Matrices and Determinants

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \quad B=A^{-1} \\ & \text { and } \quad|B|=k \Rightarrow\left|A^{-1}\right|=k \\ & \Rightarrow \quad \frac{1}{|A|}=k \Rightarrow|A|=\frac{1}{k} . \\ & \because(\operatorname{adj}(\operatorname{adj} A))^{-1}=\left(|A|^{n-2} A\right)^{-1} \text { and } n=3 \\ & \quad=\left(|A|^1 \cdot A\right)^{-1}=\left(\frac{1}{k} A\right)^{-1}=k B . \end{aligned} \end{aligned} $

$A=\left[\begin{array}{lll}2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2\end{array}\right]$

and $|A-X I|=0$

⇒ Characterstic equation of matrix $A$

$\because \operatorname{tr}(A)=2+3+2=7$

Sum of minor of diagonal elements

$ \begin{aligned} & \,\,\,\,\,\,\,\,=(4)+(3)+4=11 \\ & |A|=2(4)-2(1)+1(-1) \\ &\,\,\,\,\,\,\,\,\,\, =8-2-1=5 \\ & \because \quad|A-x I|=0 \\ & \Rightarrow \quad x^3-7 x^2+11 x-5=0 \\ & \therefore \quad \alpha+\beta+\gamma=7 \\ & \text { and } \alpha \beta+\beta \gamma+\gamma \alpha=11 \\ & \text { and } \alpha \beta \gamma=5 \\ & \therefore \alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha) \end{aligned} $

$ =49-2 \times 11=49-22=27 . $

$ \begin{aligned} &\text { Given, } 2 x-3 y+2 z=-15\\ &\begin{aligned} & 3 x+y-z=-2 \\ & x-3 y-3 z=-8 \\ & \begin{aligned} \therefore \Delta & =\left|\begin{array}{ccc} 2 & -3 & 2 \\ 3 & 1 & -1 \\ 1 & -3 & -3 \end{array}\right| \\ & =2(-3-3)+3(-9+1)+2(-9-1) \\ & =-12-24-20=-56 \\ \Delta_x & =\left|\begin{array}{ccc} -15 & -3 & 2 \\ -2 & 1 & -1 \\ -8 & -3 & -3 \end{array}\right| \\ & =-15(-3-3)+3(6-8)+2(6+8) \\ & =90-6+28=112 \\ \Delta_y & =\left|\begin{array}{ccc} 2 & -15 & 2 \\ 3 & -2 & -1 \\ 1 & -8 & -3 \end{array}\right| \\ & =2(6-8)+15(-9+1)+2(-24+2) \\ & =-4-120-44=-168 \\ \Delta_z & =\left|\begin{array}{ccc} 2 & -3 & -15 \\ 3 & 1 & -2 \\ 1 & -3 & -8 \end{array}\right| \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} & =2(-8-6)+3(-24+2)-15(-9-1) \\ & =-28-66+150=150-94=56 . \\ \alpha & =\frac{\Delta_x}{\Delta}, \beta=\frac{\Delta_y}{\Delta} \text { and } \gamma=\frac{\Delta_z}{\Delta} \\ \Rightarrow \alpha & =\frac{112}{-56}, \beta=\frac{-168}{-56} \text { and } \gamma=\frac{56}{-56} \\ \Rightarrow \alpha & =-2, \beta=3 \text { and } \gamma=-1 \\ \therefore 2 \alpha & +\beta=\gamma \end{aligned} $

Let $A=\left[\begin{array}{lll}1 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 3\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 2 & 3 \\ 4 & -3 & -1 \\ 2 & 1 & -4\end{array}\right]$

Also, $a=\operatorname{det}(\operatorname{adj}(A))$ and $b=\operatorname{det}\left(B^{-1}\right)$

Now, $a=\operatorname{det}(\operatorname{adj}(A))=(\operatorname{det}(A))^{n-1}$ for an $n \times n$ matrix.

Since $A$ is $3 \times 3$ matrix, so $a=(\operatorname{det}(A))^{3-1}=(\operatorname{det}(A))^2$

$ \begin{aligned} \operatorname{det} A & =1 \cdot\left|\begin{array}{ll} 2 & 3 \\ 3 & 3 \end{array}\right|-1 \cdot\left|\begin{array}{ll} 1 & 3 \\ 2 & 3 \end{array}\right|+2 \cdot\left|\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right| \\ & =1 \cdot(6-9)-1(3-6)+2(3-4) \\ & =-3+3-2=-2 \\ a & =(\operatorname{det}(A))^2=(-2)^2=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now, $\operatorname{det}(B)=\left|\begin{array}{ccc}1 & 2 & 3 \\ 4 & -3 & -1 \\ 2 & 1 & -4\end{array}\right|$

$ \begin{aligned} & =1 \cdot\left|\begin{array}{cc} -3 & -1 \\ 1 & -4 \end{array}\right|-2\left|\begin{array}{cc} 4 & -1 \\ 2 & -4 \end{array}\right|+3 \cdot\left|\begin{array}{cc} 4 & -3 \\ 2 & 1 \end{array}\right| \\ & =(12+1)-2(-16+2)+3(4+6) \\ & =13+28+30=71 \end{aligned} $

$ \begin{aligned} &\text { So, } \quad b=\operatorname{det}\left(B^{-1}\right)=\frac{1}{\operatorname{det}(B)}=\frac{1}{71}\\ &\begin{aligned} & \quad b=\frac{1}{71} \\ & \Rightarrow \quad b+1=\frac{1+71}{71}=\frac{72}{71} \\ & \text { Then, } \frac{b+1}{18 b}=\frac{72 / 71}{18 \times \frac{1}{71}} \\ & =\frac{72}{71} \times \frac{71}{18}=\frac{72}{18}=4=a \quad \text { [using Eq. (i)] } \\ & \therefore \frac{b+1}{18 b}=a \end{aligned} \end{aligned} $

Given, $A X=B$ and $C X=D$ and $A X=B$ has unique solution $D$ and $C X=D$ has unique solution $B$.

