Matrices and Determinants

Given that,

$ a_{i j}= \begin{cases}1, & \text { if } i+j=7 \\ 0, & \text { if } i+j \neq 7\end{cases} $

Now, $\left(A(\operatorname{adj} A) A^{-1}\right) A^2$

$ \begin{array}{lrr} =\left(A(\operatorname{Adj} A) A^{-1}\right) A^2 & \\ =\left((|A| I) A^{-1}\right) A^2 & {[\because A \operatorname{adj} A=|A| I]} \\ =-1\left(A^{-1} A\right) A & {[\because|A|=-1]} \\ =-I A & {\left[\because A^{-1} A=I\right]} \\ =-A & \end{array} $

Given system of equations has a non-trivial solution

$ \begin{aligned} & \Pi_1 \equiv x+a y+a z=0 \\ & \Pi_2 \equiv b x+y+b z=0 \\ & \Pi_3 \equiv c x+c y+z=0 \end{aligned} $

It can be written as $A X=0$

Where, $A=\left[\begin{array}{lll}1 & a & a \\ b & 1 & b \\ c & c & 1\end{array}\right]$

So, the system has non-trivial solutions hence it has infinitely many solutions or unique solutions.

$ \therefore \quad|A|=0 $

Now, let the system has infinitely many solutions

So, let $y=t ; t \in \mathbf{R}$

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

$ x+a z=-a t \Rightarrow b x+b z=-t $

Solving these, we get

$ z=\frac{t(1-a b)}{b(a-1)}, x=\frac{a}{b} t \frac{(1-b)}{(1-a)} $

Hence, if $a \neq b \neq c$, then

$ \begin{aligned} & x=\frac{a}{b} t \frac{(1-b)}{(1-a)}, y=t \\ & z=\frac{t}{b} \frac{(1-a b)}{(a-1)}, t \in \mathbf{R} \end{aligned} $

If $a=b=c$

Let $y=t_1, z=t_2, t_1, t_2 \in \mathbf{R}$

Then by Eq. (i)

$ x=-a\left(t_1+t_2\right) $

So, the system of equations has unique solution.

Given that, Rank of matrix $A=$ rank of matrix $B$

$ f(A)=f(B) \text { and } f(B)=3 $

∴ Order of matrix $B \geq$ Rank of matrix $B$.

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} &\left|\begin{array}{ccc} 2 a-2 b-4 & 4 a & 4 a \\ 4 & 2-b-a & 4 \\ 2 b & 2 b & b-a-2 \end{array}\right| \\ &= 2\left|\begin{array}{ccc} a-b-2 & 2 a & 2 a \\ 4 & 2-b-a & 4 \\ 2 b & 2 b & b-a-2 \end{array}\right| \\ &= 2\left|\begin{array}{ccc} -(a+b+2) & 0 & 2 a \\ a+b+2 & -(a+b+2) & 4 \\ 0 & a+b+2 & b-a-2 \end{array}\right| \\ & C_1 \rightarrow C_1-C_2 \text { and } C_2 \rightarrow C_2-C_3 \end{aligned} \end{aligned} $

$ \begin{aligned} & =2(a+b+2)^2\left|\begin{array}{ccc} -1 & 0 & 2 a \\ 1 & -1 & 4 \\ 0 & 1 & b-a-2 \end{array}\right| \\ & =2(a+b+2)^2(-1(-b+a+2-4)+2 a(1) \\ & =2(a+b+2)^2(a+b+2)=2(a+b+2)^3 \\ & =2\left[(a+b)^3+6(a+b)^2+12(a+b)+8\right] \end{aligned} $

We have,

$ \begin{aligned} A & =\left[\begin{array}{ccc} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{array}\right] \\ A^2 & =\left[\begin{array}{ccc} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{array}\right]\left[\begin{array}{ccc} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{array}\right] \\ A^2 & =\left[\begin{array}{ccc} 2 & -2 & -16-4 x \\ -1 & 3 & 16+4 x \\ 4+x & -8-2 x & -12+x^2 \end{array}\right] \end{aligned} $

