Matrices and Determinants

A^T = A, B^T = $-$B

Let A²B² $-$ B²A² = P

P^T = (A²B² $-$ B²A²)^T = (A²B²)^T $-$ (B²A²)^T

= (B²)^T (A²)^T $-$ (A²)^T (B²)^T

= B²A² $-$ A²B²

$ \Rightarrow $ P is skew-symmetric matrix

$\left[ {\matrix{ 0 & a & b \cr { - a} & 0 & c \cr { - b} & { - c} & 0 \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$

$ \therefore $ ay + bz = 0 ..... (1)

$-$ax + cz = 0 .... (2)

$-$bx $-$cy = 0 ..... (3)

From equation 1, 2, 3

$\Delta$ = 0 & $\Delta$₁ = $\Delta$₂ = $\Delta$₃ = 0

$ \therefore $ equation have infinite number of solution

$x - 2y + 0.z = 1$

$x - y + kz = - 2$

$0.x + ky + 4z = 6$

$\Delta = \left| {\matrix{ 1 & { - 2} & 0 \cr 1 & { - 1} & k \cr 0 & k & 4 \cr } } \right| = 4 - {k^2}$

For unique solution $4 - {k^2} \ne 0$

$ \Rightarrow $ k $ \ne $ $ \pm $ 2

For k = 2 :

$x - 2y + 0.z = 1$

$x - y + 2z = - 2$

$0.x + 2y + 4z = 6$

$\Delta x = \left| {\matrix{ 1 & { - 2} & 0 \cr 2 & { - 1} & 2 \cr 6 & 2 & 4 \cr } } \right| = ( - 8) + 2[ - 20]$

$\Delta x = - 48 \ne 0$

For k = 2, $\Delta x \ne 0$

$ \therefore $ For K = 2; The system has no solution.

$\Delta = \left| {\matrix{ 3 & { - 2} & { - k} \cr 1 & { - 4} & { - 2} \cr 1 & 2 & { - 1} \cr } } \right| = 0$

$3(4 + 4) + 2( - 2 + 2) - k(4 + 4) = 0$

$ \Rightarrow k = 3$

${\Delta _x} = \left| {\matrix{ {10} & { - 2} & { - 3} \cr 6 & { - 4} & { - 2} \cr {5m} & 2 & { - 1} \cr } } \right| \ne 0$

$10(4 + 4) + 2( - 6 + 10m) - 3(12 + 20m) \ne 0$

$80 - 12 + 20m - 36 - 60m \ne 0$

$40m \ne 32 \Rightarrow m \ne {4 \over 5}$

${\Delta _y} = \left| {\matrix{ 3 & {10} & { - 3} \cr 2 & 6 & { - 2} \cr 1 & {5m} & { - 1} \cr } } \right| \ne 0$

$3( - 6 + 10m) - 10( - 2 + 2) - 3(10m - 6) \ne 0$

$ - 18 + 30m - 30m + 18 \ne 0 \Rightarrow 0$

${\Delta _z} = \left| {\matrix{ 3 & { - 2} & {10} \cr 2 & { - 4} & 6 \cr 1 & 2 & {5m} \cr } } \right| \ne 0$

$3( - 20m - 12) + 2(10m - 6) + 10(4 + 4) - 40m + 32 \ne 0 \Rightarrow m \ne {4 \over 5}$

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.

$\operatorname{det} A=\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|=k$ (given)

$\begin{aligned} \text { Then, } \operatorname{det} B & =\left|\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2+2 a_1 & b_2+2 b_1 & c_2+2 c_1 \\ a_3 & b_3 & c_3 \end{array}\right| \\ & =\left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|+\left|\begin{array}{ccc} a_1 & b_1 & c_1 \\ 2 a_1 & 2 b_1 & 2 c_1 \\ a_3 & b_3 & c_3 \end{array}\right| \\ & =k+0 \\ \Rightarrow \operatorname{det} B & =k \end{aligned}$

$\begin{aligned} A & =\left[\begin{array}{llll}\sqrt{2020} & \sqrt{2022} & \sqrt{2022} & \sqrt{2023} \\ \sqrt{4040} & \sqrt{4042} & \sqrt{4044} & \sqrt{4046} \\ \sqrt{6060} & \sqrt{6063} & \sqrt{6066} & \sqrt{6069} \\ \sqrt{8080} & \sqrt{8084} & \sqrt{8088} & \sqrt{8092}\end{array}\right] \\ & =\left[\begin{array}{cccc}\sqrt{2020} & \sqrt{2021} & \sqrt{2022} & \sqrt{2023} \\ \sqrt{2} \cdot \sqrt{2020} & \sqrt{2} \sqrt{2021} & \sqrt{4044} & \sqrt{4046} \\ \sqrt{3} \sqrt{2020} & \sqrt{3} \sqrt{2021} & \sqrt{6066} & \sqrt{6069} \\ \sqrt{4} \sqrt{2020} & \sqrt{4} \sqrt{2021} & \sqrt{8088} & \sqrt{8092}\end{array}\right] \\ R_2 & \rightarrow R_2-\sqrt{2} R_1, R_3 \rightarrow R_3-\sqrt{3} R_1, R_4 \rightarrow R_4-\sqrt{4} R_1 \\ A & \sim\left[\begin{array}{cccc}\sqrt{2020} & \sqrt{2021} & \sqrt{2022} & \sqrt{2023} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{array}\right]\end{aligned}$

$\because$ A has one non-zero row

$\therefore$ Rank (A) = 1

Option (a) is correct.