Since, $X=D$ is the unique solution of $A X=B$

$ \therefore \quad A D=B \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

And, $X=B$ is the unique solution of $C X=D$,

$ \therefore \quad C B=D \Rightarrow B=C^{-1} D \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eq. (i) and (ii),

$ \begin{array}{r} A D=C^{-1} D \\ \left(A-C^{-1}\right) D=0 \end{array} $

So, the solution of $\left(A-C^{-1}\right) X=0$ is $D$

Given,

$ f(x)=\frac{1}{2}\left|\begin{array}{cc} f(x) & f\left(\frac{1}{x}\right)-f(x) \\ 1 & f\left(\frac{1}{x}\right) \end{array}\right| $

and $f(2)=33$

Here, $f(x)=\frac{1}{2}\left[f(x) \cdot f\left(\frac{1}{x}\right)-f\left(\frac{1}{x}\right)+f(x)\right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Let

$ \begin{array}{r} f(x)=a_0 x^n+a_1 x^{n-1}+\ldots+a_{n-1} x+a_n \\ \left(a_0 \neq 0\right) \end{array} $

Putting this value of $f(x)$ in Eq. (i),

$ \begin{aligned} & f(x)=\frac{1}{2}\left[f(x) \cdot f\left(\frac{1}{x}\right)-f\left(\frac{1}{x}\right)+f(x)\right] \\ \Rightarrow & f(x) \cdot f\left(\frac{1}{x}\right)=f\left(\frac{1}{x}\right)+f(x) \end{aligned} $

Hence, $f(x)=x^n+1$ or $-x^n+1$

Now, $f(2)=2^n+1=33$

$ \begin{array}{ll} \Rightarrow & 2^n=32 \\ \Rightarrow & 2^n=2^5 \\ \Rightarrow & n=5 \end{array} $

And $f(2)=-2^n+1=33$

$ \Rightarrow \quad-2^n=33-1=32=2^5 $

which is not possible.

$ \therefore f(x)=x^n+1=x^5+1 $

So, $f(3)=3^5+1=243+1=244$

$\because P=\left[\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right]$

$ \begin{aligned} & \because \operatorname{adj}(A)=(\operatorname{det} A) \cdot A^{-1} \\ & \Rightarrow \operatorname{det}(\operatorname{adj}(A))=(\operatorname{det} A)^{n-1} \end{aligned} $

So, for $n=3$

$ \begin{aligned} \operatorname{det}(\operatorname{adj}(A)) & =(\operatorname{det} A)^{3-1} \\ & =4^2=16 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ \begin{aligned} \text { and } \operatorname{det}(A) & =1(12-12)-\alpha(4-6)+3(4-6) \\ & =2 \alpha-6 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Hence, $2 \alpha-6=16$

$ \begin{array}{ll} \Rightarrow & 2 \alpha=22 \\ \Rightarrow & \alpha=11 \end{array} $

We are given the equation $x^3+6 x^2+5 x-42=0$.

Let's check if $x=2$ is a root by plugging it in:

$ (2)^3 + 6 \times (2)^2 + 5 \times 2 - 42 = 8 + 24 + 10 - 42 = 42 - 42 = 0 $

Since the equation equals zero, $x = 2$ is a real root. So, $\alpha = 2$.

Now, we substitute $\alpha = 2$ into the matrix:

$ \left[ \begin{array}{lll} \alpha-1 & \alpha+1 & \alpha+2 \\ \alpha-2 & \alpha+3 & \alpha-3 \\ \alpha+4 & \alpha-4 & \alpha+5 \end{array} \right] = \left[ \begin{array}{ccc} 1 & 3 & 4 \\ 0 & 5 & -1 \\ 6 & -2 & 7 \end{array} \right] $

Next, let's find the determinant of this matrix:

$ \begin{aligned} &= \left| \begin{array}{ccc} 1 & 3 & 4 \\ 0 & 5 & -1 \\ 6 & -2 & 7 \end{array} \right| \\ &= 1 \times (5 \times 7 - (-1) \times (-2)) - 3 \times (0 \times 7 - (-1) \times 6) + 4 \times (0 \times (-2) - 5 \times 6) \\ &= 1 \times (35 - 2) - 3 \times (0 + 6) + 4 \times (0 - 30) \\ &= 33 - 18 - 120 \\ &= -105 \end{aligned} $

Given matrix

$ \begin{aligned} & {\left[\begin{array}{cccc} 2 & -3 & 4 & 0 \\ 5 & -4 & 2 & 1 \\ 1 & -3 & 5 & -4 \end{array}\right]} \\ & R_2 \rightarrow R_2-\frac{5}{2} R_1 \text { and } R_3 \rightarrow R_3-\frac{1}{2} R_1 \\ & =\left[\begin{array}{cccc} 2 & -3 & 4 & 0 \\ 0 & 3.5 & -8 & 1 \\ 0 & -1.5 & 3 & -4 \end{array}\right] \\ & R_3 \rightarrow R_3+\frac{1.5}{3.5} R_2=R_3+\frac{3}{7} R_2 \\ & =\left[\begin{array}{cccc} 2 & -3 & 4 & 0 \\ 0 & 3.5 & -8 & 1 \\ 0 & 0 & -3 / 7 & 25 / 7 \end{array}\right] \end{aligned} $

$\because$ All 3 rows are non-zero

Hence, rank of the matrix $=3$

Given, $A=\left[\begin{array}{ccc}0 & k & k \\ k & -4 & -6 \\ k & -3 & -5\end{array}\right]$

$\because A$ is an singular matrix.

