Matrices and Determinants

Given, $A^3-5 A^2+7 A+I=0$

$ \Rightarrow A^5-5 A^4+7 A^3+A^2=0 $

Also, $A^5-6 A^4+12 A^3-6 A^2+2 A+2 I$

$ \begin{array}{r} =l A+m I \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ A^5-5 A^4+7 A^3+A^2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

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

$ -A^4+5 A^3-7 A^2+2 A+2 I=I A+m I \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

$ A^4-5 A^3+7 A^2+A=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

On adding Eqs. (iii) and (iv), we get

$ \begin{aligned} & 3 A+2 I=l A+m I \\ \Rightarrow \quad & I=3 \text { and } m=2 \end{aligned} $

So, $\quad l+m=5$

$ \begin{aligned} &\text { Given, }\\ &A=\left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \end{array}\right], A^{-1}=\left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & y \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array}\right] \end{aligned} $

$ \begin{aligned} & A A^{-1}=\left[\begin{array}{ccc} 1 & 0 & y+1 \\ 0 & 1 & 2 y+2 \\ 4-4 x & 3 x-3 & x y+2 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \Rightarrow y+1=0 \text { and } 4-4 x=0 \\ & \Rightarrow y=-1 \text { and } x=1 \\ & \text { So, } \log _{2 / 5}\left(2^1+2^{-1}\right)=-1 \\ & \Rightarrow(x, y) \text { satisfies } y=\log _{2 / 5}\left(2^x+2^{-x}\right) \end{aligned} $

$ \begin{aligned} & \text { Let } A=\left[\begin{array}{lll} a_1 & b_1 & 0 \\ a_2 & b_2 & 0 \\ a_3 & b_3 & 0 \end{array}\right] \\ & A X=0 \\ & {\left[\begin{array}{lll} a_1 & b_1 & 0 \\ a_2 & b_2 & 0 \\ a_3 & b_3 & 0 \end{array}\right]\left[\begin{array}{l} l \\ m \\ 0 \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]} \\ & {\left[\begin{array}{l} a_1 l+b_1 m \\ a_2 l+b_2 m \\ a_3 l+b_3 m \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]} \\ & a_1 l+b_1 m=0 \\ & a_2 l+b_2 m=0 \\ & a_3 l+b_3 m=0 \\ & \Rightarrow a_1: a_2: a_3=b_1: b_2: b_3 \Rightarrow|A|=0 \end{aligned} $

⇒ Determinant of each submatrix of order $2 \times 2$ is zero $\Rightarrow$ Rank $=1$

Since, the system of equations has non-trivial solution

$ \begin{aligned} & \Rightarrow\left|\begin{array}{ccc} 2 & 3 & a \\ 1 & a & -2 \\ 3 & 1 & 3 \end{array}\right|=0 \\ & \Rightarrow 2(3 a+2)-3(3+6)+a(1-3 a)=0 \\ & \Rightarrow 6 a+4-9-18+a-3 a^2=0 \\ & \Rightarrow 6 a-23+a-3 a^2=0 \\ & \Rightarrow 3 a^2-7 a+23=0 \\ & \Rightarrow D<0 \end{aligned} $

So, no real values of $a$ is possible.

$ \text { } \begin{aligned} 2 x+3 y+z & =-1 \\ 3 x+y+z & =4 \\ x-3 y-2 z & =1 \\ \Rightarrow \quad x-2 y & =5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ \begin{array}{lr} \Rightarrow 5 x+3 y=-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \Rightarrow 5(5+2 y)+3 y=-1 \\ \Rightarrow 25+13 y=-1 \\ \Rightarrow 13 y=-26 \\ \Rightarrow y=-2 \\ \text { So, } y=-2 \end{array} $

For non-trivial solution,

$ \begin{aligned} & \quad\left|\begin{array}{rrr} 1 & a & 1 \\ a & 2 & -1 \\ 2 & 3 & 1 \end{array}\right|=0 \\ & 1(2+3)-a(a+2)+1(3 a-4)=0 \\ & \Rightarrow \quad 5-a^2-2 a+3 a-4=0 \\ & \Rightarrow \quad a^2-a-1=0 \\ & \Rightarrow \quad a=\frac{1 \pm \sqrt{5}}{2} \end{aligned} $

