Matrices and Determinants

$A=\left|\begin{array}{ccc}-1 & x & -3 \\ 2 & 4 & z \\ y & 5 & -6\end{array}\right|$ is a symmetric

So, $A=A^T$

that means, $x=2, y=-3, z=5$

and $B=\left[\begin{array}{ccc}0 & 2 & q \\ p & 0 & -4 \\ -3 & r & s\end{array}\right]$ is a skew-matrix

So, $B=-B^T$

that means, $p=-2 q=3, r=4, s=0$

Therefore, $|A|=\left|\begin{array}{ccc}-1 & 2 & -3 \\ 2 & 4 & 5 \\ -3 & 5 & -6\end{array}\right|$

$ =-1(-49)-2(3)-3(22)=-23 $

and $|B|=\left|\begin{array}{ccc}0 & 2 & 3 \\ -2 & 0 & -4 \\ -3 & 4 & 0\end{array}\right|$

$ =-2(-12)+3(-8)=0 $

Now, $|A B|=|A||B|=-23 \times 0=-23$

$ x y z+q+r=2 \times(-3) \times(5)+3+4=-23 $

Hence, $|A|+|B|-|A B|=x y z+q+r$

$A=\left[\begin{array}{ccc}-x & 14 x & 7 x \\ 0 & 1 & 0 \\ x & -4 x & -2 x\end{array}\right]$

and $B=\left[\begin{array}{ccc}2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1\end{array}\right]$

If $A^{-1}=B \Rightarrow$ So, $A B=I$

$ \begin{aligned} & \Rightarrow\left[\begin{array}{ccc} -x & 14 x & 7 x \\ 0 & 1 & 0 \\ x & -4 x & -2 x \end{array}\right]\left[\begin{array}{ccc} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{array}\right]=5 \\ & \Rightarrow-2 x+7 x=1 \Rightarrow x=1 / 5 \end{aligned} $

Now, $\left[\begin{array}{ccc}x & x+1 & x+2 \\ x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4\end{array}\right]$

$R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_2$

$ =\left[\begin{array}{ccc} x & x+1 & x+2 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right] $

therefore determinant of the given matrix $=0$

Hence, from the given options put $x=1 / 5$ in $5 x-1$ we get, $5 \times \frac{1}{5}-1=0$

$ \begin{aligned} &\text { that means, }\\ &\left|\begin{array}{ccc} x & x+1 & x+2 \\ x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \end{array}\right|=5 x-1 \end{aligned} $

$2 x+3 y-3 z=3$

$ x+2 y+\alpha z=1 $

and $2 x-y+z=\beta$

So, $A=\left[\begin{array}{ccc}2 & 3 & -3 \\ 1 & 2 & \alpha \\ 2 & -1 & 1\end{array}\right]$

for infinite many solutions put $|A|=0$

$ \begin{aligned} & \Rightarrow 2(2+\alpha)-3(1-2 \alpha)-3(-1-4)=0 \\ & \Rightarrow 1+8 \alpha+15=0 \Rightarrow \alpha=-2 \end{aligned} $

Now, $D_y=\left|\begin{array}{ccc}2 & 3 & -3 \\ 1 & 1 & -2 \\ 2 & \beta & 1\end{array}\right|$

$ \begin{aligned} & \text { put } D_y=0 \\ & \begin{aligned} \Rightarrow 2(1+2 \beta)-3(1+4)-3(\beta-2)=0 \\ \Rightarrow \beta-7=0 \Rightarrow \beta=7 \\ \text { Therefore, } \frac{\alpha}{\beta}-\frac{\beta}{\alpha}=\frac{-2}{7}-\frac{7}{(-2)} \\ \qquad=\frac{-4+49}{14}=\frac{45}{14} \end{aligned} \end{aligned} $

We have,

$ \left|\begin{array}{ccc} 1+\sin ^2 \theta & \cos ^2 \theta & 4 \sin 4 \theta \\ \sin ^2 \theta & 1+\cos ^2 \theta & 4 \sin 4 \theta \\ \sin ^2 \theta & \cos ^2 \theta & 1+4 \sin 4 \theta \end{array}\right|=0 $

Applying $R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3$

$ \begin{aligned} & \left|\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ \sin ^2 \theta & \cos ^2 \theta & 1+4 \sin 4 \theta \end{array}\right|=0 \\ & R_3 \rightarrow R_3-\left(\sin ^2 \theta\right) R_1 \end{aligned} $