$ \begin{aligned} & \therefore \quad|A|=0 \\ & \Rightarrow \quad\left|\begin{array}{ccc} 0 & k & k \\ k & -4 & -6 \\ k & -3 & -5 \end{array}\right|=0 \\ & \Rightarrow \quad 0-k(-5 k+6 k)+k(-3 k+4 k)=0 \end{aligned} $

$\Rightarrow-k^2+k^2=0$ which is true for all real values of $k$.

$\therefore k \in R$

Given, matrix $A=\left[\begin{array}{ccc}1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6\end{array}\right]$ and the rank of $A=2$

$ \begin{aligned} & =1(6-28)-2(-24-14)+x(16+2) \\ & |A|=54+18 x \end{aligned} $

Since, the rank of $A$ is 2 then the determinant of $A$ must be zero.

So, we set the determinant equal to zero and solve for $x$.

i.e. $54+18 x=0 \Rightarrow x=-3$

$ \text { } \begin{aligned} &\left|\begin{array}{ll} 2 & 1 \\ 3 & 1 \end{array}\right|+\left|\begin{array}{cc} 1 & \frac{1}{3} \\ 3 & 1 \end{array}\right|+\left|\begin{array}{cc} \frac{1}{2} & \frac{1}{9} \\ 3 & 1 \end{array}\right| +\left|\begin{array}{cc} \frac{1}{4} & \frac{1}{27} \\ 3 & 1 \end{array}\right|+\ldots+\infty \\ &=(2-3)+\left(1-3 \times \frac{1}{3}\right)+\left(\frac{1}{2}-3 \times \frac{1}{9}\right)+\left(\frac{1}{4}-3 \times \frac{1}{27}\right)+\left(\frac{1}{8}-3 \times \frac{1}{81}\right)+\ldots+\infty \\ &=-1+0+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2^2}-\frac{1}{3^2}\right) +\left(\frac{1}{2^3}-\frac{1}{3^3}\right)+\ldots+\infty \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =-1+\left[\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\ldots \infty\right)\right. \left.-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\ldots+\infty\right)\right] \\ & =-1+\left(\frac{\frac{1}{2}}{1-\frac{1}{2}}\right)-\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right) \end{aligned}\\ &\text { (Using sum of infinite GP formula) }\\ &=-1+\frac{\frac{1}{2}}{\frac{1}{2}}-\frac{\frac{1}{3}}{\frac{2}{3}}=-1+1-\frac{1}{2}=-\frac{1}{2} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { }|A|=1(18-5)-2(6-10)+3(1-6) \\ & \quad=13+8-15=6 \\ & |\operatorname{adj}(\operatorname{adj} A)|=|A|^4=6^4 \\ & \Rightarrow(\operatorname{adj} A)^{-1}=\frac{1}{6} A \\ & \Rightarrow|\operatorname{adj}(\operatorname{adj} A)|(\operatorname{adj} A)^{-1}=k A \\ & \Rightarrow 6^4 \cdot \frac{1}{6} A=k A \\ & \therefore k=216 \end{aligned} \end{aligned} $

$\alpha+2 \beta+3 \gamma=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned} & 3 \alpha+\beta+\gamma=3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \alpha+3 \beta+3 \gamma=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

From Eqs. (ii), $3 \alpha+\gamma=3-\beta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

Substracting Eq. (iii) from Eq. (i)

$ \begin{aligned} & \alpha+2 \beta+3 \gamma=4 \\ & \alpha+3 \beta+3 \gamma=2 \\ & -\quad-\quad- \,\,\,\,\,\,\,\,-\\ & \hline-\beta=2 \Rightarrow \beta=-2 \end{aligned} $

Bu Eq. (iv)

$ \begin{aligned} 3 \alpha+\gamma & =3-(-2)=5 \\ \Rightarrow 3 \alpha+\gamma & =1-2 \beta \end{aligned} $

$ \begin{aligned} &\text { } 2 x+y-z=7, x-3 y+2 z=1 \text {, }\\ &\begin{aligned} & x+4 y-3 z=5 \\ & A=\left[\begin{array}{rrr} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & -3 \end{array}\right] \\ & \Rightarrow|A|=\left|\begin{array}{rrr} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & -3 \end{array}\right| \\ & \quad=2(9-8)-1(-3-2)-1(4+3) \\ & \quad=2+5-7=0 \\ & \Rightarrow \text { Number of solution = 0 } \end{aligned} \end{aligned} $

$2 x+p y+6 z=8, x+2 y+q z=5$ and $x+y+3 z=4$ may have no solution.