The given equation is:

$ \left| \begin{array}{lll} x & x^2 & 1 + x^3 \\ y & y^2 & 1 + y^3 \\ z & z^2 & 1 + z^3 \end{array} \right| = 0 $

We can rewrite the third column as (1) + (the cube term), so this is the same as:

$ \left| \begin{array}{lll} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{array} \right| + \left| \begin{array}{lll} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{array} \right| = 0 $

Let’s call the first determinant as $A$ and the second as $B$.

Step 1: Find the value of $|A|$

Let: $ A = \left| \begin{array}{lll} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{array} \right| $

Subtract first row from second and third, $ R_2 \rightarrow R_2 - R_1,\quad R_3 \rightarrow R_3 - R_1 $ so, $ \left| \begin{array}{ccc} x & x^2 & 1 \\ y-x & y^2 - x^2 & 0 \\ z-x & z^2 - x^2 & 0 \end{array} \right| $

Now, since last column for $R_2$ and $R_3$ is 0, expand along third column:

$|A| = (y-x)(z^2 - x^2) - (z-x)(y^2 - x^2)$

$= (y-x)(z-x)(z+x) - (z-x)(y-x)(y+x)$

$= (y-x)(z-x)(z+x - y-x)$

$= (y-x)(z-x)(z-y)$

Step 2: Find the value of $|B|$

$ B = \left| \begin{array}{ccc} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{array} \right| $

Factor $x, y, z$ from each row: $ = x y z \left| \begin{array}{ccc} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{array} \right| $

Now, do the same row operation as before: $ R_2 \rightarrow R_2 - R_1,\quad R_3 \rightarrow R_3 - R_1 $

So determinant becomes: $ x y z \left| \begin{array}{ccc} 1 & x & x^2 \\ 0 & y-x & y^2 - x^2 \\ 0 & z-x & z^2 - x^2 \end{array} \right| $

Expand this along the first column:

$|B| = x y z \left\{ (y-x)(z^2 - x^2) - (z-x)(y^2 - x^2) \right\}$

$= x y z (y-x)(z-x)(z-y)$

Step 3: Use the original equation

We have: $ |A| + |B| = 0 $ So, $ (y-x)(z-x)(z-y) + x y z (y-x)(z-x)(z-y) = 0 $

Factor out $(y-x)(z-x)(z-y)$: $ (y-x)(z-x)(z-y)(1 + x y z) = 0 $

Because $x, y, z$ are all different, $(y-x)(z-x)(z-y) \neq 0$. So, $ 1 + x y z = 0 $

Therefore, $ x y z = -1 $

Let $n=2$ (say) and $A=\left[\begin{array}{ll}a & b \\ 0 & c\end{array}\right]$, Then $\operatorname{adj} A=\left[\begin{array}{cc}c & -b \\ 0 & a\end{array}\right]$ which is a upper triangular matrix. Then, for any $n \times n$ matrix $A$, $\operatorname{adj} A$ is also upper triangular matrix.

$f(x)=\left|\begin{array}{ccc}x & x^2 & x^3 \\ 1 & 2 x & 3 x^2 \\ 0 & 2 & 6 x\end{array}\right|$

Expand along column 1

$\begin{aligned} f(x) & =x\left(12 x^2-6 x^2\right)-\left(6 x^3-2 x^3\right) \\ & =x\left(6 x^2\right)-4 x^3=6 x^3-4 x^3=2 x^3 \\ \therefore \quad f(x) & =2 x^3 \Rightarrow f^{\prime}(x)=6 x^2 \\ f^{\prime \prime}(x) & =12 x \end{aligned}$

Then, $f^{\prime \prime}(x): f^{\prime}(x)=12 x: 6 x^2=2: x$

Trace of A = Sum of diagonal elements of A

Given, $A=\left[\begin{array}{ccc}1 & -5 & 7 \\ 0 & 7 & 9 \\ 11 & 8 & 9\end{array}\right]$

$=1+7+9=17$

$\begin{aligned} & \text { }-x+y \cos C+z \cos B=0 \\ & x \cos C-y+z \cos A=0 \\ & x \cos B+y \cos A-Z=0 \\ & \Delta=\left|\begin{array}{ccc}-1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1\end{array}\right| \\ & =-1\left(1-\cos ^2 A\right)+\cos C(\cos A \cos B+\cos C) \\ & +\cos B(\cos A \cos C+\cos B) \\ & \end{aligned}$

$\begin{aligned} & =-1+\cos ^2 A+2 \cos A \cos B \cos C \\ & \qquad+\cos ^2 C+\cos ^2 B \\ & =-\sin ^2 A+\cos ^2 B+\cos ^2 C+2 \cos A \cos B \cos C \\ & \text { Clearly, } \Delta>0 . \\ & \Rightarrow \text { System of equations has a non-zero solution. }\end{aligned}$

$\begin{aligned} \text { Let } A & =\left[\begin{array}{cc} 1 & \tan \theta \\ -\tan \theta & 1 \end{array}\right] \\ A^{-1} & =\frac{\operatorname{adj}(A)}{|A|} \\ & =\left(\frac{1}{\sec ^2 \theta}\right)\left[\begin{array}{cc} 1 & -\tan \theta \\ \tan \theta & 1 \end{array}\right] \end{aligned}$

So, $\frac{1}{\sec ^2 \theta}\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]$