So, $\quad a=\frac{1+\sqrt{5}}{2}$

$ \text { } \begin{aligned} |\operatorname{adj} A| & =|A|^{n-1} \\ \left|\operatorname{adj}\left(A^2\right)\right| & =\left|A^2\right|^{n-1} \\ & =\left|A^2\right|^2 \\ \Rightarrow \quad A^2 & =\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}\right] \\ & =\left[\begin{array}{lll} 7 & 8 & 6 \\ 5 & 7 & 6 \\ 6 & 6 & 5 \end{array}\right] \end{aligned} $

$ \begin{aligned} & \Rightarrow|A|^2=7(35-36)-8(25-36)+6(30-42) \\ & \quad=-7+88-72=9 \\ & \therefore\left|\operatorname{adj}\left(A^2\right)\right|=\left|A^2\right|^2=9^2=81 \end{aligned} $

$ \begin{array}{r} x-2 y+z=0 \\ 2 x+3 y+z=6 \\ x+2 y+p z=q \end{array} $

$ \begin{aligned} &\text { System has infinite solution, }\\ &\begin{array}{ll} \therefore & \Delta=\Delta_x=\Delta_y=\Delta_z=0 \\ & \Delta=\left|\begin{array}{ccc} 1 & -2 & 1 \\ 2 & 3 & 1 \\ 1 & 2 & p \end{array}\right|=0 \\ \Rightarrow & 1(3 p-2)+2(2 p-1)+1(4-3)=0 \\ \Rightarrow & 3 p-2+4 p-2+1=0 \\ \Rightarrow & 7 p=3 \Rightarrow p=\frac{3}{7} \\ & \Delta_z=\left|\begin{array}{ccc} 1 & -2 & 0 \\ 2 & 3 & 6 \\ 1 & 2 & q \end{array}\right|=0 \\ \Rightarrow & 1(3 q-12)+2(2 q-6)=0 \\ \Rightarrow & 3 q-12+4 q-12=0 \\ \Rightarrow & 7 q=24 \Rightarrow q=\frac{24}{7} \\ \therefore & q-p=\frac{24}{7}-\frac{3}{7}=3 \end{array} \end{aligned} $

We have,

$ \begin{aligned} & (\sin \theta) x-y+z=0 \\ & x-(\cos \theta) y+z=0 \\ & x+y+(\sin \theta) z=0 \end{aligned} $

System has non-trivial solution.

$ \therefore \quad \Delta=0 $

$ \begin{aligned} & \left|\begin{array}{ccc} \sin \theta & -1 & 1 \\ 1 & -\cos \theta & 1 \\ 1 & 1 & \sin \theta \end{array}\right|=0 \\ & \Rightarrow \quad \sin \theta(-\sin \theta \cos \theta-1) +1(\sin \theta-1)+1(1+\cos \theta)=0 \\ & \Rightarrow \quad-\sin ^2 \theta \cos \theta-\sin \theta+\sin \theta-1+1 +\cos \theta=0 \\ & \Rightarrow \quad \cos \theta\left(1-\sin ^2 \theta\right)=0 \\ & \Rightarrow \quad \cos ^3 \theta=0 \Rightarrow \cos \theta=0 \Rightarrow \theta=\frac{\pi}{2} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} A & =\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{array}\right] \\ |A| & =1(1-3)-2(2-1)+3(6-1) \\ & =-2-2+15=11 \\ B & =\left[\begin{array}{lll} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{array}\right] \\ \Rightarrow \quad & |B|=2(4-8)-3(6-4)+4(12-4) \\ & =-8-6+32=18 \\ \therefore \quad & \sqrt{|\operatorname{adj} A B|}=\sqrt{|\operatorname{adj} A||\operatorname{adj} B|} \\ & =\sqrt{|A|^{3-1}|B|^{3-1}} \\ & =|A||B|=11(18)=198 \end{aligned} \end{aligned} $

