gamma \neq 28 \\\\ & \Rightarrow \Delta \neq 0=\text { unique solution } \\\\ & \begin{aligned} & \text { (S) } \beta \neq \frac{1}{2}(7 \alpha-3), \alpha=1, \gamma=28 \\\\ & \therefore \Delta \neq 0, \Delta=4-2 \beta \\\\ & \Delta x=44-22 \beta \\\\ & \Delta y=4 \beta-8 \\\\ & \Delta z=0 \end{aligned} \end{aligned} $

$\therefore x=11, y=-2, z=0$ is the solution.

$A=\left(\begin{array}{cc}\frac{5}{2} & \frac{3}{2} \\ -\frac{3}{2} & -\frac{1}{2}\end{array}\right)$

$A^{2}=\left(\begin{array}{cc}4 & 3 \\ -3 & -2\end{array}\right)$

$A^{3}=\left(\begin{array}{cc}\frac{11}{2} & \frac{9}{2} \\ -\frac{9}{2} & -\frac{7}{2}\end{array}\right)$

$A^{4}=\left(\begin{array}{cc}7 & 6 \\ -6 & -5\end{array}\right)$

and so on

$A^n=\left(\begin{array}{cc}\frac{3 n}{2}+1 & \frac{3 n}{2} \\ -\frac{3 n}{2} & -\frac{3 n}{2}+1\end{array}\right)$

Now

$ \begin{aligned} A^{2022} & =\left(\begin{array}{ll} \frac{3 \times 2022}{2}+1 & \frac{3 \times 2022}{2} \\ \frac{-3 \times 2022}{2} & \frac{-3 \times 2022}{2}+1 \end{array}\right) \\\\ & =\left(\begin{array}{cc} 3034 & 3033 \\ -3033 & -3032 \end{array}\right) \end{aligned} $

$\therefore$ Option $(\mathrm{A})$ is correct

Given:

$ \begin{gathered} x+y+z=1 ..........(1)\\\\ 10 x+100 y+1000 z=0 .........(2)\\\\ q r x+p r y+p q z=0 ............(3) \end{gathered} $

Now equation (3) can be re-written as

$ \frac{x}{p}=\frac{y}{q}=\frac{z}{r}=0 \quad \because\{p, q, r \neq 0\} $

Now given $p, q$ and $r$ are $10^{\text {th }}, 100^{\text {th }}$ and $1000^{\text {th }}$ term of an. H.P.,

So let $p=\frac{y}{a+9 d}, q=\frac{1}{a+99 d}, r=\frac{1}{a+999 d}$

Now from equation (3)

$ (a+9 d) x+(a+99 d) y+(a+999 d) z=0 $

Now from and (1), (2) and (3) we get

$ \begin{aligned} \Delta & =\left|\begin{array}{ccc} 1 & 1 & 1 \\ 10 & 100 & 1000 \\ a+9 d & a+99 d & a+999 d \end{array}\right|=0 \\\\ \Delta x & =\left|\begin{array}{ccc} 1 & 1 & 1 \\ 0 & 100 & 1000 \\ 0 & a+99 d & a+999 d \end{array}\right|=900(d-a) \\\\ \Delta y & =\left|\begin{array}{ccc} 10 & 0 & 1000 \\ 1+9 d & 0 & a+999 d \end{array}\right|=990(a-d) \\\\ \Delta z & =\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 10 & 100 & 0 \\ a+9 d & a+99 d & 0 \end{array}\right|=90(d-a) \end{aligned} $

(I) If $\frac{q}{r}=10 \Rightarrow a=d$

$ \Delta=\Delta_x=\Delta_y=\Delta_z=0 $

And equation (1) and equation (2) represents non-parallel plane equation (2) and equation (3) represents same plane

$\Rightarrow$ Infinitely many solutions.

Now finding solution by taking $z=\lambda$ then

$ \begin{aligned} & x+y =1-\lambda \\\\ & x+10 y =-100 \lambda \\\\ & x =\frac{10}{9}+100 \lambda \\\\ & y =\frac{-1}{9}-11 \lambda \\\\ & \Rightarrow (x, y, z) \in\left(\frac{10}{9}+10 \lambda, \frac{-1}{9}-11 \lambda, \lambda\right) \end{aligned} $

So $\mathrm{P}$ is not valid for any value of $\lambda \rightarrow \mathrm{Q}$

(II) $\frac{p}{r} \neq 100 \Rightarrow a \neq d$

$ \Delta=0 \,\text{and}\, \Delta_{x^{\prime}} \Delta_{y^{\prime}} \Delta_z \neq 0 $

So no solution.