$\begin{aligned} &=\frac{1}{\sec ^2 \theta}\left[\begin{array}{cc} 1-\tan ^2 \theta & -2 \tan \theta \\ 2 \tan \theta & -\tan ^2 \theta+1 \end{array}\right] \\ & {\left[\begin{array}{ll} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{array}\right]=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right] } \\ & \Rightarrow \quad a=\cos 2 \theta, b=\sin 2 \theta \end{aligned}$

$\begin{aligned} & {\left[\begin{array}{ccc}a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b\end{array}\right] \quad\left[R_1 \rightarrow R_1+R_3\right]} \\ & =(a+b+c)\left[\begin{array}{ccc}1 & 1 & 1 \\ a-b & b-c & c-a \\ b+c & c+a & a+b\end{array}\right] \\ & =(a+b+c)\left[\begin{array}{ccc}1 & 1 & 1 \\ a+c & b+a & b+c \\ b+c & c+a & a+b\end{array}\right] \quad {\left[R_2 \rightarrow R_2+R_3\right]} \\ & \end{aligned}$

$\begin{aligned} C_2 & \rightarrow C_2-C_1 \text { and } C_3 \rightarrow C_3-C_1 \\ & =(a+b+c)\left[\begin{array}{ccc}1 & 0 & 0 \\ a+c & b-c & b-a \\ b+c & a-b & a-c\end{array}\right] \\ & =(a+b+c)[(b-c)(a-c)-(b-a)(a-b)] \\ & =(a+b+c)\left(a b-b c-c a+c^2-b a+b^2+a^2-a b\right) \\ & =(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right) \\ & =a^3+b^3+c^3-3 a b c\end{aligned}$

$\begin{aligned} & \text { }\left|\begin{array}{ccc} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{array}\right| \\ & =(b+c)[(c+a)(a+b)-b c]-a\left(a b+b^2-b c\right) \\ & =(b+c)\left(a c+b c+a^2+a b-b c\right) \quad+a\left(b c-c^2-a c\right) \\ & -a^2 b-a b^2+a b c+a b c-a c^2-a^2 c \\ & =a b c+a^2 b+a b^2+a c^2+a^2 c+a b c -a^2 b-a b^2+a b c+a b c-a c^2-a^2 c\\ & =4 a b c \end{aligned}$

Option (c) is correct.

${D^T} = ABC$

and $S = ABCD$

${S^2} = (ABCD)(ABCD) = ({D^T}\,.\,D)({D^T}D)$

$ = {({D^T}D)^2}$ ..... (i)

$\because$ $S = ABCD = {D^T}D$

${S^T} = {({D^T}D)^T} = {D^T}D$

$\therefore$ From Eq. (i), we get

${S^2} = {({S^T})^2}$

$\therefore$ ${S^2} = {({S^T})^2} = {({S^2})^T}$

$\begin{aligned} A & =\left[\begin{array}{lll} a^2 & 15 & 31 \\ 12 & b^2 & 41 \\ 35 & 61 & c^2 \end{array}\right], \\ B & =\left[\begin{array}{ccc} 2 a & 3 & 5 \\ 2 & 2 b & 8 \\ 1 & 4 & 2 c-3 \end{array}\right] \end{aligned}$

Given, trace of $A=\operatorname{trace}$ of $B$

$\begin{aligned} & \Rightarrow a^2+b^2+c^2=2 a+2 b+2 c-3 \\ & \Rightarrow\left(a^2-2 a+1\right)+\left(b^2-2 b+1\right)+\left(c^2-2 c+1\right)=0 \\ & \Rightarrow(a-1)^2+(b-1)^2+(c-1)^2=0 \\ & \Rightarrow a=1, b=1, c=1 \end{aligned}$

Then product of diagonal elements of $B$

$=(2 a)(2 b)(2 c-3)=(2)(2)(2-3)=-4$

$\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|$

$ +\left|\begin{array}{ccc} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n-1} b & (-1)^n c \end{array}\right|=0$

$\begin{aligned} & \because\left.|A|=\mid A^T\right] \\ & \Rightarrow\left|\begin{array}{ccc} a & -b & c \\ a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \end{array}\right| \\ &+(-1)^n\left|\begin{array}{ccc} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ a & -b & c \end{array}\right|=0 \end{aligned}$

$\begin{aligned} & R_2 \leftrightarrow R_3 \text { and } R_3 \leftrightarrow R_1 \text { in 2nd determinant } \\ & \Rightarrow\left|\begin{array}{ccc} a & -b & c \\ a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \end{array}\right| \\ & +(-1)^n\left|\begin{array}{ccc} a & -b & c \\ a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \end{array}\right|=0 \\ \end{aligned}$

Both determinants are equal and their sum is zero when $n$ is an odd integer.