$ \begin{aligned} & x=\frac{\Delta x}{\Delta}, y=\frac{\Delta y}{\Delta}, z=\frac{\Delta z}{\Delta} \\ & \Delta=\left|\begin{array}{lll} 2 & p & 6 \\ 1 & 2 & q \\ 1 & 1 & 3 \end{array}\right|=(p-2)(q-3) \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Delta_x=\left|\begin{array}{lll} 8 & p & 6 \\ 5 & 2 & q \\ 4 & 1 & 3 \end{array}\right|=(15-4 q)(2-p) \\ & \Delta_y=\left|\begin{array}{lll} 2 & 8 & 6 \\ 1 & 5 & q \\ 1 & 4 & 3 \end{array}\right|=0 \\ & \Delta_z=\left|\begin{array}{lll} 2 & p & 8 \\ 1 & 2 & 5 \\ 1 & 1 & 4 \end{array}\right|=p-2 \end{aligned}\\ &\text { For no solution }\\ &\begin{aligned} & (p-2)(q-3)=0 \\ & p-2 \neq 0 \\ & q=3 \\ & p \neq 2 \end{aligned} \end{aligned} $

A is det of order 3 with entries 0 or 1 .

$\therefore B$ and $C$ has same number of elements.

(c) $A=\left[\begin{array}{ccc}x & y & 0 \\ -3 & 1 & 2 \\ 1 & -2 & z\end{array}\right]$ and

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

Cofactor of $z=9$, cofactor of 1 in $3^{\text {rd }}$ row of $A=4$

$ \begin{aligned} & & x+3 y & =9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow & & x & =3 \\ & & 2 y & =4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \Rightarrow & & y & =2 \end{aligned} $

Cofactor of $x=3$

$ z+4=3 \Rightarrow z=-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

$ \begin{aligned} \therefore A B & =\left[\begin{array}{ccc} 3 & 2 & 0 \\ -3 & 1 & 2 \\ 1 & -2 & -1 \end{array}\right]\left[\begin{array}{ccc} 1 & -2 & -2 \\ 2 & 0 & 1 \\ 2 & 1 & 0 \end{array}\right] \\ & =\left[\begin{array}{ccc} 7 & -6 & -4 \\ 3 & 8 & 7 \\ -5 & -3 & -4 \end{array}\right] \end{aligned} $

$ \begin{aligned} &\text { We have, } A=\left[\begin{array}{ccc} 1 & 2 & -2 \\ 2 & -1 & 2 \\ -1 & 1 & -2 \end{array}\right]\\ &\begin{aligned} |A| & =|(2-2)-2(-4+2)-2(2-1)|=4-2=2 \\ \operatorname{adj}(A) & =\left[\begin{array}{ccc} 0 & 2 & 2 \\ 2 & -4 & -6 \\ 1 & -3 & -5 \end{array}\right] \\ A^{-1} & =\frac{\operatorname{adj}(A)}{|A|}=\frac{1}{2}\left[\begin{array}{ccc} 0 & 2 & 2 \\ 2 & -4 & -6 \\ 1 & -3 & -5 \end{array}\right] \\ \text { Then, } A+2 A^{-1} & =\left[\begin{array}{ccc} 1 & 4 & 0 \\ 4 & -5 & -4 \\ 0 & -2 & -7 \end{array}\right] \end{aligned} \end{aligned} $

We have, $A=\left[\begin{array}{lll}a & b & c \\ d & e & f \\ l & m & n\end{array}\right]$ and $\operatorname{adj}(A)=\left[\begin{array}{rrr}0 & 4 & -6 \\ 10 & 8 & 0 \\ 2 & 4 & -4\end{array}\right] |\operatorname{adj} A|=|A|^2 \Rightarrow|A|^2=16 \Rightarrow|A|=4$

We know that $\operatorname{adj}(\operatorname{adj} A)=|A|^{n-2} A=|A|^{3-2} A=|A| A$

$ \begin{aligned} & \Rightarrow \quad A=\frac{1}{|A|} \operatorname{adj}(\operatorname{adj} A) \\ & \text { Here, } \operatorname{adj}(\operatorname{adj} A)=\left[\begin{array}{rrr} -32 & -8 & 48 \\ 40 & 12 & -60 \\ 24 & 8 & -40 \end{array}\right] \\ & \therefore \quad A=\frac{1}{4}\left[\begin{array}{rrr} -32 & -8 & 48 \\ 40 & 12 & -60 \\ 24 & 8 & -40 \end{array}\right]=\left[\begin{array}{rrr} -8 & -2 & 12 \\ 10 & 3 & -15 \\ 6 & 2 & -10 \end{array}\right] \\ & \begin{array}{l} \Rightarrow a=-8, b=-2, c=12, d=10, e=3, f=-15, l=6, m=2, \\ n=-10 \\ \therefore \quad \frac{c d}{f b}+\frac{\ln }{e m}=\frac{12 \times 10}{(-15) \times(-2)}+\frac{6 \times(-10)}{3 \times 2} \\ \quad=\frac{120}{30}-\frac{60}{6}=4-10=-6 \end{array} \end{aligned} $

Also, (a) $2 a=-16$

(b) $a+m=-8+2=-6$

(c) $a+b=-8-2=-10$

(d) $a=-8$

$ \text { } \begin{aligned} Given, \Delta_1 & =\left|\begin{array}{ccc} -11 & 1 & -7 \\ -4 & 1 & -2 \\ 5 & 1 & 1 \end{array}\right| \\ \because \Delta \Delta_1 & =-11(3)-1(6)-7(-9)=-33-6+63=24 \\ \Delta_3 & =\left|\begin{array}{ccc} 4 & 1 & -11 \\ 1 & 1 & -4 \\ 4 & 1 & 5 \end{array}\right|=4(9)-1(21)-11(-3)=48 \\ \Delta & =\left|\begin{array}{ccc} 4 & 1 & -7 \\ 1 & 1 & -2 \\ 4 & 1 & 1 \end{array}\right|=4(3)-1(9)-7(-3)=24 \end{aligned} $