$ \text { II } \rightarrow \mathrm{S} $

(III) If $ \frac{p}{q} \neq 10 \Rightarrow a \neq d \text { then } \Delta_z \neq 0 $

So no solution.

$ \text { III } \rightarrow \mathrm{S} $

(IV) If $ \Rightarrow a=d \text { then } \Delta_z=0 \Rightarrow \Delta_x=\Delta_y=0 $

So infinitely many solutions.

$ \mathrm{IV} \rightarrow \mathrm{R} $

It is given that matrix

$M = \left[ {\matrix{ {{{\sin }^4}\theta } \cr {1 + {{\cos }^2}\theta } \cr } \matrix{ { - 1 - {{\sin }^2}\theta } \cr {{{\cos }^4}\theta } \cr } } \right] = \alpha I + \beta {M^{ - 1}}$, where


$\alpha $ = $\alpha $($\theta $) and $\beta $ = $\beta $($\theta $) are real numbers and I is the 2 $ \times $ 2 identity matrix.

Now,

$(M) = \left| M \right| = {\sin ^4}\theta {\cos ^4}\theta + 1 + {\sin ^2}\theta + {\cos ^2}\theta + {\sin ^2}\theta {\cos ^2}\theta$

= ${\sin ^4}\theta {\cos ^4}\theta + {\sin ^2}\theta {\cos ^2}\theta + 2$

and$\left[ {\matrix{ {{{\sin }^4}\theta } \cr {1 + {{\cos }^2}\theta } \cr } \matrix{ { - 1 - {{\sin }^2}\theta } \cr {{{\cos }^4}\theta } \cr } } \right] = \left[ {\matrix{ \alpha \cr 0 \cr } \,\matrix{ 0 \cr \alpha \cr } } \right]$$ + {\beta \over {\left| M \right|}}(adj\,M)$

$ \because $ $\left[ {{M^{ - 1}} = {{adj\,M} \over {\left| M \right|}}} \right]$

$ \Rightarrow $ $\left[ {\matrix{ {{{\sin }^4}\theta } \cr {1 + {{\cos }^2}\theta } \cr } \matrix{ { - 1 - {{\sin }^2}\theta } \cr {{{\cos }^4}\theta } \cr } } \right] = \left[ {\matrix{ \alpha \cr 0 \cr } \,\matrix{ 0 \cr \alpha \cr } } \right] + {\beta \over {\left| M \right|}}\left[ {\matrix{ {{{\cos }^4}\theta } \cr { - 1 - {{\cos }^2}\theta } \cr } \matrix{ {1 + {{\sin }^2}\theta } \cr {{{\sin }^4}\theta } \cr } } \right]$

$ \because $ $\left\{ {adj\left[ {\matrix{ a & b \cr c & d \cr } } \right] = \left[ {\matrix{ d & { - b} \cr { - c} & a \cr } } \right]} \right\}$

$ \Rightarrow \beta = - \left| M \right|$ and $\alpha = {\sin ^4}\theta + {\cos ^4}\theta $

$ \Rightarrow \alpha = \alpha (\theta ) = 1 - {1 \over 2}{\sin ^2}(2\theta )$, and $\beta = \beta (\theta ) = - \left\{ {{{\left( {{{\sin }^2}\theta {{\cos }^2}\theta + {1 \over 2}} \right)}^2} + {7 \over 4}} \right\} = - \left\{ {{{\left( {{{{{\sin }^2}(2\theta )} \over 4} + {1 \over 2}} \right)}^2} + {7 \over 4}} \right\}$

Now, ${\alpha ^*} = {\alpha _{\min }} = {1 \over 2}$

and ${\beta ^*} = {\beta _{\min }} = - {{37} \over {16}}$

$ \because $ $\alpha $ is minimum at sin²(2$\theta $) = 1 and $\beta $ is minimum at sin²(2$\theta $) = 1

So, ${\alpha ^*} + {\beta ^*} = {1 \over 2} - {{37} \over {16}} = - {{29} \over {16}}$