$ \begin{aligned} &\left|\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 1 & 1+4 \sin 4 \theta \end{array}\right|=0\\ &\text { Expand w.r.t along } C_1 \text {, we get }\\ &\begin{aligned} & 1+4 \sin 4 \theta+1=0 \\ & 4 \sin 4 \theta=-2 \\ & \sin 4 \theta=\frac{-1}{2}=\sin \left(\pi+\frac{\pi}{6}\right) \\ & 4 \theta=\frac{7 \pi}{6} \Rightarrow \theta=\frac{7 \pi}{24} \end{aligned} \end{aligned} $

Given, $2 x+p y+6 z=8$

$ \begin{aligned} & x+2 y+q z=5 \\ & x+y+3 z=4 \end{aligned} $

For infinite solution

$ \begin{aligned} & \Delta=\Delta_x=\Delta_y=\Delta_z=0 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Delta_z=0 \\ & \Rightarrow\left|\begin{array}{lll} 2 & p & 8 \\ 1 & 2 & 5 \\ 1 & 1 & 4 \end{array}\right|=0 \\ & \Rightarrow \quad 2(8-5)-p(4-5)+8(1-2)=0 \\ & \Rightarrow \quad 6+p-8=0 \Rightarrow p=2 \end{aligned} $

$ \begin{aligned} &\text { We have, } x^a y^b=e^m\\ &x^a=\frac{e^m}{y^b} \Rightarrow x=\frac{e^{\frac{m}{a}}}{y^{\frac{b}{a}}} \Rightarrow x^c y^d=e^n \end{aligned} $

$ \Rightarrow \quad \frac{e^{\frac{m c}{a}}}{y^{\frac{b c}{a}}} \cdot y^d=e^n $

$ \begin{aligned} \Rightarrow & & y^{d-\frac{b c}{a}}=e^{n-\frac{m c}{a}} \\ \Rightarrow & & y^{a d-k c}=e^{a n-m c} \\ \Rightarrow & & y=e^{\frac{a n-m c}{a d-k c}}=e^{\frac{\Delta_2}{\Delta_3}} \end{aligned} $

$ \begin{aligned} &\text { and } &x=e^{\frac{\Delta_1}{\Delta_3}} \end{aligned} $

$ \quad \begin{aligned}\because & \Delta_1=m d-b n \\ & \Delta_2=a n-m c \\ & \Delta_3=a d-b c \end{aligned} $

$ \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 { We have, } A=\left[\begin{array}{lll} 83 & 74 & 41 \\ 93 & 96 & 31 \\ 24 & 15 & 79 \end{array}\right]\\ &\begin{aligned} A^T & =\left[\begin{array}{lll} 83 & 93 & 24 \\ 74 & 96 & 15 \\ 41 & 31 & 79 \end{array}\right] . \\ \therefore A-A^T & =\left[\begin{array}{lll} 83 & 74 & 41 \\ 93 & 96 & 31 \\ 24 & 15 & 79 \end{array}\right]-\left[\begin{array}{lll} 83 & 93 & 24 \\ 74 & 96 & 15 \\ 41 & 31 & 79 \end{array}\right] \\ & =\left[\begin{array}{ccc} 0 & -19 & 17 \\ 19 & 0 & 16 \\ -17 & -16 & 0 \end{array}\right] \end{aligned}\\ &\begin{aligned} & \therefore \operatorname{det}\left(A-A^T\right)=\left|\begin{array}{ccc} 0 & -19 & 17 \\ 19 & 0 & 16 \\ -17 & -16 & 0 \end{array}\right| \\ & 0(0+16 \times 16)+19(9 \times 0+16 \times 17) \\ & +17(-19 \times 16-0) \\ & =19 \times 16 \times 17-17 \times 19 \times 16 \\ & =0 \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { We have, }\left|\begin{array}{lll} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{array}\right|>0 \\ & =\{(a b c-a)-1(c-1)+1(1-b)\}>0 \\ & =(a b c-a-c+1+1-b)>0 \\ & =(a b c-(a+b+c)+2)>0 \\ & =a b c+2>(a+b+c) \\ & \begin{array}{l} a b c+2>3(a b c)^{\frac{1}{3}} \end{array} \\ & {\left[\because A m>G m \Rightarrow \frac{a+b+c}{3}>(a+b)^{1 / 3}\right]} \\ & \text { let } x=(a b c)^{1 / 3} \Rightarrow a b c=x^3 \\ & \qquad x^3+2>3 x \\ & \qquad \begin{array}{l} x^3-3 x+2>0 \\ (x-1)^2(x+2)>0 \\ (x+2)>0=x>-2 \\ \quad 1 \\ \quad=(a b c)^{\frac{1}{3}}>-2=(a b c)>-8 \end{array} \end{aligned} \end{aligned} $