$ \begin{aligned} &\text { By inspection }\\ &\begin{aligned} & \Delta_2=\left|\begin{array}{ccc} 4 & -11 & -7 \\ 1 & -4 & -2 \\ 4 & 5 & 1 \end{array}\right|=4(6)+11(9)-7(21)=-24 \\ & x=\frac{\Delta_1}{\Delta}=\frac{24}{24}=1, y=\frac{\Delta_2}{\Delta}=\frac{-24}{24}=-1 \text { and } z=\frac{\Delta_3}{\Delta}=\frac{48}{24}=2 \\ & X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right] \end{aligned} \end{aligned} $

(i) $A B=0 \Rightarrow A=0$ or $B=0$

Which is false.

(ii) If $A B=B A=I_3 \Rightarrow A^{-1}=B$

$\because$ It is true.

$ \begin{aligned}(iii)\,\, (A-B)^2 & =(A-B)(A-B) \\ & =A^2-A B-B A+B^2 \end{aligned} $

$\Rightarrow A^2-2 A B+B^2$ is true only when

$ A B=B A $

$\because$ It is false.

Given, $A=\left[\begin{array}{ccc}1 & -1 & 2 \\ -2 & 3 & -3 \\ 4 & -4 & 5\end{array}\right]$ then

$A^T=\left[\begin{array}{ccc}1 & -2 & 4 \\ -1 & 3 & -4 \\ 2 & -3 & 5\end{array}\right]$

Now, $A A^T=\left[\begin{array}{ccc}6 & -11 & 18 \\ -11 & 22 & -35 \\ 18 & -35 & 57\end{array}\right]$

Then, $A A^T-A-A^T=\left[\begin{array}{ccc}4 & -8 & 12 \\ -8 & 16 & -28 \\ 12 & -28 & 47\end{array}\right]$

We have, $A=\left[\begin{array}{ccc}x & 2 & 1 \\ -2 & y & 0 \\ 2 & 0 & -1\end{array}\right]$

$ \begin{aligned}trace\,\, of\, (A) & =x+y-1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \operatorname{det} A & =-2 y-x y-4=-6 \\ 2 y+x y & =2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

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

$ \begin{aligned} & 2 y+y(1-y)=2 \\ \Rightarrow \quad & y^2-3 y+2=0 \end{aligned} $

$ y=1 \text { and } 2 $

and $x=0$ and -1

Then, the minor of the element 1 is

$ \begin{aligned} & \left|\begin{array}{cc} -2 & y \\ 2 & 0 \end{array}\right| \\ & =-2 y=-2(2)=-4 \end{aligned} $

$ \begin{aligned} & \text { } \Delta=\left|\begin{array}{ccc} 2 \sin A \cos A & \sin C & \sin B \\ \sin C & 2 \sin B \cos B & \sin A \\ \sin B & \sin A & 2 \sin C \cos C \end{array}\right| \\ & =\left|\begin{array}{ccc} 2 k a \cos A & k c & k b \\ k c & 2 k b \cos B & k a \\ k b & k a & 2 k c \cos C \end{array}\right| \\ & {\left[\because \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k\right]} \end{aligned} $

$ =k^3\left|\begin{array}{ccc} 2 a \cos A & c & b \\ c & 2 b \cos B & a \\ b & a & 2 c \cos c \end{array}\right| $

$ =k^3 \left\lvert\, \begin{array}{ll} a \cos A+a \cos A & a \cos B+b \cos A \\ a \cos B+b \cos A & b \cos B+b \cos B \\ c \cos A+a \cos C & b \cos C+c \cos B \end{array}\right. $

$ \left.\begin{aligned} & c \cos A+a \cos C \\ & b \cos C+c \cos B \\ & c \cos C+c \cos C \end{aligned} \right\rvert\, $

$ \begin{aligned} & \left.=k^3\left|\begin{array}{ccc} \cos A & a & 0 \\ \cos B & b & 0 \\ \cos C & c & 0 \end{array}\right|| | \begin{array}{ccc} a & \cos A & 0 \\ b & \cos B & 0 \\ c & \cos C & 0 \end{array} \right\rvert\, \\ & =k^3 \times 0=0 \end{aligned} $

$\begin{aligned} & B=\left[\begin{array}{ll} 1 & 3 \\ 1 & 5 \end{array}\right] \\ & A B^{-1}=A^{-1} \\ & \Rightarrow A^2=B \end{aligned}$

Also, $B C B^{-1}=A$

$\begin{aligned} \Rightarrow C & =B^{-1} A B \\ \Rightarrow C^4 & =\left(B^{-1} A B\right)\left(B^{-1} A B\right)\left(B^{-1} A B\right)\left(B^{-1} A B\right) \\ & =B^{-1} A^4 B \\ & =B^{-1} B^2 B \\ \Rightarrow \quad C^4 & =B^2 \end{aligned}$