Sum of diagonal entries of M^TM is $\sum\limits_{}^{} {a_i^2} $

$\sum\limits_{i = 1}^9 {a_i^2} = 5$

Possibilities

I. 2, 1, 0, 0, 0, 0, 0, 0, 0, which gives ${{9!} \over {7!}}$ matrices

II. 1, 1, 1, 1, 1, 0, 0, 0, 0, which gives ${{9!} \over {4!\, \times 5!}}$ matrices

Total matrices = 9 $ \times $ 8 + 9 $ \times $ 7 $ \times $ 2 = 198

Here, $P = \left[ {\matrix{ 1 & 0 & 0 \cr 4 & 1 & 0 \cr {16} & 4 & 1 \cr } } \right]$

$\therefore$ ${P^2} = \left[ {\matrix{ 1 & 0 & 0 \cr 4 & 1 & 0 \cr {16} & 4 & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 4 & 1 & 0 \cr {16} & 4 & 1 \cr } } \right]$

$ = \left[ {\matrix{ 1 & 0 & 0 \cr {4 + 4} & 1 & 0 \cr {16 + 32} & {4 + 4} & 1 \cr } } \right]$

$ = \left[ {\matrix{ 1 & 0 & 0 \cr {4 \times 2} & 1 & 0 \cr {16(1 + 2)} & {4 \times 2} & 1 \cr } } \right]$ ...... (i)

and ${P^3} = \left[ {\matrix{ 1 & 0 & 0 \cr {4 \times 2} & 1 & 0 \cr {16(1 + 2)} & {4 \times 2} & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 4 & 1 & 0 \cr {16} & 4 & 1 \cr } } \right]$

$ = \left[ {\matrix{ 1 & 0 & 0 \cr {4 \times 3} & 1 & 0 \cr {16(1 + 2 + 3)} & {4 \times 3} & 1 \cr } } \right]$ ..... (ii)

From symmetry,

${P^{50}} = \left[ {\matrix{ 1 & 0 & 0 \cr {4 \times 50} & 1 & 0 \cr {16(1 + 2 + 3 + .... + 50)} & {4 \times 50} & 1 \cr } } \right]$

$\because$ ${P^{50}} - Q = I$ [given]

$\therefore$ $\left[ {\matrix{ {1 - {q_{11}}} & { - {q_{12}}} & { - {q_{13}}} \cr {200 - {q_{21}}} & {1 - {q_{22}}} & { - {q_{23}}} \cr {16 \times {{50} \over 2}(51) - {q_{31}}} & {200 - {q_{32}}} & {1 - {q_{33}}} \cr } } \right]$

$ = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$

$ \Rightarrow 200 - {q_{21}} = 0$, ${{16 \times 50 \times 51} \over 2} - {q_{31}} = 0$, $200 - {q_{32}} = 0$

$\therefore$ ${q_{21}} = 200$, ${q_{32}} = 200$, ${q_{31}} = 20400$

Thus, ${{{q_{31}} + {q_{32}}} \over {{q_{21}}}} = {{20400 + 200} \over {200}}$

$ = {{20600} \over {200}} = 103$

We have ${P^T} = 2P + I$

We get ${P^T} - 2P = I$ ..... (i)

Taking transpose, we have ${({P^T} - 2P)^T} = {I^T}$

$ \Rightarrow P - 2{P^T} = I$ ..... (ii)

From (i) and (ii) on eliminating P^T we have

$ - 4P + P = 3I \Rightarrow P = - I$ $\therefore$ $P + I = 0$

Thus, $(P + 1)X = 0 \Rightarrow PX = - X$

Here, $P = {[{a_{ij}}]_{3 \times 3}} = \left[ {\matrix{ {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr } } \right]$

$Q = {[{b_{ij}}]_{3 \times 3}} = \left[ {\matrix{ {{b_{11}}} & {{b_{12}}} & {{b_{13}}} \cr {{b_{21}}} & {{b_{22}}} & {{b_{23}}} \cr {{b_{31}}} & {{b_{32}}} & {{b_{33}}} \cr } } \right]$

where, ${b_{ij}} = {2^{i + j}}{a_{ij}}$

$\therefore$ $\left| Q \right| = \left| {\matrix{ {4{a_{11}}} & {8{a_{12}}} & {16{a_{13}}} \cr {8{a_{21}}} & {16{a_{22}}} & {32{a_{23}}} \cr {16{a_{31}}} & {32{a_{32}}} & {64{a_{33}}} \cr } } \right|$