Given, $A^2=A$

$ \begin{array}{ll} \because & 4+x=1 \Rightarrow x=-3 \\ \therefore & A=\left[\begin{array}{ccc} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{array}\right] \\ & |A|=2(-9+8)+2(3-4)-4(2-3) \\ & =-2-2+4=0 \end{array} $

Rank of $A=2 \Rightarrow r+x=2-3=-1$

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} \Delta_1 & =\left|\begin{array}{ll} m & b \\ n & d \end{array}\right|=m d-b n \\ \Delta_2 & =\left|\begin{array}{ll} a & m \\ c & n \end{array}\right|=a n-m c \\ \Delta_3 & =\left|\begin{array}{ll} a & b \\ c & d \end{array}\right|=a d-b c \\ x^a y^b & =e^m, x^c y^d=e^n \\ y & =\left(\frac{e^m}{x^a}\right)^{1 / b} \text { and } y=\left(\frac{e^n}{x^c}\right)^{1 / d} \end{aligned} \end{aligned} $

$ \begin{aligned} & \therefore & \frac{e^{m / b}}{x^{a / b}} & =\frac{e^{n / d}}{x^{c / d}} \\ \Rightarrow & & x^{c / d-a / b} & =e^{n / d-m / b} \\ \Rightarrow & & x & =\left(e^{n / d-m / b}\right)^{\frac{1}{c}-\frac{a}{b}} \\ \Rightarrow & & x & =e^{\left(\frac{b n-m d}{b d} \times \frac{b d}{b c-a d}\right)} \\ \Rightarrow & & x & =e^{\frac{b n-m d}{b c-a d}} \\ \Rightarrow & & x & =e^{\frac{m d-b n}{a d-b c}}=e^{\frac{\Delta_1}{\Delta_3}} \end{aligned} $

Putting the value of $x$ in $y=\frac{e^{m / b}}{x^{a / b}}$ we get $\quad y=e^{\frac{a n-m c}{a d-b c}}=e^{\frac{\Delta_2}{\Delta_3}}$

We have, $A=\left[\begin{array}{ccc}1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6\end{array}\right]$ and

$ \begin{aligned} & B=\left[\begin{array}{lll} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array}\right]=\left[\begin{array}{ccc} 2 & 0 & -2 \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array}\right] \\ & \therefore A B=\left[\begin{array}{ccc} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6 \end{array}\right]\left[\begin{array}{ccc} 2 & 0 & -2 \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array}\right] \end{aligned} $

$ =\left[\begin{array}{cc} 2+4 b_{21}+2 b_{31} & 0+4 b_{22}+2 b_{32} \\ 4-b_{21}+4 b_{31} & 0+b_{22}+4 b_{32} \\ -6+7 b_{21}-6 b_{31} & 0+7 b_{22}-6 b_{32} \\ & -2+4 b_{23}+2 b_{33} \\ & -4-b_{23}+4 b_{33} \\ & 6+7 b_{23}-6 b_{33} \end{array}\right] $

$ =\left[\begin{array}{ccc} 2 & 14 & -4 \\ 4 & 1 & -8 \\ -6 & 15 & 12 \end{array}\right] $

On solving above equal matrices with corresponding elements, we get

$ \begin{aligned} & b_{21}=b_{31}=0, b_{22}=3, b_{32}=1, b_{23}=0 \\ & \text { and } b_{33}=-1 \\ & \therefore B=\left[\begin{array}{ccc} 2 & 0 & -2 \\ 0 & 3 & 0 \\ 0 & 1 & -1 \end{array}\right] \\ & \therefore|B|=2(-3-0)+(-2)(0-0)=-6 \\ & \text { and Trace }(B)=2+3-1=4 \\ & \therefore|B|+\operatorname{trace}(B)=-6+4=-2 \end{aligned} $

Given, $A$ is $m \times n$ matrix of rank 4

$A$ contains $m$ th order of non-singular sub matrix.