Step 1: Find the determinant of $ A $

Matrix $ A $ is: $ A=\left[\begin{array}{lll} 1 & 5 & 2 \\ 4 & 1 & 3 \\ 2 & 6 & 3 \end{array}\right] $

So, $ |A| = 1(3-18) - 5(12-6) + 2(24-2) $

Calculate each part:

• $1 \times (3 - 18) = 1 \times (-15) = -15$
• $-5 \times (12 - 6) = -5 \times 6 = -30$
• $2 \times (24 - 2) = 2 \times 22 = 44$

Add them up: $ |A| = -15 - 30 + 44 = -1 $

Step 2: Find the value of $ \left| (\operatorname{adj} A)^{-1} \right| $

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

The order of matrix $A$ is 3, so: $ |\operatorname{adj} A| = |A|^{3-1} = |A|^2 $

So, $ \left|(\operatorname{adj} A)^{-1}\right| = \frac{1}{|A|^2} $

Since $ |A| = -1 $, $ \left|(\operatorname{adj} A)^{-1}\right| = \frac{1}{(-1)^2} = \frac{1}{1} = 1 $

Given, $x+\lambda y-2 z=1$

$ x-y+\lambda z=2 $

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

∵ Given, system is inconsistent

$ \begin{aligned} & D=0 \Rightarrow\left|\begin{array}{rrr} 1 & \lambda & -2 \\ 1 & -1 & \lambda \\ 1 & -2 & 3 \end{array}\right|=0 \\ & \Rightarrow 1(-3+2 \lambda)-\lambda(3-\lambda)-2(-2+1)=0 \\ & \Rightarrow 2 \lambda-3-3 \lambda+\lambda^2+2=0 \end{aligned} $

$ \therefore \lambda_1+\lambda_2=1 $

Given, $(\sin \theta) x+y-2 z=0$

$ 2 x-y+(\cos \theta) z=0 $

And $-3 x+(\sec \theta) y+3 z=0$

∵ system has non-trivial solution.

$ \begin{aligned} & \Delta=0 \\ & \left|\begin{array}{ccc} \sin \theta & 1 & -2 \\ 2 & -1 & \cos \theta \\ -3 & \sec \theta & 3 \end{array}\right|=0 \\ & \sin \theta(-3-1)-1(6+3 \cos \theta)-2(2 \sec \theta-3) =0\\ & \Rightarrow-4 \sin \theta-6-3 \cos \theta-4 \sec \theta+6=0 \\ & \Rightarrow 4 \sec \theta+3 \cos \theta+4 \sin \theta=0 \\ & \Rightarrow 4+3 \cos ^2 \theta+4 \sin \theta \cos \theta=0 \\ & \Rightarrow 4+3\left(\frac{1+\cos 2 \theta}{2}\right)+2 \sin 2 \theta=0 \\ & \Rightarrow 8+3+3 \cos 2 \theta+4 \sin 2 \theta=0 \\ & \Rightarrow 11+3 \cos 2 \theta+4 \sin 2 \theta=0 \end{aligned} $

This equation has no solution because the maximum value of $4 \sin 2 \theta+3 \cos 2 \theta$ is 5 . Since $5+11 \neq 0$, there is no value of $\theta$ that satisfies the equation.

$ \begin{aligned} &\\ &\begin{aligned} & \text { } A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \\ & \because \operatorname{adj}(\operatorname{adj} A)=|A|^{n-2} A \\ & \operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))=|\operatorname{adj} A|^{2-2} \cdot \operatorname{adj} A \\ & \quad=\operatorname{adj} A=|A| A^{-1}\left(\because A^{-1}=\frac{1}{|A|} \operatorname{adj} A\right) \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \left|\begin{array}{ccc} x & -3 & 2 \\ -1 & -2 & x-1 \\ 1 & x-2 & 3 \end{array}\right|=0 \\ & \Rightarrow x(-6-(x-1)(x-2))+3(-3-(x-1)) +2(-(x-2)+2)=0 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow x\left(-6-x^2+3 x-2\right)+3(-3-x+1) +2(-x+2+2)=0 \\ & \Rightarrow x\left(-x^2+3 x-8\right)+3(-2-x) +2(4-x)=0 \\ & \Rightarrow-x^3+3 x^2-8 x-6-3 x+8-2 x=0 \\ & \therefore \text { Sum of roots }=\frac{-3}{-1}=3 . \end{aligned} $