$ = 4 \times 8 \times 16\left| {\matrix{ {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr {2{a_{21}}} & {2{a_{22}}} & {2{a_{23}}} \cr {4{a_{31}}} & {4{a_{32}}} & {4{a_{33}}} \cr } } \right|$

$ = {2^9} \times 2 \times 4\left| {\matrix{ {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr } } \right|$

$ = {2^{12}}.\left| P \right| = {2^{12}}.2 = {2^{13}}$

Given, ${[\matrix{ a & b & c \cr } ]_{1 \times 3}}{\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right]_{3 \times 3}} = [\matrix{ 0 & 0 & 0 \cr } ]$

$ \Rightarrow \left[ {\matrix{ {a + 8b + 7c} \cr {9a + 2b + 3c} \cr {7a + 7b + 7c} \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$

$ \Rightarrow a + 8b + 7c = 0$ ..... (i)

$ \Rightarrow 9a + 2b + 3c = 0$ .... (ii)

$ \Rightarrow a + b + c = 0$ ....... (iii)

On multiplying Eq. (iii) by 2, then subtract from Eq. (ii), we get

$7a + c = 0$ ..... (iv)

Again multiplying Eq. (iii) by 3, then subtract from Eq. (ii), we get

$6a - b = 0$ ..... (v)

$\therefore$ $b = 6a$ and $c = - 7a$

As (a, b, c) lies on $2x + y + z = 1$

$ \Rightarrow 2a + b + c = 1$

$ \Rightarrow 2a + 6a - 7a = 1$

$\Rightarrow$ a = 1, b = 6 and c = $-$7

$\therefore$ $7a + b + c = 7 + 6 - 7 = 6$

Given, ${[\matrix{ a & b & c \cr } ]_{1 \times 3}}{\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right]_{3 \times 3}} = [\matrix{ 0 & 0 & 0 \cr } ]$

$ \Rightarrow \left[ {\matrix{ {a + 8b + 7c} \cr {9a + 2b + 3c} \cr {7a + 7b + 7c} \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$

$ \Rightarrow a + 8b + 7c = 0$ ..... (i)

$ \Rightarrow 9a + 2b + 3c = 0$ .... (ii)

$ \Rightarrow a + b + c = 0$ .... (iii)

On multiplying Eq. (iii) by 2, then subtract from Eq. (ii), we get

$7a + c = 0$ ..... (iv)

Again multiplying Eq. (iii) by 3, then subtract from Eq. (ii), we get

$6a - b = 0$ ..... (v)

$\therefore$ b = 6a and c = $-$ 7a

If a = 2, b = 12 and c = $-$14

$\therefore$ ${3 \over {{\omega ^a}}} + {1 \over {{\omega ^b}}} + {3 \over {{\omega ^c}}}$

$ \Rightarrow {3 \over {{\omega ^2}}} + {1 \over {{\omega ^{12}}}} + {3 \over {{\omega ^{ - 14}}}} = {3 \over {{\omega ^2}}} + 1 + 3{\omega ^2}$

$ = 3\omega + 1 + 3{\omega ^2}$

$ = 1 + 3(\omega + {\omega ^2})$

$ = 1 - 3 = - 2$

Given, ${[\matrix{ a & b & c \cr } ]_{1 \times 3}}{\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right]_{3 \times 3}} = [\matrix{ 0 & 0 & 0 \cr } ]$

$ \Rightarrow \left[ {\matrix{ {a + 8b + 7c} \cr {9a + 2b + 3c} \cr {7a + 7b + 7c} \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$

$ \Rightarrow a + 8b + 7c = 0$ ..... (i)

$ \Rightarrow 9a + 2b + 3c = 0$ .... (ii)

$ \Rightarrow a + b + c = 0$ .... (iii)

On multiplying Eq. (iii) by 2, then subtract from Eq. (ii), we get

$7a + c = 0$ ..... (iv)

Again multiplying Eq. (iii) by 3, then subtract from Eq. (ii), we get

$6a - b = 0$ ..... (v)

$\therefore$ b = 6a and c = $-$ 7a

If b = 6, a = 1 and c = $-$7

$\therefore$ $a{x^2} + bx + c = 0$

$ \Rightarrow {x^2} + 6x - 7 = 0$

$ \Rightarrow (x + 7)(x - 1) = 0$

$\therefore$ $x = 1,7$

$ \Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {{1 \over 1} - {1 \over 7}} \right)}^n} \Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {{6 \over 7}} \right)}^n}} } $

$ \Rightarrow 1 + {6 \over 7} + {\left( {{6 \over 7}} \right)^n} + ....\infty $

$ = {1 \over {1 - {6 \over 7}}} = {1 \over {1/7}} = 7$

$\left| A \right| \ne 0$, as non-singular.