$A$ is non-singular matrix

$\therefore A$ is a square matrix of order $M$

∴ Rank of $A$ is $M$

∴ order of $A=4$       [rank of $A=4$ ]

Hence, rows of $A=4$.

Given, $C$ and $D$ are non-singular matrix of order $n$

$ \quad \begin{aligned} \because\,\,\,\,\,\,\, |C| & \neq 0,|D| \neq 0 \\ C D & =-D C \\ |C D| & =|-D C| \\ |C||D| & =(-1)^n|D||C| \\ 1 & =(-1)^n \end{aligned} $

$\therefore n$ is even integer.

It is given that the matrices $A$ and $B$ are non-singular of order 3 and $|B|=k, a$ positive integer, so

$ \begin{aligned} \left|k^{-1} A^{-1}\right|=\left(k^{-1}\right)^3\left|A^{-1}\right| & =\frac{1}{k^3|A|} \\ & {\left[\because\left|A^{-1}\right|=\frac{1}{|A|}\right] } \end{aligned} $

$ \begin{gathered} =\frac{1}{|B|^3|A|} \\ \because\left|\operatorname{adj}\left(A^{-1}\right)\right|=\left|A^{-1}\right|^{3-1}=\left|A^{-1}\right|^2=\frac{1}{|A|^2} \end{gathered} $

$\left[\because|\operatorname{adj}(A)|=|A|^{n-1}\right.$, where $n$ is the order of matrix $A$ ]

Now, since it is given that $B A B^{-1}=I$

$ \begin{aligned} & \Rightarrow \quad A B^{-1}=B^{-1} \\ & \therefore B A^k B^{-1}=B A^{k-1}\left(A B^{-1}\right)=B A^{k-1} B^{-1} \\ & =B A^{k-2}\left(A B^{-1}\right)=B A^{k-2} B^{-1} \\ & \therefore B A^k B^{-1}=B A B^{-1}=I \\ & \text { and }, B \frac{\operatorname{adj}(B)}{|B|}=B B^{-1}=I \end{aligned} $

Therefore, $B A^k B^{-1}=B \frac{\operatorname{adj}(B)}{|B|}$, if

$ B A B^{-1}=I $

Now, the adj $\left(\operatorname{adj}\left(A^{-1}\right)\right)$

$ =\left|A^{-1}\right|^{3-2}\left(A^{-1}\right) $

$\left[\because \operatorname{adj}(\operatorname{adj} A)=|A|^{n-2} A\right.$ where $n$ is the order of matrix $A$ ]

$ =\frac{1}{|A|}\left(A^{-1}\right) $

We have,

$ \begin{array}{r} 2 x+p y+6 z=8 \\ x+2 y+q z=5 \\ x+y+3 z=4 \end{array} $

Now, if we express above equation in matrix form, We have,

$ A X=B $

Where,

$ A=\left[\begin{array}{lll} 2 & p & 6 \\ 1 & 2 & q \\ 1 & 1 & 3 \end{array}\right], B=\left[\begin{array}{l} 8 \\ 5 \\ 4 \end{array}\right], X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] $

For no solution $|A|=0$

$ \begin{aligned} & \Rightarrow \quad\left|\begin{array}{lll} 2 & p & 6 \\ 1 & 2 & q \\ 1 & 1 & 3 \end{array}\right|=0 \\ & \Rightarrow \quad 2(6-q)-p(3-q)+6(1-2)=0 \\ & \Rightarrow \quad 12-2 q-3 p+p q-6=0 \\ & \Rightarrow \quad 3 p+2 q-p q-6=0 \\ & \Rightarrow \quad 3 p-6+2 q-p q=0 \\ & \Rightarrow \quad 3(p-2)-q(p-2)=0 \\ & \Rightarrow \quad(p-2)(3-q)=0 \Rightarrow \\ & \vec{p}=2,3 \end{aligned} $

But for $p=2$ equations $2 x+p y+6 z=8$ and $x+y+3 z=4$ are same.