$\begin{array}{lr}\text { }\left|\begin{array}{ccc}1 & -3 & 1 \\ 1 & 6 & 4 \\ 1 & 3 x & x^2\end{array}\right|=0 \\ \Rightarrow 1\left(6 x^2-12 x\right)+3\left(x^2-4\right)+1(3 x-6)=0 \\ \Rightarrow 6 x^2-12 x+3 x^2-12+3 x-6=0 \\ \Rightarrow 9 x^2-9 x-18=0 \\ \Rightarrow x^2-x-2=0\end{array}$

$\begin{aligned} & \text { } A=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right] \\ & \text { Then, } A^T=\left[\begin{array}{cc} a & b \\ -b & a \end{array}\right] \\ & \because \quad A A^T=20 I \\ & \Rightarrow \quad\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]\left[\begin{array}{cc} a & b \\ -b & a \end{array}\right]=20\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{cc} a^2+b^2 & 0 \\ 0 & a^2+b^2 \end{array}\right]=\left[\begin{array}{cc} 20 & 0 \\ 0 & 20 \end{array}\right] \\ & \therefore \quad a^2+b^2=20 \\ & \Rightarrow \quad(a+b)^2-2 a b=20 \\ & \Rightarrow \quad(a+b)^2=20+2 \times \frac{5}{2} \quad\left[\because a b=\frac{5}{2}\right] \\ & \text { or } \quad(a+b)^2=25 \\ & or\quad a+b= \pm 5 \\ \end{aligned}$

Equation whose roots are $a$ and $b$ is given by $x^2+(a+b) x+a b=0$

or $\quad x^2 \pm 5 x+\frac{5}{2}=0$ or $2 x^2 \pm 10 x+5=0$

$\begin{aligned} & \text { (c) } A=\left[\begin{array}{rrr} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{array}\right] \\ & \therefore|A|=1(1+3)+1(2+3)+1(2-1)=10 \quad ... (\mathrm{i})\\ & \text { and } 10 B=\left[\begin{array}{rrr} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{array}\right] \\ & \therefore|10 B|=4(2 \alpha)-2(-15-\alpha)+2(10) \\ & \Rightarrow 10^3|B|=10 \alpha+50 \quad\left[\because|k A|=k^n|A|\right] \ldots \mathrm{(ii)}\\ & \because B=A^{-1}\\ & \Rightarrow|B|=\left|A^{-1}\right|=|A|^{-1} \\ & \Rightarrow|B|=\frac{1}{|A|}=\frac{1}{10} \quad [\text { [rom Eq. (i)} \end{aligned}$

[from Eq. (i)] From Eq. (ii), we get

$1000~.~\frac{1}{10}=10\alpha+50$

or $\alpha=\frac{50}{10}=5$

Let $A=\left[\begin{array}{ccc}4 & 2 & 1-x \\ 5 & k & 1 \\ 6 & 3 & 1+x\end{array}\right]$

Given that, rank of matrix $A$ is 1 .

$\Rightarrow$ All the minors of order 2 are zero.

$\therefore {M_{33}} = 0$

$ \Rightarrow \left| {\matrix{ 4 & 2 \cr 5 & k \cr } } \right| = 0$

or $4 k-10=0$ or $k=\frac{5}{2}$

and $M_{11}=\left|\begin{array}{cc}k & 1 \\ 3 & 1+x\end{array}\right|=0$

or $\left|\begin{array}{cc}\frac{5}{2} & 1 \\ 3 & 1+x\end{array}\right|=0$

or $\frac{5}{2}+\frac{5}{2} x-3=0$

or $\frac{5}{2} x =\frac{1}{2}$

or $x =\frac{1}{5}$

$a_1, a_2, \ldots, a_9$, are in GP.

Then, $\frac{a_2}{a_1}=\frac{a_3}{a_2}=\ldots \frac{a_9}{a_8}=r$ [common ratio]

Now, $\left|\begin{array}{lll}\log a_1 & \log a_2 & \log a_3 \\ \log a_4 & \log a_5 & \log a_6 \\ \log a_7 & \log a_8 & \log a_9\end{array}\right|$

$C_3 \rightarrow C_3-C_2, C_2 \rightarrow C_2-C_1 $

$\left| {\matrix{ {\log {a_1}} & {\log {{{a_2}} \over {{a_1}}}} & {\log {{{a_3}} \over {{a_2}}}} \cr {\log {a_4}} & {\log {{{a_5}} \over {{a_4}}}} & {\log {{{a_6}} \over {{a_5}}}} \cr {\log {a_7}} & {\log {{{a_8}} \over {{a_7}}}} & {\log {{{a_9}} \over {{a_8}}}} \cr } } \right|$

or $\left| {\matrix{ {\log {a_1}} & {\log r} & {\log r} \cr {\log {a_4}} & {\log r} & {\log r} \cr {\log {a_7}} & {\log r} & {\log r} \cr } } \right|=0$

[$\because$ C$_2$ and C$_3$ are identical]

$\begin{aligned} & \text { } \mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{c}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ & \mathbf{a} \cdot \mathbf{a}=4+1+9=14 \\ & \mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{a}=2+3-3=2\end{aligned}$

$\begin{aligned} & \mathbf{a} \cdot \mathbf{c}=\mathbf{c} \cdot \mathbf{a}=6-1-6=-1 \\ & \text { b. } \mathbf{b}=1+9+1=11 \\ & \mathbf{b} \cdot \mathbf{c}=\mathbf{c} \cdot \mathbf{b}=3-3+2=2 \\ & \mathbf{c} \cdot \mathbf{c}=9+1+4=14 \\ & \text { Now, }\left|\begin{array}{lll}\mathbf{a} \cdot \mathbf{a} & \mathbf{a} \cdot \mathbf{b} & \mathbf{a} \cdot \mathbf{c} \\ \mathbf{b} \cdot \mathbf{a} & \mathbf{b} \cdot \mathbf{b} & \mathbf{b} \cdot \mathbf{c} \\ \mathbf{c} \cdot \mathbf{a} & \mathbf{c} \cdot \mathbf{b} & \mathbf{c} \cdot \mathbf{c}\end{array}\right|=\left|\begin{array}{rrr}14 & 2 & -1 \\ 2 & 11 & 2 \\ -1 & 2 & 14\end{array}\right| \\ & =14(154-4)-2(28+2)-1(4+11) \\ & =2100-60-15 \\ & =2025 \\ & \end{aligned}$