$\therefore$ $\left| {\matrix{ 1 & a & b \cr \omega & 1 & c \cr {{\omega ^2}} & \omega & 1 \cr } } \right| \ne 0$

$ \Rightarrow 1(1 - c\omega ) - a(\omega - c{\omega ^2}) + b({\omega ^2} - {\omega ^2}) \ne 0$

$ \Rightarrow 1 - c\omega - a\omega + ac{\omega ^2} \ne 0$

$ \Rightarrow (1 - c\omega )(1 - a\omega ) \ne 0$

$ \Rightarrow a \ne {1 \over \omega },c \ne {1 \over \omega } \Rightarrow a = \omega ,c = \omega $

and $b \in \{ \omega ,{\omega ^2}\} \Rightarrow 2$ solutions

Given,

A $\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$ has two distinct solution we know that three Planes cannot intersect at two distinct point.

Hence, number of $3 \times 3$ matrix $A$ is zero.

We must have $a^2-b^2=1 < p$

$ (a+b)(a-b)=1 < p $

Either $a-b=0$ or $a+b$ is a multiple of $p$ when $a=b$ number of matrices is $p$ and when $a+b$ $=$ multiple of $p$.

$ \Rightarrow a, b \text { has } p-1 $

$ \begin{aligned} \text { Total number of matrices } & =p+p-1 \\\\ & =2 p-1 \end{aligned} $

We have an odd prime $p$. Consider the set

$ T_p \;=\; \Bigl\{ A \;=\;\begin{pmatrix} a & b \\[6pt] c & a \end{pmatrix} \;:\; a,b,c \in \{0,1,2,\dots,p-1\} \Bigr\}. $

We want to count the number of such matrices $A$ that satisfy:

$\text{trace}(A) \not\equiv 0 \pmod{p}$.

Since $\text{trace}(A) = a + a = 2a$, this means

$ 2a \;\not\equiv\; 0 \pmod{p}. $

Because $p$ is an odd prime, $2$ is invertible $\bmod\,p$.

Hence $2a \equiv 0 \pmod{p}$ if and only if $a \equiv 0 \pmod{p}$.

Thus the condition

$\text{trace}(A) \not\equiv 0 \pmod{p}$

is equivalent to

$ a \;\neq\; 0 \pmod{p}. $

In other words, $a$ must be a nonzero element modulo $p$; so

$a\in\{1,2,\dots,p-1\}.$

$\det(A) \equiv 0 \pmod{p}$.

For the matrix

$\begin{pmatrix} a & b \\ c & a \end{pmatrix},$ we have

$ \det(A) \;=\; a^2 - bc. $

The condition $\det(A)\equiv 0\pmod{p}$ means

$ a^2 \;-\; bc \;\equiv\; 0 \pmod{p} \quad\Longleftrightarrow\quad bc \;\equiv\; a^2 \pmod{p}. $

Putting these two pieces together:

$a$ is nonzero ($a \in \{1,\dots,p-1\}$).

$bc \equiv a^2 \pmod{p}.$

---

Counting the solutions

We must count the number of triples $(a,b,c)$ with $a\neq 0$ and $bc \equiv a^2$.

Range for $a$

Since $a\in \{1,2,\dots,p-1\}$, there are $p-1$ possible values of $a$.

Condition $bc \equiv a^2$ for given $a$

Since $a \neq 0$, $a^2$ is also nonzero $\bmod\,p$.

If $b = 0$, then $bc \equiv 0$, which would only be equal to $a^2$ if $a^2 \equiv 0$. But $a^2\neq 0$ since $a\neq 0$. So $b=0$ gives no solutions.

Similarly, if $c=0$, then $bc=0$, which again cannot equal the nonzero $a^2$. So $c=0$ gives no solutions either.