Also, $C^2=\left(B^{-1} A B\right)\left(B^{-1} A B\right)$

$\begin{aligned} & =B^{-1} A^2 B \\ & =B^{-1} B B \end{aligned}$

$\begin{aligned} & \Rightarrow C^2=B \\ & \Rightarrow C^4+\alpha C^2+\beta I=0 \\ & \Rightarrow B^2+\alpha B+\beta I=0 \\ & B^2=\left[\begin{array}{ll} 1 & 3 \\ 1 & 5 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 1 & 5 \end{array}\right]=\left[\begin{array}{ll} 4 & 18 \\ 6 & 28 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ll} 4 & 18 \\ 6 & 28 \end{array}\right]+\alpha\left[\begin{array}{ll} 1 & 3 \\ 1 & 5 \end{array}\right]+\left[\begin{array}{ll} \beta & 0 \\ 0 & \beta \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ & \Rightarrow 4+\alpha+\beta=0 \end{aligned}$

and $18+3 \alpha=0$

$\begin{aligned} & \Rightarrow \alpha=-6 \\ & \Rightarrow \beta=2 \\ & \Rightarrow 2 \beta-\alpha=10 \end{aligned}$

$\begin{aligned} & 3 x+5 y+\lambda z=3 \\ & 7 x+11 y-9 z=2 \\ & 97 x+155 y-189 z=\mu \end{aligned}$

$\begin{aligned} & {\left[\begin{array}{ccc} 3 & 5 & \lambda \\ 7 & 11 & -9 \\ 97 & 155 & -189 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \\ \mu \end{array}\right]} \\ & A X=B \\ & X=A^{-1} B \\ & X=\frac{\operatorname{adj} A}{|A|} B \end{aligned}$

For Infinitely many solution

$\begin{aligned} & |A|=0 \text { and }(\operatorname{adj} A) B=0 \\ & \left|\begin{array}{ccc} 3 & 5 & \lambda \\ 7 & 11 & -9 \\ 97 & 155 & -189 \end{array}\right|=0 \\ & -2052+2250+18 \lambda=0 \\ & \Rightarrow \lambda=-11 \\ & \operatorname{adj} A=\left[\begin{array}{ccc} -684 & -760 & 76 \\ 450 & 500 & -50 \\ 18 & 20 & -2 \end{array}\right] \\ & (\operatorname{adj} A) B=\left[\begin{array}{ccc} 684 & -760 & 76 \\ 450 & 500 & -50 \\ 18 & 20 & -2 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \\ \mu \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \\ & \Rightarrow 54+40-2 \mu=0 \\ & \Rightarrow 2 \mu=94 \\ & \Rightarrow \mu=47 \\ & \Rightarrow \mu+2 \lambda=25 \end{aligned}$

$\left|\begin{array}{lll} \alpha & b & c \\ a & \beta & c \\ a & b & \gamma \end{array}\right|=0$

$\begin{aligned} & R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3 \\ & \Rightarrow\left|\begin{array}{ccc} \alpha-a & b-\beta & 0 \\ 0 & \beta-b & c-\gamma \\ a & b & \gamma \end{array}\right|=0 \end{aligned}$

Take $\alpha$-a, $\beta$-b, $\gamma$-c common from column-1, 2 and 3 respectively

$\begin{aligned} & (\alpha-a)(\beta-b)(\gamma-c)\left|\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ \frac{a}{\alpha-a} & \frac{b}{\beta-b} & \frac{\gamma}{\gamma-c} \end{array}\right|=0 \\ & \Rightarrow \frac{\gamma}{\gamma-c}+\frac{b}{\beta-b}+\frac{a}{\alpha-a}=0 \end{aligned}$

$\begin{aligned} & x+4 y-z=\lambda \\ & 7 x+9 y+\mu z=-3 \\ & 5 x+y+2 z=-1 \\ & {\left[\begin{array}{ccc} 1 & 4 & -1 \\ 7 & 9 & \mu \\ 5 & 1 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} \lambda \\ -3 \\ -1 \end{array}\right]} \\ & A=\left[\begin{array}{lll} 1 & 4 & -1 \\ 7 & 9 & \mu \\ 5 & 1 & 2 \end{array}\right], B=\left[\begin{array}{c} \lambda \\ -3 \\ -1 \end{array}\right] \\ & A X=B \\ & X=A^{-1} B \\ & \quad=\frac{\operatorname{adj} A}{|A|} B \end{aligned}$

If $|A|=0$ and $(\operatorname{adj} A) \cdot B=0$, system has infinitely many solutions.

$\begin{aligned} & |A|=18-\mu-4(14-5 \mu)-1(7-45)=0 \\ & \Rightarrow 18-\mu-56+20 \mu+38=0 \\ & \Rightarrow 19 \mu=0 \\ & \Rightarrow \mu=0 \\ & \text { Also adjA }=\left[\begin{array}{ccc} 18 & -9 & 9 \\ -14 & 7 & -7 \\ -38 & 19 & -19 \end{array}\right] \\ & (\text { adj } A) \cdot B=0 \\ & {\left[\begin{array}{ccc} 18 & -9 & 9 \\ -14 & 7 & -7 \\ -38 & 19 & -19 \end{array}\right]\left[\begin{array}{c} \lambda \\ -3 \\ -1 \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]} \\ & 18 \lambda+27-9=0 \\ & \Rightarrow 18 \lambda=-18 \\ & \Rightarrow \lambda=-1 \\ & \Rightarrow 2 \mu+3 \lambda=3(-1)=-3 \end{aligned}$