Given, $\left| {\matrix{ 0 & {x - p} & {x - q} \cr {x + p} & 0 & {x - r} \cr {x + q} & {x - r} & 0 \cr } } \right| = 0$

$ - (x - p)\left| {\matrix{ {x + p} & {x - r} \cr {x + q} & 0 \cr } } \right| + (x - q)\left| {\matrix{ {x + p} & 0 \cr {x + q} & {x - r} \cr } } \right| = 0$

$ \Rightarrow (x - p)(x + q)(x - r) + (x - q)(x + p)(x - r) = 0$

$ \Rightarrow (x - r)[(x - p)(x + q) + (x - q)(x + p)] = 0$

$ \Rightarrow (x - r)[{x^2} - px + qx - pq + {x^2} - qx + px - pq] = 0$

$ \Rightarrow (x - r)[2{x^2} - 2pq] = 0 \Rightarrow x - r = 0$ or ${x^2} - pq = 0$

either x = r or pq

A . adj A = | A | I

| A | = xyz $-$ 8x $-$ 3(z $-$ 8) + 2(2 $-$ 2y)

| A | = xyz $-$ (8x + 3z + 4y) + 28

$\Rightarrow$ 60 $-$ 20 + 28 = 68

$\Rightarrow$ (adj A)^$-$1 always exists whenever (A)^$-$1 exists.

$\therefore$ A . adj A = | A | I

$ = 68\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] = \left[ {\matrix{ {68} & 0 & 0 \cr 0 & {68} & 0 \cr 0 & 0 & {68} \cr } } \right]$

$A = \left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$

A² = $\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$

$ \Rightarrow $ A² = $\left[ {\matrix{ {\cos 2\theta } & {\sin 2\theta } \cr { - \sin 2\theta } & {\cos 2\theta } \cr } } \right]$

Similarly, Aⁿ = $\left[ {\matrix{ {\cos n\theta } & {\sin n\theta } \cr { - \sin n\theta } & {\cos n\theta } \cr } } \right]$

$ \therefore $ B = A + A⁴

= $\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$ + $\left[ {\matrix{ {\cos 4\theta } & {\sin 4\theta } \cr { - \sin 4\theta } & {\cos 4\theta } \cr } } \right]$

= $\left[ {\matrix{ {\cos 4\theta + \cos \theta } & {\sin 4\theta + \sin \theta } \cr { - \sin 4\theta - \sin \theta } & {\cos 4\theta + \cos \theta } \cr } } \right]$

detB = (cos4$\theta $ + cos$\theta $)² + (sin4$\theta $ + sin$\theta $)²

= cos²4$\theta $ + cos²$\theta $ + 2cos4$\theta $ cos$\theta $
+ sin²4$\theta $ + sin2$\theta $ + 2sin4$\theta $ –sin$\theta $

= 2 + 2 ( cos4$\theta $ cos$\theta $ + sin4$\theta $ sin$\theta $)

$ \Rightarrow $ detB = 2 + 2 cos3$\theta $

at $\theta $ = ${\pi \over 5}$

detB = 2 + 2cos ${{3\pi } \over 5}$

= 2(1 - sin18)

= 2(1 - ${{\sqrt 5 - 1} \over 4}$)

= 2$\left( {{{5 - \sqrt 5 } \over 4}} \right)$

= ${{{5 - \sqrt 5 } \over 2}}$ $ \simeq $ 1.385

$ \therefore $ detB $ \in $ (1, 2)

For infinite many solutions

D = D₁ = D₂ = D₃ = 0

Now D = $\left| {\matrix{ 1 & 1 & 1 \cr 1 & 2 & 3 \cr 1 & 3 & \lambda \cr } } \right|$ = 0

$ \Rightarrow $ 1. (2$\lambda $ – 9) –1.($\lambda $ – 3) + 1.(3 – 2) = 0

$ \Rightarrow $ $\lambda $ = 5

Now D₁ = $\left| {\matrix{ 2 & 1 & 1 \cr 5 & 2 & 3 \cr \mu & 3 & 5 \cr } } \right|$ = 0

$ \Rightarrow $ 2(10 – 9) –1(25 – 3$\mu $) + 1(15 – 2$\mu $) = 0

$ \Rightarrow $ $\mu $ = 8

$\left| {\matrix{ {{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \cr {1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right|$

R₁ $ \to $ R₁ – R₂, R₂ $ \to $ R₂ – R₃

$\left| {\matrix{ { - 1} & 1 & 0 \cr 1 & 0 & { - 1} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right|$

= –1(sin² x) – 1(1 + sin2x + cos² x)

= - sin2x - 2

$ \therefore $ minimum value when sin2x = 1

m = - 2 - 1 = -3

$ \therefore $ Maximum value when sin2x = –1

M = -2 + 1 = -1

$ \therefore $ (m, M) = (–3, –1)

x + y + 3z = 0 .....(i)
x + 3y + k²z = 0 .........(ii)
3x + y + 3z = 0 ......(iii)