Therefore $b$ and $c$ must both be nonzero modulo $p$.

Once $b$ is chosen to be any nonzero element in $\{1,\dots,p-1\}$, there is a unique $c \equiv a^2 \,b^{-1} \pmod{p}$. Hence:

For each nonzero $b$ (that is, $b\in\{1,\dots,p-1\}$), there is exactly one $c\in\{1,\dots,p-1\}$ satisfying $bc \equiv a^2$.

Consequently, for each fixed nonzero $a$, the number of $(b,c)$-pairs is $p-1$.

Putting it all together

We have $p-1$ choices for $a\neq 0$.

For each such $a$, there are $p-1$ pairs $(b,c)$ satisfying $bc\equiv a^2$.

Therefore, in total, the number of matrices is

$ (p-1)\,\times\,(p-1) \;=\; (p-1)^2. $

Hence the correct answer matches option C, which is

$ \boxed{(p-1)^2.} $

Let $p$ be an odd prime number, and consider the set

$ T_{p} \;=\; \Bigl\{ A \;=\;\begin{pmatrix} a & b\\[6pt] c & a \end{pmatrix} \;:\; a,b,c \in \{0,1,2,\dots,p-1\} \Bigr\}. $

We want to count the number of matrices $A \in T_p$ whose determinant is not divisible by $p$.

Recall that for

$ A \;=\; \begin{pmatrix} a & b \\ c & a \end{pmatrix}, $

the determinant is

$ \det(A) \;=\; a^2 \;-\; b\,c. $

Hence $\det(A)$ is not divisible by $p$ precisely when

$ a^2 \;-\; b\,c \;\not\equiv\; 0 \pmod{p}, \quad\text{i.e.,}\quad a^2 \;\neq\; b\,c \pmod{p}. $

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Total number of matrices

Since $a,b,c$ each range over $\{0,1,\dots,p-1\}$, there are

$ p^3 $

total possible matrices in $T_p$.

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Step 1: Count how many $(a,b,c)$ do satisfy $a^2 \equiv b\,c \pmod{p}$

It will be easier to first count the number of triples $(a,b,c)$ for which

$ a^2 \;\equiv\; b\,c \pmod{p}, $

and then subtract that from $p^3$.

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Case A: $a = 0$

Then $a^2 = 0$. We need

$ b\,c \;\equiv\; 0 \pmod{p}. $

Over the field $\mathbf{F}_p$, the product $b\,c\equiv 0$ if and only if at least one of $b$ or $c$ is $0$ (since $p$ is prime).

If $b=0$, then $c$ can be anything in $\{0,\dots,p-1\}$.

This gives $p$ possibilities.

If $c=0$, then $b$ can be anything in $\{0,\dots,p-1\}$.

This also gives $p$ possibilities.

However, the pair $(b=0, c=0)$ is counted in both cases, so we must subtract it out once.

Hence, for $a=0$, the number of $(b,c)$ such that $b\,c \equiv 0$ is

$ p \;+\; p \;-\; 1 \;=\; 2p \;-\; 1. $

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Case B: $a \neq 0$

Then $a$ can be any nonzero element in $\{1,2,\dots,p-1\}$; there are $p-1$ such choices. In that scenario, $a^2 \neq 0\pmod{p}$. We want

$ b\,c \;\equiv\; a^2 \pmod{p}. $

Over $\mathbf{F}_p$, a nonzero right-hand side $a^2$ can be written as $b\,c$ in the following way:

If $b=0$, then $b\,c = 0$, which cannot equal $a^2\neq 0$.

No solutions in that subcase.

If $b \neq 0$, then for each nonzero $b$ there is exactly one $c$ satisfying $b\,c \equiv a^2\pmod{p}$, namely $c \equiv a^2 b^{-1}\pmod{p}$.

Since $b$ can be any of the $p-1$ nonzero elements, we get $p-1$ solutions $(b,c)$ for each nonzero $a$.

Thus, for each $a\neq 0$, there are $p-1$ pairs $(b,c)$. Hence, for $p-1$ values of $a$, the total number of solutions in this case is

$ (p-1)\,\times\,(p-1) \;=\; (p-1)^2. $

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Total number of solutions to $a^2 = b\,c$

Summing these two cases:

Case A $(a=0)$: $2p - 1$ solutions.