$\begin{aligned} & |A-\lambda I|=0 \\ & \left|\begin{array}{ccc} 2-\lambda & a & 0 \\ 1 & 3-\lambda & 1 \\ 0 & 5 & b-\lambda \end{array}\right|=0 \\ & (2-\lambda)[(3-\lambda)(b-\lambda)-5]-a[b-\lambda-0]+0=0 \\ & (2-\lambda)\left[3 b-3 \lambda-b \lambda+\lambda^2-5\right]-a b+a \lambda=0 \\ & \lambda^3-(b+5) \lambda^2+(1-a+5 b) \lambda+(10-6 b+a b)=0 \\ & A^3-(b+5) A^2+(1-a+5 b) A+(10-6 b+a b) I=0 \\ & \Rightarrow \mathrm{b}+5=4,1-a+5 b=1,10-6 b+a b=21 \\ & \Rightarrow a=-5, b=-1 \\ & \Rightarrow 2 a+3 b=-13 \end{aligned}$

$\begin{aligned} & |A|=3 \\ & \left|\operatorname{adj}\left(-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}(2 A)^{-1}\right)\right)\right)\right| \\ & =\left|-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}\left((2 A)^{-1}\right)\right)\right)\right|^2 \end{aligned}$

$\begin{aligned} & =4^6\left|-3 \operatorname{adj}\left(\operatorname{aadj}\left((2 A)^{-1}\right)\right)\right|^4 \\ & =4^6 \cdot 3^{12}\left|\operatorname{aadj}\left((2 A)^{-1}\right)\right|^8 \\ & =4^6 \cdot 3^{12} \cdot 3^{24}\left|(2 A)^{-1}\right|^{16} \\ & =4^6 \cdot 3^{36} 2^{-48}\left|A^{-1}\right|^{16} \\ & =\frac{2^{-36} 3^{36}}{3^{16}}=2^{-36} 3^{20} \\ & m=-36 \\ & n=20 \\ & m+2 n=4 \end{aligned}$

We are given a determinant $A_r$ defined as:

$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$

We need to find the value of $2 A_{10} - A_8$.

Let's calculate the determinant $A_r$. We can use properties of determinants or expand it directly. A helpful trick is to use row operations to simplify the determinant before expanding. Let's perform the following row operations:

Replace Row 2 with (Row 2) - 2 $\times$ (Row 1)

Replace Row 3 with (Row 3) - 3 $\times$ (Row 1)

The determinant becomes:

$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r - 2r & 2 - 2(1) & (n^2-\beta) - 2(\frac{n^2}{2}+\alpha) \\ (3 r-2) - 3r & 3 - 3(1) & \frac{n(3 n-1)}{2} - 3(\frac{n^2}{2}+\alpha)\end{array}\right|$

Let's simplify the entries in the new rows:

Row 2:

First element: $2r - 2r = 0$

Second element: $2 - 2 = 0$

Third element: $(n^2-\beta) - 2(\frac{n^2}{2}+\alpha) = n^2 - \beta - n^2 - 2\alpha = -2\alpha - \beta$

Row 3:

First element: $(3r-2) - 3r = -2$

Second element: $3 - 3 = 0$

Third element: $\frac{n(3 n-1)}{2} - 3(\frac{n^2}{2}+\alpha) = \frac{3n^2 - n}{2} - \frac{3n^2}{2} - 3\alpha = \frac{3n^2 - n - 3n^2}{2} - 3\alpha = -\frac{n}{2} - 3\alpha$

So the determinant simplifies to:

$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 0 & 0 & -2\alpha - \beta \\ -2 & 0 & -\frac{n}{2} - 3\alpha\end{array}\right|$

Now, we can easily calculate this determinant by expanding along the second column, since it has two zeros. The expansion along the second column is:

$A_r = -1 \cdot \left|\begin{array}{cc} 0 & -2\alpha - \beta \\ -2 & -\frac{n}{2} - 3\alpha \end{array}\right|$

$A_r = -1 \cdot \left( (0) \times (-\frac{n}{2} - 3\alpha) - (-2) \times (-2\alpha - \beta) \right)$

$A_r = -1 \cdot \left( 0 - (4\alpha + 2\beta) \right)$

$A_r = -1 \cdot (-4\alpha - 2\beta)$

$A_r = 4\alpha + 2\beta$

Notice that the value of the determinant $A_r$ does not depend on $r$ at all! It's a constant value determined by $\alpha$ and $\beta$.

Now we need to find the value of $2 A_{10} - A_8$.

Since $A_r = 4\alpha + 2\beta$ for any value of $r$, we have:

$A_{10} = 4\alpha + 2\beta$

$A_8 = 4\alpha + 2\beta$

So, $2 A_{10} - A_8 = 2 (4\alpha + 2\beta) - (4\alpha + 2\beta)$

$2 A_{10} - A_8 = 8\alpha + 4\beta - 4\alpha - 2\beta$

$2 A_{10} - A_8 = (8\alpha - 4\alpha) + (4\beta - 2\beta)$

$2 A_{10} - A_8 = 4\alpha + 2\beta$

The value of the expression $2 A_{10} - A_8$ is $4\alpha + 2\beta$.