$\left| {\matrix{ 1 & 1 & 3 \cr 1 & 3 & {{k^2}} \cr 3 & 1 & 3 \cr } } \right|$ = 0

$ \Rightarrow $ 9 + 3 + 3k² – 27 – k² – 3 = 0

$ \Rightarrow $ k² = 9

Perform (i) – (iii),

–2x = 0 $ \Rightarrow $ x = 0

Now from (i), y + 3z = 0

$ \Rightarrow $ ${y \over z} = - 3$

$ \therefore $ x + $\left( {{y \over z}} \right)$ = -3

$\left| {\matrix{ x & {a + y} & {x + a} \cr y & {b + y} & {y + b} \cr z & {c + y} & {z + c} \cr } } \right|$

C₃ $ \to $ C₃ – C₁

= $\left| {\matrix{ x & {a + y} & a \cr y & {b + y} & b \cr z & {c + y} & c \cr } } \right|$

C₂ $ \to $ C₂ – C₃

= $\left| {\matrix{ x & y & a \cr y & y & b \cr z & y & c \cr } } \right|$

R₃ $ \to $ R₃ – R₁, R₂ $ \to $ R₂ – R₁

= $\left| {\matrix{ x & y & a \cr {y - x} & 0 & {b - a} \cr {z - x} & 0 & {c - a} \cr } } \right|$

= (–y)[(y – x) (c – a) – (b – a) (z – x)]

Given, a + x = b + y = c + z + 1

= (–y)[(a – b) (c – a) + (a – b) (a – c – 1)]

= (–y)[(a – b) (c – a) + (a – b) (a – c) + b – a)

= –y(b – a) = y(a – b)

D = $\left| {\matrix{ 2 & { - 4} & \lambda \cr 1 & { - 6} & 1 \cr \lambda & { - 10} & 4 \cr } } \right|$ = 0

$ \Rightarrow $ $\lambda $ = 3, $ - {2 \over 3}$

D₁ = $\left| {\matrix{ 1 & { - 4} & \lambda \cr 2 & { - 6} & 1 \cr 3 & { - 10} & 4 \cr } } \right|$

= 14 + 4(5) + $\lambda $(–2)

= –2$\lambda $ + 6

D₂ = $\left| {\matrix{ 2 & 1 & \lambda \cr 1 & 2 & 1 \cr \lambda & 3 & 4 \cr } } \right|$

= –2($\lambda $ – 3)($\lambda $ + 1)

D₃ = $\left| {\matrix{ 2 & { - 4} & 1 \cr 1 & { - 6} & 2 \cr \lambda & { - 10} & 3 \cr } } \right|$

= – 2$\lambda $ + 6

When , $\lambda $ = 3 then

D = D₁ = D₂ = D₃ = 0

$ \Rightarrow $ Infinite many solution

When $\lambda $ = $ - {2 \over 3}$ then D₁, D₂, D₃ none of them is zero so equations are inconsistant.

$ \therefore $ $\lambda $ = $ - {2 \over 3}$

Given
$f\left( \theta \right) = \left| {\matrix{ { - {{\sin }^2}\theta } & { - 1 - {{\sin }^2}\theta } & 1 \cr { - {{\cos }^2}\theta } & { - 1 - {{\cos }^2}\theta } & 1 \cr {12} & {10} & { - 2} \cr } } \right|$

C₁ $ \to $ C₁ – C₂ , C₃ $ \to $ C₃ + C₂

= $\left| {\matrix{ 1 & { - 1 - {{\sin }^2}\theta } & { - {{\sin }^2}\theta } \cr 1 & { - 1 - {{\cos }^2}\theta } & { - {{\cos }^2}\theta } \cr 2 & {10} & 8 \cr } } \right|$

C₂ $ \to $ C₂ – C₃

= $\left| {\matrix{ 1 & { - 1} & { - {{\sin }^2}\theta } \cr 1 & { - 1} & { - {{\cos }^2}\theta } \cr 2 & 2 & 8 \cr } } \right|$

= 1(2cos²$\theta $ – 8) + (8 + 2cos²$\theta $) – 4sin²$\theta $

= 4cos²$\theta $ - 4cos²$\theta $

= 4 cos 2$\theta $

$\theta $ $ \in $ $\left[ {{\pi \over 4},{\pi \over 2}} \right]$

$ \Rightarrow $ 2$\theta $ $ \in $ $\left[ {{\pi \over 2},{\pi }} \right]$

$ \Rightarrow $ cos 2$\theta $ $ \in $ [-1, 0]

$ \Rightarrow $ 4cos 2$\theta $ $ \in $ [-4, 0]

$ \Rightarrow $ $f\left( \theta \right)$ $ \in $ [-4, 0]

$ \therefore $ (m, M) = (–4, 0)

Let A = $\left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right]$

For Ax₁ = b₁ :

$ \Rightarrow $ $\left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right]\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right] = \left[ {\matrix{ 1 \cr 0 \cr 0 \cr } } \right]$

$ \therefore $ ${a_1} + {a_2} + {a_3} = 1$ ....(1)

${a_4} + {a_5} + {a_6} = 0$ ......(2)

${a_7} + {a_8} + {a_9} = 0$ .....(3)

For Ax₂ = b₂ :

$\left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right]\left[ {\matrix{ 0 \cr 2 \cr 1 \cr } } \right] = \left[ {\matrix{ 0 \cr 2 \cr 0 \cr } } \right]$

$ \therefore $ $2{a_2} + {a_3} = 0$ .....(4)