Case B $(a\neq 0)$: $(p-1)^2$ solutions.

Hence the total count of $(a,b,c)$ satisfying $a^2 \equiv b\,c$ is

$ (2p -1) \;+\; (p-1)^2 \;=\; (2p -1) \;+\; \bigl(p^2 \;-\;2p \;+\;1\bigr) \;=\; p^2. $

(A classic result: there are exactly $p^2$ solutions to $b\,c = a^2$ over $\mathbf{F}_p^3$.)

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Step 2: Count how many $(a,b,c)$ satisfy $a^2 \neq b\,c$

Since there are $p^3$ total triples $(a,b,c)$, the number that satisfy

$ a^2 \;\neq\; b\,c \pmod{p} $

is simply

$ p^3 \;-\; p^2. $

This directly corresponds to matrices $A$ whose determinant $a^2 - b\,c$ is $\not\equiv 0\pmod{p}$.

Hence, the number of matrices in $T_p$ with $\det(A)$ not divisible by $p$ is

$ \boxed{p^3 - p^2}. $

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The expression $p^3 - p^2$ matches Option D.

Here,

Matrix [A]_{3 × 3} is symmetric

Here, 5 be the number of 1’s and 4 be the 0’s entry in matrix A

Let x be the number of 1’s on main diagonal.

Since A is symmetric so, y be the number of 1’s below the main diagonal

So , x + 2y = 5

x = 1 and y = 2

or x = 3 and y = 1

Case I : x = 1 and y = 2

Here, Main diagonal entries in 0, 0 and 1. Hence, we can choose main diagonal in 3 ways, and element above the diagonal is 1, 1, 0 . Hence, we can choose it element above the main diagonal in 3 ways , element of below diagonal depends on above the main diagonal.

Hence, total way = 3 × 3 = 9

Case II : x = 3 and y = 1

Here, Main diagonal entries in 1, 1 and 1.
Hence, we can choose main diagonal 1 ways and element above the diagonal is 1, 0, 0. Hence, we can choose it element above the main diagonal in 3 ways, element of below diagonal depends on above the main diagonal.

Hence, total ways = 3 × 4 = 12

So, Total matrices = 9 + 3 = 12

We have

$\left[ {\matrix{ 0 & a & b \cr a & 0 & c \cr b & c & 1 \cr } } \right]$

We can see that either $b = 0$ or $c = 0 \Rightarrow |A| \ne 0$; therefore two matrices.

$\left[ {\matrix{ 0 & a & b \cr a & 1 & c \cr b & c & 0 \cr } } \right]$

Now, either $a = 0$ or $c = 0 \Rightarrow |A| \ne 0$; therefore, two matrices

$\left[ {\matrix{ 1 & a & b \cr a & 0 & c \cr b & c & 0 \cr } } \right]$

Now, either $a = 0$ or $b = 0 \Rightarrow |A| \ne 0$; therefore two matrices.

$\left[ {\matrix{ 1 & a & b \cr a & 1 & c \cr b & c & 1 \cr } } \right]$

When $a = b = 0 \Rightarrow |A| = 0$

When $a = c = 0 \Rightarrow |A| = 0$

When $b = c = 0 \Rightarrow |A| = 0$

Therefore, there are only six matrices.

The six matrices A for which $|A| = 0$ are as follows:

$\left[ {\matrix{ 0 & 0 & 1 \cr 0 & 0 & 1 \cr 1 & 1 & 1 \cr } } \right] \Rightarrow $ inconsistent.

$\left[ {\matrix{ 0 & 1 & 0 \cr 1 & 1 & 1 \cr 0 & 1 & 0 \cr } } \right] \Rightarrow $ inconsistent.

$\left[ {\matrix{ 1 & 1 & 1 \cr 1 & 0 & 0 \cr 1 & 0 & 0 \cr } } \right] \Rightarrow $ infinite solutions.

$\left[ {\matrix{ 1 & 1 & 0 \cr 1 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] \Rightarrow $ inconsistent.

$\left[ {\matrix{ 1 & 0 & 1 \cr 0 & 1 & 0 \cr 1 & 0 & 1 \cr } } \right] \Rightarrow $ inconsistent.

$\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 1 \cr 0 & 1 & 1 \cr } } \right] \Rightarrow $ infinite solutions.