Matrices and Determinants

We can write the matrix as

$ M = \begin{bmatrix}2 & -1 \\ 1 & 0\end{bmatrix} = \begin{bmatrix}1 & -1 \\ 1 & -1\end{bmatrix} + \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} $

So, let $ B = \begin{bmatrix}1 & -1 \\ 1 & -1\end{bmatrix} $. Then $ M = B + I $.

Find $ B^2 $

$ B^2 = \begin{bmatrix}1 & -1 \\ 1 & -1\end{bmatrix} \begin{bmatrix}1 & -1 \\ 1 & -1\end{bmatrix} = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} $

This means $ B $ is a nilpotent matrix, i.e., its square is a zero matrix.

Expand $ M^{26} $ using the Binomial Theorem

$ M^{26} = (B + I)^{26} = {^{26}C_0} I + {^{26}C_1} B + {^{26}C_2} B^2 + \ldots + {^{26}C_{26}} B^{26} $

As $ B^2 = 0 $, all higher powers of $ B $ are zero. So,

$ M^{26} = I + 26B $

Substitute the values

$ M^{26} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} + 26 \begin{bmatrix}1 & -1 \\ 1 & -1\end{bmatrix} = \begin{bmatrix}27 & -26 \\ 26 & -25\end{bmatrix} $

So, $ p = 27, \, q = -26, \, r = 26, \, s = -25. $

Find the sum $ D = \sum_{k=1}^{26} M^k $

$ D = M + M^2 + M^3 + \ldots + M^{26} $

Substitute $ M = B + I $:

$ D = (B + I) + (B + I)^2 + (B + I)^3 + \ldots + (B + I)^{26} $

Since $ (B + I)^k = I + kB $, we have

$ D = (I + B) + (I + 2B) + (I + 3B) + \ldots + (I + 26B) $

Combine like terms:

$ D = 26I + B(1 + 2 + 3 + \ldots + 26) $

The sum $ 1 + 2 + 3 + \ldots + 26 = 351 $, so

$ D = 26I + 351B $

Substitute the matrices:

$ \sum_{k=1}^{26} M^k = \begin{bmatrix}377 & -351 \\ 351 & -325\end{bmatrix} = \begin{bmatrix}a & b \\ c & d\end{bmatrix} $

Hence, $ a = 377, \, b = -351, \, c = 351, \, d = -325. $

So, option (B) is incorrect.

Check option (C)

Given system of equations:

$ 27x - 26y = m $

$ 26x - 25y = n $

Determinant of coefficients:

$ \left|\begin{array}{rr} 27 & -26 \\ 26 & -25 \end{array}\right| \neq 0 $

Since the determinant is non-zero, a unique solution exists.

On solving, we get:

$ x = -24n - 25m, \quad y = -23n - 26m $

Both $ x $ and $ y $ are integers. Hence, option (C) is correct.

Check option (D)

We have:

$ \left|\begin{array}{cc} a + t & b \\ c & d + t \end{array}\right| = ad + at + dt + t^2 - bc $

Substitute $ a = 377, b = -351, c = 351, d = -325 $:

$ = (377)(-325) + t(52) + t^2 + (351)(351) = t^2 + 52t + 676 = (t + 26)^2 $

Since $ (t + 26)^2 \neq 0 $ for all $ t > 0 $, the determinant is non-zero, so a unique solution exists. Option (D) is correct.

Check option (A)

Let $ N = \begin{bmatrix}j & k \\ l & m\end{bmatrix} $.

Compute $ MN $:

$ MN = \begin{bmatrix}2 & -1 \\ 1 & 0\end{bmatrix} \begin{bmatrix}j & k \\ l & m\end{bmatrix} = \begin{bmatrix}2j - l & 2k - m \\ j & k\end{bmatrix} $

Compute $ N\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix} $:

$ N\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}j & j + k \\ l & l + m\end{bmatrix} $

Comparing both, we get $ j = l $ and $ 2k - m = j + k $ ⟹ $ m - k = j $.

Determinant of $ N $:

$ |N| = jm - lk = j(m - k) $

This is not always zero if $ j \neq 0 $ and $ m \neq k $, which means $ N $ can be invertible. Hence, option (A) is correct.

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$