$2{a_5} + {a_6} = 2$ ....(5)

$2{a_8} + {a_9} = 0$ ....(6)

For Ax₃ = b₃ :

$\left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right]\left[ {\matrix{ 0 \cr 0 \cr 1 \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 2 \cr } } \right]$

$ \therefore $ ${{a_3} = 0}$

${{a_6} = 0}$

${{a_9} = 2}$

Putting value of ${{a_3}}$ in equation (4), we get

${{a_2}}$ = 0

Putting value of ${{a_6}}$ in equation (5), we get

${{a_5}}$ = 1

Putting value of ${{a_9}}$ in equation (6), we get

${{a_8}}$ = -1

Putting value of ${{a_2}}$ and ${{a_3}}$ in equation (1), we get

${{a_1}}$ = 1

Putting value of ${{a_5}}$ and ${{a_6}}$ in equation (6), we get

${{a_4}}$ = -1

Putting value of ${{a_5}}$ and ${{a_6}}$ in equation (6), we get

${{a_4}}$ = -1

Putting value of ${{a_8}}$ and ${{a_9}}$ in equation (6), we get

${{a_7}}$ = -1

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

So, |A| = 2(1) = 2

$D = 0\,\left| {\matrix{ 1 & 1 & 1 \cr 2 & 4 & { - 1} \cr 3 & 2 & \lambda \cr } } \right| = 0$

$ \Rightarrow $ $(4\lambda + 2) - 1(2\lambda + 3) + 1(4 - 12) = 0$

$ \Rightarrow $ $4\lambda + 2$ $ - 2\lambda - 3$$ - $8$ = 0$

$ \Rightarrow $ $2\lambda = 9 \Rightarrow \lambda = {9 \over 2}$

${D_x} = \left| {\matrix{ 2 & 1 & 1 \cr 6 & 4 & { - 1} \cr \mu & 2 & { - 9/2} \cr } } \right| = 0$

$ \Rightarrow \mu = 5$

By checking all the options we find (C) is correct

$2\lambda + \mu = 14$

$ \because $ $A = \left[ {\matrix{ {\cos \theta } & {i\sin \theta } \cr {i\sin \theta } & {\cos \theta } \cr } } \right]$

$ \therefore $ ${A^n} = \left[ {\matrix{ {\cos \,n\theta } & {i\sin \,n\theta } \cr {i\sin \,n\theta } & {\cos \,n\theta } \cr } } \right],n \in N$

$ \therefore $${A^5} = \left[ {\matrix{ {\cos \,5\theta } & {i\sin \,5\theta } \cr {i\sin \,5\theta } & {\cos \,5\theta } \cr } } \right] = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$

$ \therefore $ $a = \cos 5\theta ,\,b = i\sin 5\theta = c,\,d = \cos 5\theta $

$ \therefore $ ${a^2} - {b^2} = {\cos ^2}5\theta + {\sin ^2}5\theta = 1$

${a^2} - {c^2} = {\cos ^2}5\theta + {\sin ^2}5\theta = 1$

${a^2} - {d^2} = {\cos ^2}5\theta - {\cos ^2}5\theta = 1$

${a^2} + {b^2} = {\cos ^2}5\theta - {\sin ^2}5\theta = \cos 10\theta = \cos {{10\pi } \over {24}}$

and $0 < \cos {{5\pi } \over {12}} < 1 $

$\Rightarrow 0 \le {a^2} + {b^2} \le 1$

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

$B = adj\,(adj\,A)$

$ = |A{|^{n - 2}}A$

$ = |A{|^{3 - 2}}.A$ [As here n = 3]

$ = |A|.A$ .....(1)

Now, $|adj\,A| = \left[ {\matrix{ 2 & { - 1} & 1 \cr { - 1} & 0 & 2 \cr 1 & { - 2} & { - 1} \cr } } \right]$

$ = + 2(0 + 4) + 1(1 - 2) + 1(2 - 0)$

$ = 8 - 1 + 2$

$ = 9$

Also we know, |adj A| = |A|^n$-$1

$ \therefore $ Here |adj A| = |A|^{3 $-$ 1} = |A|²

$ \therefore $ |A|² = 9

$ \Rightarrow $ |A| = $ \pm $3

Given, |A| = $\lambda $

$ \therefore $ $\lambda $ = $ \pm $3

$ \Rightarrow $ |$\lambda $| = 3

From (1)

B = ($ \pm $3)A

Given, |(B^$-$1)^T| = $\mu $

$ \Rightarrow $ |(B^T)^$-$1| = $\mu $

$ \Rightarrow $ ${1 \over {|{B^T}|}} = \mu $

$ \Rightarrow $ ${1 \over {|B|}} = \mu $ [As |B^T| = |B|]

$ \Rightarrow $ ${1 \over {| \pm 3A|}} = \mu $

$ \Rightarrow $ ${1 \over {{{\left( { \pm 3} \right)}^3}\left| A \right|}} = \mu $ [As |KA| = Kⁿ |A|]

$ \Rightarrow {1 \over {( \pm 27)|A|}} = \mu $

$ \Rightarrow \mu = {1 \over {( \pm 27)( \pm 3)}} = \pm {1 \over {81}}$

$ \therefore $ $(|\lambda |,\,\mu ) = \left( {3,\, \pm {1 \over {81}}} \right)$

Here correct option will be $\left( {3,\,{1 \over {81}}} \right)$

$\Delta $ = $\left| {\matrix{ {x - 2} & {2x - 3} & {3x - 4} \cr {2x - 3} & {3x - 4} & {4x - 5} \cr {3x - 5} & {5x - 8} & {10x - 17} \cr } } \right|$