The given system of equations are homogeneous and has only trival solution, then

$ \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|=0 $

$\operatorname{Rank}[A]=n$, where $n$ is the order of square matrix and

$ 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] $

$\therefore \operatorname{Rank}[A]=3$

Given,

$ A=\left[\begin{array}{lll} 0 & 1 & 2 \\ 2 & 3 & 0 \\ 4 & 0 & 3 \end{array}\right] $

and $B$ is a matrix such that $A B=B A$. And $A B$ is not an identity matrix.

Let

$ B=\left[\begin{array}{lll} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array}\right] $

Now, $A B=B A$

$ \begin{aligned} & {\left[\begin{array}{lll} 0 & 1 & 2 \\ 2 & 3 & 0 \\ 4 & 0 & 3 \end{array}\right]\left[\begin{array}{lll} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array}\right]}=\left[\begin{array}{lll} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array}\right]\left[\begin{array}{lll} 0 & 1 & 2 \\ 2 & 3 & 0 \\ 4 & 0 & 3 \end{array}\right] \\ & {\left[\begin{array}{lll} b_1+2 c_1 & b_2+2 c_2 & b_3+2 c_3 \\ 2 a_1+3 b_1 & 2 a_2+3 b_2 & 2 a_3+3 b_3 \\ 4 a_1+3 c_1 & 4 a_2+3 c_2 & 4 a_3+3 c_3 \end{array}\right]} =\left[\begin{array}{llll} 2 a_2+4 a_3 & a_1+3 a_2 & 2 a_1+3 a_3 \\ 2 b_2+4 b_3 & b_1+3 b_2 & 2 b_1+3 b_3 \\ 2 c_2+4 c_3 & c_1+3 c_2 & 2 c_1+3 c_3 \end{array}\right] \end{aligned} $

On comparing two matrix and solve for $a_1, a_2, a_3, b_1, b_2, b_3, c_1, c_2$ and $c_3$, we get

$ \begin{aligned} & a_1=9, a_2=-3 \text { and } a_3=-6 \\ & b_1=-6, b_2=-8 \text { and } b_3=4 \\ & c_1=-12 c_2=4 \text { and } c_3=-2 \end{aligned} $

So, $B=\left[\begin{array}{ccc}9 & -3 & -6 \\ -6 & -8 & 4 \\ -12 & 4 & -2\end{array}\right]$

Let $A=\left[\begin{array}{ccc}x-2 & 0 & 1 \\ 1 & x+3 & 2 \\ 2 & 0 & 2 x-1\end{array}\right]$ is a singular matrix.

So, $|A|=0$

$ \left|\begin{array}{ccc} x-2 & 0 & 1 \\ 1 & x+3 & 2 \\ 2 & 0 & 2 x-1 \end{array}\right|=0 $

$ \begin{aligned} & \Rightarrow(x-2)[(x+3)(2 x-1)-0]-0 \\ & \Rightarrow \quad(x-2)(x+3)(2 x-1)-2(x+3)]=0 \\ & \Rightarrow \quad(x+3)[(x-2)(2 x-1)-2]=0 \\ & \Rightarrow \quad(x+3)\left[2 x^2-x-4 x+2-2\right]=0 \\ & \Rightarrow \quad(x+3)\left[2 x^2-5 x\right]=0 \\ & \Rightarrow \quad x(x+3)(2 x-5)=0 \\ & \Rightarrow \quad x=-3,0,5 / 2 \end{aligned} $

So, $\alpha=-3, \beta=0$ and $\gamma=5 / 2$

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ [given } \alpha<\beta<\gamma \text { ] } $

Now,

$ \begin{aligned} 2 \alpha+3 \beta+4 \gamma & =2 \times(-3)+3 \times 0+4 \times \frac{5}{2} \\ & =-6+10=4 \end{aligned} $

To determine the nature of the solutions for the system of linear equations:

$ \begin{aligned} x + 2y + z &= -3, \\ 3x + 3y - 2z &= -1, \\ 2x + 7y + 7z &= -4, \end{aligned} $

we begin by calculating the determinant $\Delta$ of the coefficient matrix:

$ \Delta = \left| \begin{array}{ccc} 1 & 2 & 1 \\ 3 & 3 & -2 \\ 2 & 7 & 7 \end{array} \right| $

Calculating this determinant:

$ \Delta = 1(21 + 14) - 2(21 + 4) + 1(21 - 6) $

$ = 1 \times 35 - 2 \times 25 + 1 \times 15 $

$ = 35 - 50 + 15 = 0 $

Since $\Delta = 0$, the system of equations is either inconsistent (no solution) or dependent (infinitely many solutions). To resolve this, we check $\Delta_x$, the determinant of the matrix when the first column is replaced by the constants from the right side of the equations:

$ \Delta_x = \left| \begin{array}{ccc} -3 & 2 & 1 \\ -1 & 3 & -2 \\ -4 & 7 & 7 \end{array} \right| $

Calculating $\Delta_x$:

$ \Delta_x = (-3)(35) - 2(-15) + 1(5) $

$ = -105 + 30 + 5 = -70 $

Since $\Delta = 0$ and $\Delta_x \neq 0$, the system has no solution. Hence, the equations are inconsistent.

To determine the condition for $a$ and $b$ so, that the set of equations

$ \begin{aligned} & x+2 y+3 z=6 \\ & x+3 y+5 z=9 \end{aligned} $

$2 x+5 y+a z=b$ has a unique solution. We need to ensure that the coefficient matrix has a non zero determinant.

Let $A=\left(\begin{array}{lll}1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & a\end{array}\right)$

Then, $\operatorname{det}(A)$

$ \begin{aligned} & =1(3 a-25)-2(\cdot a-10)+3(5-6) \\ & =1(3 a-25)-2(a-10)+3(-1) \\ & =3 a-25-2 a+20-3 \\ & =a-8 \end{aligned} $

For system to have a unique solution. $\operatorname{det}(A)$ must be non-zero.

i.e. $a-8 \neq 0$

Thus, $a \neq 8$

So, $b$ can be other number which should be a non-zero real number and therefore $b$ must be a real number.

Thus, $a \neq 8, b \in R$.

Given, $ |PQ|=1 $ and $ |P|=9 $.

For any $ n \times n $ matrix $ A $, the determinant of $\operatorname{adj}(A)$ is $|A|^{n-1}$, where $ n $ is the dimension of the square matrix.

If $ B = kA $, where $ k $ is a scalar, then $|B| = k^n|A|$ for an $ n \times n $ matrix $ A $.

Also, $|PQ| = |P||Q|$.

Utilizing these properties, for a $ 3 \times 3 $ matrix we have:

$ |3Q| = 3^3|Q| = 27|Q|. $

Now, considering the matrix $ \operatorname{adj}(3Q) $:

$ \operatorname{Det}(\operatorname{adj}(3Q)) = (27|Q|)^2. $

Since $ |PQ| = 1 $ and $ |P| = 9 $, we have:

$ \begin{aligned} 1 & = 9 \cdot |Q|, \\ |Q| & = \frac{1}{9}. \end{aligned} $

Finally, compute $\operatorname{det}(\operatorname{adj}(3Q))$:

$ \begin{aligned} \operatorname{det}(\operatorname{adj}(3Q)) & = (27|Q|)^2 \\ & = (27 \times \frac{1}{9})^2 \\ & = 3^2 \\ & = 9. \end{aligned} $

For $\operatorname{adj}(P \cdot \operatorname{adj}(3Q))$:

$ \begin{aligned} \operatorname{adj}(P \cdot \operatorname{adj}(3Q)) & = (9 \times 9)^{3-1} \\ & = (9^2)^2 \\ & = 9^4. \end{aligned} $

We know that for a square matrix the product $A \cdot \operatorname{adj}(A)=\operatorname{det}(A)$. I where I is the identity matrix of the same order as $A$.

If $A$ is a $3 \times 3$ matrix, then the product should be : $A \cdot \operatorname{adj}(A)=\operatorname{det}(A) \cdot I_3$

Given matrix

$ \begin{aligned} A & =\left[\begin{array}{lll} a & 1 & 2 \\ 1 & 2 & b \\ c & 1 & 3 \end{array}\right] \\ \operatorname{adj}(A) & =\left[\begin{array}{ccc} 7 & -1 & -5 \\ -3 & 9 & 5 \\ 1 & -3 & 5 \end{array}\right] \end{aligned} $