So, $p \neq 2, q=3$

We have,

$ \begin{array}{r} (\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0 \\ (\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0 \\ 2 x+(3 \lambda+1) y+3(\lambda-1) z=0 \end{array} $

Now, it can be express as $\quad A X=0$

where, $A=\left[\begin{array}{ccc}\lambda-1 & 3 \lambda+1 & 2 \lambda \\ \lambda-1 & 4 \lambda-2 & \lambda+3 \\ 2 & 3 \lambda+1 & 3(\lambda-1)\end{array}\right]$ and $X=\left[\begin{array}{c}x \\ y \\ z\end{array}\right]$

For non-zero solution $|A|=0$

$ \left[\begin{array}{ccc} \lambda-1 & 3 \lambda+1 & 2 \lambda \\ \lambda-1 & 4 \lambda-2 & \lambda+3 \\ 2 & 3 \lambda+1 & 3(\lambda-1) \end{array}\right]=0 $

On applying $R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3-R_1$,

$ \left[\begin{array}{ccc} \lambda-1 & 3 \lambda+1 & 2 \lambda \\ 0 & \lambda-3 & -\lambda+3 \\ -\lambda+3 & 0 & \lambda-3 \end{array}\right]=0 $

$ \begin{aligned} & \Rightarrow(\lambda-1)\left[(\lambda-3)^2\right]+(-\lambda+3)((3 \lambda+1) \\ & \quad(-\lambda+3)-2 \lambda(\lambda-3))=0 \\ & \Rightarrow(\lambda-1)(\lambda-3)^2-(\lambda-3) \\ & \quad[(\lambda-3)(-3 \lambda-1-2 \lambda))=0 \\ & \Rightarrow(\lambda-3)^2[\lambda-1+5 \lambda+1]=0 \\ & \Rightarrow(\lambda-3)^2(6 \lambda)=0 \Rightarrow \lambda=3,0 \\ & \therefore p=3, q=0 \\ & \therefore p^2+q^2-p q=9+0-0=9 \end{aligned} $

We have,

$(1 + \alpha )x + \beta y + z = 2$

$\alpha x + (1 + \beta )y + z = 3$

$\alpha x + \beta y + 2z = 2$

For unique solution $\left| {\matrix{ {1 + \alpha } & \beta & 1 \cr \alpha & {1 + \beta } & 1 \cr \alpha & \beta & 2 \cr } } \right| \ne 0$

Apply R₁ $\to$ R₁ - R₂ and R₂ $\to$ R₂ - R₃, we get

$\left| {\matrix{ 1 & { - 1} & 0 \cr 0 & 1 & { - 1} \cr \alpha & \beta & 2 \cr } } \right| \ne 0$

Expanding along R₁, we get

$1(2 + \beta ) + 1(0 + \alpha ) \ne 0$

$ = \alpha + \beta + 2 \ne 0$

Only option (c) satisfied above equation.

We have,

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

We know that, $\left| {A \to \lambda I} \right| = 0$

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

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

$ \Rightarrow {(2 - \lambda )^2} - 1 = 0$

$ \Rightarrow 4 - 4\lambda + {\lambda ^2} - 1 = 0$

$ \Rightarrow {\lambda ^2} - 4\lambda + 3 = 0$

$\therefore$ ${A^2} - 4A + 3I = 0$

Multiply by A^$-$1, we get

$A - 4I + 3{A^{ - 1}} = 0$

$ \Rightarrow {A^{ - 1}} = {{ - 1} \over 3}A + {4 \over 3}I$

$\therefore$ $\alpha = - {1 \over 3}$ and $\beta = {4 \over 3}$

$\alpha + \beta = - {1 \over 3} + {4 \over 3} = {3 \over 3} = 1$