The given system of linear equations can be represented as,

$\begin{aligned} & \left(\begin{array}{ccc|c} 1 & 1 & 1 & 4 \\ 2 & 5 & 5 & 17 \\ 1 & 2 & m & n \end{array}\right) \\ & \sim\left(\begin{array}{ccc|c} 1 & 1 & 1 & 4 \\ 0 & 3 & 3 & 9 \\ 0 & 1 & m-1 & n-4 \end{array}\right) \\ & \sim\left(\begin{array}{ccc|c} 1 & 1 & 1 & 4 \\ 0 & 1 & 1 & 3 \\ 0 & 0 & m-2 & n-7 \end{array}\right) \end{aligned}$

$\because$ System of equations has infinitely many solutions

$\therefore m=2 \& n=7$

Which satisfy equation given in option (1).

$\text { (i.e., } 2^2+7^2-14=39 \text { ) }$

$A=\left[\begin{array}{ccc} \beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2 \alpha \end{array}\right], B=\left[\begin{array}{ccc} 3 \alpha & -9 & 3 \alpha \\ -\alpha & 7 & -2 \alpha \\ -2 \alpha & 5 & -2 \beta \end{array}\right]$

Cofactor of $A$-matrix is

$=\left[\begin{array}{ccc} 2 \alpha^2-\alpha \beta & -\left(2 \alpha^2+\beta^2\right) & \alpha^2+\alpha \beta \\ -\left(2 \alpha^2-3 \alpha\right) & (2 \alpha \beta+3 \beta) & -(2 \alpha \beta) \\ \alpha \beta-3 \alpha & -\left(\beta^2-3 \alpha\right) & \beta \alpha-\alpha^2 \end{array}\right]$

which is equal to matrix $B$

So, by comparing elements of two matrix

$\begin{aligned} & \Rightarrow \alpha \beta-3 \alpha=-2 \alpha \\ & \Rightarrow \alpha \beta-\alpha=0 \\ & \Rightarrow \alpha(\beta-1)=0 \\ & \Rightarrow \alpha=0 \text { or } \beta=1[\because \alpha \text { cannot be } 0] \\ & \Rightarrow \beta=1 \\ & \text { and }-\beta^2+3 \alpha=5 \\ & \Rightarrow 3 \alpha=6 \\ & \Rightarrow \alpha=2 \\ & A=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 2 & 1 \\ -1 & 2 & 4 \end{array}\right] \\ & \operatorname{Det}(A B)=|A||B|=|A|\left|(\operatorname{adj} A)^{\top}\right| \\ & =|A| \cdot|A|^2 \\ & =|A|^3 \\ & =(6-18+18)^3 \\ & =6^3 \\ & =216 \\ \end{aligned}$

$\begin{aligned} & |A|=3 \\ & |B|=2 \\ & \left.\left|A^T\right||A| \mid(\operatorname{adj}(2 A))^{-1}\|\operatorname{adj}(4 B)\|(\operatorname{adj}(A B))^{-1}\right)|A|\left|A^T\right| \\ & 3 \cdot 3 \frac{1}{64 \cdot 9}(64)^2 \cdot 4 \cdot \frac{1}{9 \cdot 4} 3 \cdot 3 \\ & =64 \end{aligned}$

$\begin{aligned} & 11 x+y+\lambda z=-5 \\ & 2 x+3 y+5 z=3 \\ & 8 x-19 y-39 z=\mu \\ & \Delta=0 \Rightarrow\left|\begin{array}{ccc} 11 & 1 & \lambda \\ 2 & 3 & 5 \\ 8 & -19 & -39 \end{array}\right|=0 \\ & 11(-39.3+19.5)-1(-39.2-40)+\lambda(-38-24)=0 \\ & =11(-117+95)-1(-118)-62 \lambda=0 \\ & =-242+118=62 \lambda \\ & \Rightarrow \lambda=-2 \\ & \Delta 2=0 \\ & \Rightarrow\left|\begin{array}{ccc} 11 & 1 & -5 \\ 2 & 3 & 3 \\ 8 & -19 & \mu \end{array}\right|=0 \end{aligned}$

$\begin{aligned} & 11(3 \mu+57)-1(2 \mu-24)-5(-38-24)=0 \\ & 33 \mu+627-2 \mu+24+310=0 \\ & \mu=-31 \\ & \Rightarrow \lambda^4-31 \\ & =16+31 \\ & =47 \end{aligned}$

$\begin{aligned} & \operatorname{adj}(A)=\left[\begin{array}{ll} 1 & -2 \\ 0 & 1 \end{array}\right] \\ & (\operatorname{adj} A)^2=\left[\begin{array}{ll} 1 & -4 \\ 0 & 1 \end{array}\right] \\ & (\operatorname{adj} A)^3=\left[\begin{array}{cc} 1 & -6 \\ 0 & 1 \end{array}\right] \\ & (\operatorname{adj} A)^4=\left[\begin{array}{cc} 1 & -8 \\ 0 & 1 \end{array}\right] \\ & (\operatorname{adj} A)^r=\left[\begin{array}{cc} 1 & (-2 r) \\ 0 & 1 \end{array}\right] \end{aligned}$

$\begin{aligned} & B=\sum_{r=0}^{10}(\operatorname{adj} A)^r=\left[\begin{array}{ll} \sum_\limits{r=0}^{10} 1 & \sum_\limits{r=0}^{10}(-2 r) \\ \sum_\limits{r=0}^{10}(0) & \sum_\limits{r=0}^{10}(1) \end{array}\right] \\ & B=\left[\begin{array}{cc} 11 & -110 \\ 0 & 11 \end{array}\right] \end{aligned}$

$\text { Sum of elements }=-110+11+11=-88$