Let's multiply $A$ with adj $(A)$

$ A \cdot \operatorname{adj}(A)=\left[\begin{array}{lll} a & 1 & 2 \\ 1 & 2 & b \\ c & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} 7 & -1 & -5 \\ -3 & 9 & 5 \\ 1 & -3 & 5 \end{array}\right] $

$ =\left[\begin{array}{ccc} 7 a-1 & b+1 & 7c \\ -a+3 & -3 b+17 & -c \\ -5 a+15 & 5 b+5 & -5+20 \end{array}\right] $

Setting these elements equal to identity matrix adjusted by determinant

We get, $-a+3=0$

$ \begin{aligned} \Rightarrow \quad a & =3, b+1=0 \Rightarrow b=-1 \\ 7 c & =0 \Rightarrow c=0 \end{aligned} $

Thus, $\quad a=3, b=-1$ and $c=0$

Hence, $a^2+b^2+c^2$

$ =3^2+(-1)^2+0=9+1=10 $

$ \begin{aligned} &\text { Given that } 3 A=\left[\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{array}\right]\\ &\begin{aligned} \Rightarrow A & =\left[\begin{array}{ccc} \frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\ \frac{a}{3} & \frac{2}{3} & \frac{b}{3} \end{array}\right] \\ \therefore A^T & =\left[\begin{array}{ccc} \frac{1}{3} & \frac{2}{3} & \frac{a}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & -\frac{2}{3} & \frac{b}{3} \end{array}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Now, } A A^T=I \Rightarrow\left[\begin{array}{ccc} \frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\ \frac{a}{3} & \frac{2}{3} & \frac{b}{3} \end{array}\right]\left[\begin{array}{ccc} \frac{1}{3} & \frac{2}{3} & \frac{a}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & -\frac{2}{3} & \frac{b}{3} \end{array}\right] =\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ccc} \frac{1}{9}+\frac{4}{9}+\frac{4}{9} & \frac{2}{9}+\frac{2}{9}-\frac{4}{9} & \frac{a}{9}+\frac{4}{9}+\frac{2 b}{9} \\ \frac{2}{9}+\frac{2}{9}-\frac{4}{9} & \frac{4}{9}+\frac{1}{9}+\frac{4}{9} & \frac{2 a}{9}+\frac{2}{9}-\frac{2 b}{9} \\ \frac{a}{9}+\frac{4}{9}+\frac{2 b}{9} & \frac{2 a}{9}+\frac{2}{9}-\frac{2 b}{9} & \frac{a^2}{9}+\frac{4}{9}+\frac{b^2}{9} \end{array}\right] =\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned} $

$ \begin{aligned} & \Rightarrow\left[\begin{array}{ccc} 1 & 0 & \frac{a+4+2 b}{9} \\ \frac{a+4+2 b}{9} & \frac{2 a+2-2 b}{9} & \frac{\frac{a^2+4+b^2}{9}}{0} \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \therefore \quad a^2+b^2+4=9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\\ & \Rightarrow \quad a^2+b^2=5 \\ & \text { and } 2 a+2-2 b=0 \\ & \Rightarrow \quad b=a+1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \text { and } a+4+2 b=0 \\ & \Rightarrow \quad a+2 b=-4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned}\\ &\text { On solving Eqs. (ii) and (iii), we get }\\ &a+2 b=-4 \end{aligned} $

$ \begin{aligned} \Rightarrow & & a+2(a+1) & =-4 \\ \Rightarrow & & a+2 a+2 & =-4 \\ \Rightarrow & & 3 a & =-6 \\ \Rightarrow & & a & =-2 \text { and } b=-2+1 \\ \Rightarrow & & b & =-1 \end{aligned} $

$ \begin{aligned} &\text { }\\ &Now,\,\,\,\,\,\,\,\,\,\,\frac{a}{b}+\frac{b}{a}=\frac{-2}{-1}+\frac{-1}{-2}=2+\frac{1}{2}=\frac{5}{2} \end{aligned} $

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

$ \begin{aligned} &=2(a+b+c)\left|\begin{array}{ccc} 1 & a & b \\ 1 & b+c+2 a & b \\ 1 & a & c+a+2 b \end{array}\right|\\ &\text { Applying } R_1 \rightarrow R_1-R_2 \text { and } R_2 \rightarrow R_2-R_3\\ &\begin{aligned} & =2(a+b+c)\left|\begin{array}{ccc} 0 & -(a+b+c) & 0 \\ 0 & a+b+c & -(a+b+c) \\ 1 & a & c+a+2 b \end{array}\right| \\ & =2(a+b+c)^3\left|\begin{array}{ccc} 0 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & a & c+a+2 b \end{array}\right| \\ & =2(a+b+c)^3[1(1-0)]=2(a+b+c)^3 \end{aligned} \end{aligned} $

To clarify:

When we have a matrix $ B $ of order 3 with a determinant $ |B| = 6 $, the formula to find the determinant of its adjugate matrix, $|\operatorname{adj}(B)|$, is given by:

$ |\operatorname{adj}(B)| = |B|^{n-1} $

where $ n $ is the order of the matrix $ B $.