R₂ $ \to $ R₂ – R₁
R₃ $ \to $ R₃ – R₂

= $\left| {\matrix{ {x - 2} & {2x - 3} & {3x - 4} \cr {x - 1} & {x - 1} & {x - 1} \cr {x - 2} & {2\left( {x - 2} \right)} & {6\left( {x - 2} \right)} \cr } } \right|$

= $\left( {x - 1} \right)\left( {x - 2} \right)\left| {\matrix{ {x - 2} & {2x - 3} & {3x - 4} \cr 1 & 1 & 1 \cr 1 & 2 & 6 \cr } } \right|$

C₁ $ \to $ C₁ - C₂
C₂ $ \to $ C₂ - C₃

= $\left( {x - 1} \right)\left( {x - 2} \right)\left| {\matrix{ { - x + 1} & { - x + 1} & {3x - 4} \cr 0 & 0 & 1 \cr { - 1} & { - 4} & 6 \cr } } \right|$

= -(x - 1)(x - 2)[-4(1 - x) + 1(1 - x)]

= -(x² - 3x + 2)[3x - 3]

= -3x³ + 9x² - 6x + 3x² - 9x + 6

= -3x³ + 12x² - 15x + 6 = Ax³ + Bx² + Cx + D

$ \therefore $ A = -3, B = 12, C = -15

$ \therefore $ B + C = 12 – 15 = – 3

Given,
${a^3} + {b^3} + {c^3} = 2$

${A^T}A = I$

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

$ \Rightarrow {a^2} + {b^2} + {c^2} = 1$

and ab + bc + ca = 0

Now (a + b + c)² = 1

$ \Rightarrow (a + b + c) = \pm 1$

So, ${a^3} + {b^3} + {c^3} - 3abc$

$ \Rightarrow (a + b + c)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$

$ \Rightarrow \pm 1(1 - 0) = \pm 1$

$ \therefore 2 - 3abc = \pm 1$

$ \Rightarrow 3abc = 2 \pm 1 = 3,1$

$ \Rightarrow abc = 1,{1 \over 3}$

Let $X = \left[ {\matrix{ x \cr y \cr z \cr } } \right]$

PX = O

$\left[ {\matrix{ 1 & 2 & 1 \cr { - 2} & 3 & { - 4} \cr 1 & 9 & { - 1} \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$

x + 2y + z = 0........(1)

-2x + 3y – 4z = 0....(2)

x + 9y - z = 0..........(3)

from (1) & (3)
$ \Rightarrow $ 2x+11y =0

from (1) & (2)
$ \Rightarrow $ 2x + 11y = 0

from (2) & (3)
–6x –33y = 0
$ \Rightarrow $ 2x +11y = 0

putting value of x in (1), we get
–7y + 2z = 0

Now ${\left( {{{11y} \over 2}} \right)^2} + {y^2} + {\left( {{{7y} \over 2}} \right)^2} = 1$

y²(121 + 1 + 49) = 4

y²(171) = 4

$y = \pm {2 \over {\sqrt {171} }}$

$ \Rightarrow x = \pm {7 \over {\sqrt {171} }}$

$ \Rightarrow z = \pm {{11} \over {\sqrt {171} }}$

$ \therefore $ So, there are 2 solution set of (x, y, z)

For no solution :

$\Delta $ = 0 and $\Delta $₁/$\Delta $₂/$\Delta $₃ $ \ne $ 0

$\Delta $ = $\left| {\matrix{ 2 & { - 1} & 2 \cr 1 & { - 2} & \lambda \cr 1 & \lambda & 1 \cr } } \right|$ = 0

$ \Rightarrow $ 2(–2 – $\lambda $²) + 1 (1 – $\lambda $) + 2($\lambda $ + 2) = 0

$ \Rightarrow $ –2$\lambda $² + $\lambda $ + 1 = 0

$ \Rightarrow $ $\lambda $ = 1, $ - {1 \over 2}$

When $\lambda $ = 1

2x – y + 2z = 2 ...(1)

x – 2y + z = –4 ...(2)

x + y + z = 4 ...(3)

Adding (2) and (3), we get

2x – y + 2z = 0 (contradiction) hence no solution.

$ \therefore $ $\lambda $ = 1 belongs to set S.

When $\lambda $ = $ - {1 \over 2}$

2x – y + 2z = 2 ...(1)

x – 2y $ - {1 \over 2}$z = –4 ...(2)

x $ - {1 \over 2}$y + z = 4 ...(3)

(1) and (3) contradict each other, hence no solution.

$ \therefore $ $\lambda $ = $ - {1 \over 2}$ belongs to set S.

Let A = $\left[ {\matrix{ a & b \cr c & d \cr } } \right]$, where a, b, c, d $ \in $ {0, 1}

$ \Rightarrow $ |A| = ad – bc

$ \therefore $ ad = 0 or 1 and bc = 0 or 1

So possible values of |A| are 1, 0 or –1

(P) If A $ \ne $ I₂ then |A| is either 0 or –1

(Q) If |A| = 1 then ad = 1 and bc = 0

$ \Rightarrow $ a = d = 1 $ \Rightarrow $ Tr(A) = 2