In this case, since $ n = 3 $, we have:

$ |\operatorname{adj}(B)| = 6^{3-1} = 6^2 = 36 $

This confirms that the assertion is true.

However, the reason provided, $ |\operatorname{adj}(B)| = |B|^n $, is incorrect because it states that the determinant of the adjugate matrix is raised to the power of $ n $ rather than $ n-1 $. The correct relationship involves raising the determinant of $ B $ to the power of $ n-1 $.

Thus, the assertion is true, but the given reason is false.

Given the matrix $ A = \left[\begin{array}{lll} 2 & 3 & 4 \\ 1 & k & 2 \\ 4 & 1 & 5 \end{array}\right] $, which is singular, we need to find the quadratic equation with roots $ k $ and $ \frac{1}{k} $.

Since $ A $ is singular, its determinant is zero:

$ |A| = \left|\begin{array}{lll} 2 & 3 & 4 \\ 1 & k & 2 \\ 4 & 1 & 5 \end{array}\right| = 2(5k - 2) - 3(5 - 8) + 4(1 - 4k) = 0 $

Expanding this:

$ 2(5k - 2) = 10k - 4 $

$ -3(5 - 8) = -3 \cdot 3 = -9 $

$ 4(1 - 4k) = 4 - 16k $

Putting it all together, we get:

$ 10k - 4 - 9 + 4 - 16k = 0 $

Simplifying:

$ -6k + 9 = 0 $

Solving for $ k $:

$ k = \frac{3}{2} $

Thus, the roots of the quadratic equation are $ \frac{3}{2} $ and $ \frac{2}{3} $. Using the standard form of a quadratic equation $ x^2 - (\alpha + \beta)x + \alpha \beta = 0 $ where the roots are $ \alpha = \frac{3}{2} $ and $ \beta = \frac{2}{3} $, we calculate:

$ \alpha + \beta = \frac{3}{2} + \frac{2}{3} = \frac{9}{6} + \frac{4}{6} = \frac{13}{6} $

$ \alpha \beta = \frac{3}{2} \cdot \frac{2}{3} = 1 $

Substitute these into the quadratic equation:

$ x^2 - \left(\frac{13}{6}\right)x + 1 = 0 $

To eliminate fractions, multiply through by 6:

$ 6x^2 - 13x + 6 = 0 $

This is the quadratic equation with the specified roots.

Given that $|P| = \left|\frac{A}{2}\right|$, we start by using the property of the adjoint matrix, where $|\text{adj} (A)| = |A|^{n-1}$ for an $ n \times n $ matrix. Here, since $ A $ is a $ 4 \times 4 $ matrix, $|\text{adj} (A)| = |A|^{3}$.

Substituting the given condition, we have:

$ |A|^{3} = \left(\frac{1}{2}\right)^{4} |A| $

Simplifying, this becomes:

$ |A|^{3} = \frac{1}{16} |A| $

Assuming $|A| \neq 0$, divide both sides by $|A|$:

$ |A|^{2} = \frac{1}{16} $

Taking the square root on both sides gives:

$ |A| = \pm \frac{1}{4} $

To find $|A^{-1}|$, we use the property $|A^{-1}| = \frac{1}{|A|}$:

$ |A^{-1}| = \pm 4 $

Therefore, the absolute determinant of the inverse of $ A $ is $\pm 4$.

We have,

$ \begin{array}{r} x+2 y+3 z=4 \\ 4 x+5 y+3 z=5 \\ 3 x+4 y+3 z=\lambda \end{array} $

Write the equation in matrix form $\left[A_1^1 B\right]$

$ \left[\begin{array}{lll:l} 1 & 2 & 3 & 4 \\ 4 & 5 & 3 & 5 \\ 3 & 4 & 3 & \lambda \end{array}\right] $

$R_2 \rightarrow R_2-4 R_1$ and $R_3 \rightarrow R_3-3 R_1$

$ \sim\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & -3 & -9 & -11 \\ 0 & -2 & -6 & \lambda-12 \end{array}\right] $

$ \sim\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & -3 & -9 & -11 \\ 0 & 0 & 0 & 3 \lambda-14 \end{array}\right] $

$ R_3 \leftrightarrow 3 R_3-2 R_2 $

For consistency $3 \lambda-14=0 \Rightarrow 3 \lambda=14$

Given, $3 \lambda=n+100$

Therefore, $n=14-100$

$ n=-86 $

So, the value of $n$ is -86 .

Let $\Delta=\left|\begin{array}{ccc}a & b & c \\ a^2 & b^2 & c^2 \\ 1 & 1 & 1\end{array}\right|\quad \ldots \text { (i) }$

Applying, $R_1 \rightarrow R_1+R_3, R_2 \rightarrow R_2+R_3$ we get

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

Applying $C_1 \rightarrow C_1-C_2$ and $C_2 \rightarrow C_2-C_3$, we get

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

$ =a b^2-a^2 b-a c^2+a^2 c-b^2 c+b c^2 $

Thus, option (a) and (b) are correct. Now, option (c),

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

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

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

$ =-1\left[\left(b^2+b-a^2-a\right)(c-a)-\left(c^2+c-a^2-a\right)(b-a)\right] $

$ =\left(c^2+c-a^2-a\right)(b-a)-\left(b^2+b-a^2-a\right)(c-a) $

$ =c^2 b-a c^2+b c-a c-a^2 b+a^3-a b+a^2-b^2 c+a b^2-b c+a b+a^2 c -a^3+a c-a^2 $

Hence, option (d) is correct required answer.

Given, $A B C D E=I$

$\Rightarrow A^{-1}(A B C D E) E^{-1}=A^{-1} I E^{-1}$

( $\because$ Pre multiply by $A^{-1}$ and post multiply by $E^{-1}$, respectively)

$ \begin{array}{ll} \Rightarrow & I B C D I=A^{-1} E^{-1} \\ \Rightarrow & B C D=A^{-1} E^{-1} \end{array} $

Now, pre multiply by $B^{-1}$ and post multiply by $D^{-1}$, we get

$ \begin{array}{l} \Rightarrow & B^{-1}(B C D) D^{-1} & =B^{-1} A^{-1} E^{-1} D^{-1} \\ \Rightarrow & B^{-1} B C D D^{-1} & =B^{-1} A^{-1} E^{-1} D^{-1} \\ \Rightarrow & I C I & =B^{-1} A^{-1} E^{-1} D^{-1} \\ \Rightarrow & C & =B^{-1} A^{-1} E^{-1} D^{-1} \\ \therefore & C^{-1} & =\left(B^{-1} A^{-1} E^{-1} D^{-1}\right)^{-1} \\ & & =\left(D^{-1}\right)^{-1}\left(E^{-1}\right)^{-1}\left(A^{-1}\right)^{-1}\left(B^{-1}\right)^{-1} \\ & & \text { (using reversal law) } \end{array} $

$\begin{array} & &=D E A B \\ \therefore \quad C^{-1} & =D E A B \end{array}$

Given, $A=\left[a_{i j}\right], 1 \leq i, j \leq n$ with $n \geq 2$ and $a_{i j}=i+j$

$\therefore$ The matrix is

$A=\left[\begin{array}{ccccc}2 & 3 & 4 & \cdots & n+1 \\ 3 & 4 & 5 & \cdots & n+2 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ n+1 & n+2 & n+3 & \cdots & n+n\end{array}\right]$

$ =\left[\begin{array}{cccc} 2 & 3 & \cdots & n+1 \\ 3 & 4 & \cdots & n+2 \\ 4 & 5 & \cdots & n+3 \\ \vdots & \vdots & \cdots & \vdots \\ n+1 & n+2 & \cdots & 2 n \end{array}\right] $

Use elementary row operation to subtract 1st row from the $i$ th row for $2 \leq i \leq n$, we get

$ A=\left[\begin{array}{ccccc} 2 & 3 & 4 & \cdots & n+1 \\ 1 & 1 & 1 & \cdots & 1 \\ 2 & 2 & 2 & \cdots & 2 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ n-1 & n-1 & n-1 & \cdots & n-1 \end{array}\right] $

Applying, $R_3 \rightarrow R_3-2 R_2$ and $R_4 \rightarrow R_4-3 R_2$ and so on

$ A=\left[\begin{array}{ccccc} 2 & 3 & 4 & \cdots & n+1 \\ 1 & 1 & 1 & \cdots & 1 \\ 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 \end{array}\right] $

$\therefore$ Rank $A=2$

Given the matrix $ A $:

$ A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} $

We are tasked to find $ A^2 - 5A + 6I $.

Calculate $ A^2 $:

$ A^2 = A \cdot A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} $

Perform the matrix multiplication:

$ A^2 = \begin{bmatrix} 1\cdot1 + 0\cdot2 + 2\cdot3 & 1\cdot0 + 0\cdot1 + 2\cdot2 & 1\cdot2 + 0\cdot3 + 2\cdot4 \\ 2\cdot1 + 1\cdot2 + 3\cdot3 & 2\cdot0 + 1\cdot1 + 3\cdot2 & 2\cdot2 + 1\cdot3 + 3\cdot4 \\ 3\cdot1 + 2\cdot2 + 4\cdot3 & 3\cdot0 + 2\cdot1 + 4\cdot2 & 3\cdot2 + 2\cdot3 + 4\cdot4 \end{bmatrix} $

$ = \begin{bmatrix} 1 + 0 + 6 & 0 + 0 + 4 & 2 + 0 + 8 \\ 2 + 2 + 9 & 0 + 1 + 6 & 4 + 3 + 12 \\ 3 + 4 + 12 & 0 + 2 + 8 & 6 + 6 + 16 \end{bmatrix} $

$ = \begin{bmatrix} 7 & 4 & 10 \\ 13 & 7 & 19 \\ 19 & 10 & 28 \end{bmatrix} $

Calculate $ 5A $:

$ 5A = 5 \cdot \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 10 \\ 10 & 5 & 15 \\ 15 & 10 & 20 \end{bmatrix} $

Calculate $ 6I $:

$ 6I = 6 \cdot \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} $

Combine the results to find $ A^2 - 5A + 6I $:

$ A^2 - 5A + 6I = \begin{bmatrix} 7 & 4 & 10 \\ 13 & 7 & 19 \\ 19 & 10 & 28 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 10 \\ 10 & 5 & 15 \\ 15 & 10 & 20 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} $

$ = \begin{bmatrix} 7 - 5 + 6 & 4 - 0 + 0 & 10 - 10 + 0 \\ 13 - 10 + 0 & 7 - 5 + 6 & 19 - 15 + 0 \\ 19 - 15 + 0 & 10 - 10 + 0 & 28 - 20 + 6 \end{bmatrix} $

$ = \begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 14 \end{bmatrix} $

Thus, the resulting matrix is:

$ \begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 14 \end{bmatrix} $

We have, $\left|\begin{array}{ccc}x^2+2 x & x+2 & 1 \\ 2 x+1 & x-1 & 1 \\ x+2 & -1 & 1\end{array}\right|=0$

$ \begin{aligned} & \Rightarrow\left(x^2+2 x\right)\left|\begin{array}{cc} x-1 & 1 \\ -1 & 1 \end{array}\right|-(x+2)\left|\begin{array}{cc} 2 x+1 & 1 \\ x+2 & 1 \end{array}\right|+1 \\ & \Rightarrow\left(x^2+2 x\right)(x-1+1)-(x+2)(2 x+1-x-2) \\ & \quad+1\left(-2 x-1-x^2-x+2\right)=0 \\ & \Rightarrow x\left(x^2+2 x\right)-(x+2)(x-1)+1 \\ & \quad\left(-x^2-3 x+1\right)=0 \\ & \Rightarrow x^3+2 x^2-x^2-x+2-x^2-3 x+1=0 \\ & \Rightarrow x^3-4 x+3=0 \end{aligned} $

On putting, $x=1$, we get

$ (1)^3-4 \times 1+3=0 $

Thus, $x=1$ is a root of the given equation.

Thus, $(x-1)\left(x^2+x-3\right)=0$

Solve $x^2+x-3=0$

$ \begin{aligned} x & =\frac{-1 \pm \sqrt{1^2-4 \times 1 \times(-3)}}{2 \times 1} \\ & =\frac{-1 \pm \sqrt{1+12}}{2}=\frac{-1 \pm \sqrt{13}}{2} \end{aligned} $

Hence, positive roots are 1 and $\frac{-1+\sqrt{13}}{2}$

Sum of positive roots $=1+\left(\frac{-1+\sqrt{13}}{2}\right)$

$ =\frac{1+\sqrt{13}}{2} $