We have, $\left|\begin{array}{ccc}1 & 2 & 3-\lambda \\ 0 & -1-\lambda & 2 \\ 1-\lambda & 1 & 3\end{array}\right|=A \lambda^3+B \lambda^2+C \lambda+D$

Now, $\left|\begin{array}{ccc}1 & 2 & 3-\lambda \\ 0 & -1-\lambda & 2 \\ 1-\lambda & 1 & 3\end{array}\right|$

$ \begin{aligned} & =1(-3-3 \lambda-2)+(1-\lambda)(4+(1+\lambda)(3-\lambda)) \\ & =(-3 \lambda-5)+(1-\lambda)\left(4+3-\lambda+3 \lambda-\lambda^2\right) \\ & =(-3 \lambda-5)+(1-\lambda)\left(7+2 \lambda-\lambda^2\right) \\ & =(-3 \lambda-5)+\left(7+2 \lambda-\lambda^2-7 \lambda-2 \lambda^2+\lambda^3\right) \\ & =\lambda^3-3 \lambda^2-8 \lambda+2 \end{aligned} $

$ \begin{aligned} &\therefore \lambda^3-3 \lambda^2-8 \lambda+2=A \lambda^3+B \lambda^2+C \lambda+D\\ &\text { On comparing both sides, we get }\\ &\begin{aligned} A & =1, B=-3, C=-8, D=2 \\ \therefore \quad D+A & =2+1=3 \end{aligned} \end{aligned} $

We have

$ A+2 B=\left[\begin{array}{rrr} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{array}\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

And $\quad 2 A-B=\left[\begin{array}{rrr}2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2\end{array}\right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now, $2 A+4 B=\left[\begin{array}{rcr}2 & 4 & 0 \\ 12 & -6 & 6 \\ -10 & 6 & 2\end{array}\right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Substract Eqs. (ii) from (iii), we get

$ \begin{aligned} &\begin{aligned} & 5 B=\left[\begin{array}{rrr} 2 & 4 & 0 \\ 12 & -6 & 6 \\ -10 & 6 & 2 \end{array}\right]-\left[\begin{array}{rrr} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{array}\right]=\left[\begin{array}{rrr} 0 & 5 & -5 \\ 10 & -5 & 0 \\ -10 & 5 & 0 \end{array}\right] \\ \Rightarrow & B=\left[\begin{array}{rrr} 0 & 1 & -1 \\ 2 & -1 & 0 \\ -2 & 1 & 0 \end{array}\right] \end{aligned}\\ &\therefore \text { From Eqs. (ii), we get }\\ &\begin{aligned} & 2 A=\left[\begin{array}{rrr} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{array}\right]+\left[\begin{array}{rrr} 0 & 1 & -1 \\ 2 & -1 & 0 \\ -2 & 1 & 0 \end{array}\right] \\ \Rightarrow & 2 A=\left[\begin{array}{rrr} 2 & 0 & 4 \\ 4 & -2 & 6 \\ -2 & 2 & 2 \end{array}\right] \Rightarrow A=\left[\begin{array}{rrr} 1 & 0 & 2 \\ 2 & -1 & 3 \\ -1 & 1 & 1 \end{array}\right] \\ \therefore & \operatorname{tr}(A)-\operatorname{tr}(B) \\ & =(1-1+1)-(0-1+0)=1+1=2 \end{aligned} \end{aligned} $

We have,

$A, C$ are $3 \times 3$ matrices

$B \times D$ are $3 \times 1$ matrices

$A X=B$ has unique solution

$C X=D$ has infinite solution.

Since, $A X=B$ have a unique solution.

$\Rightarrow \operatorname{rank}(A)=$ number of unknowns $=3$

$\therefore \quad \operatorname{rank}([A: B])=\operatorname{rank}(A)=3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Since, $C X=D$ have infinite many solutions

$\therefore \quad \operatorname{Rank}(C)=$ not unique $\Rightarrow \operatorname{rank}(C)<3$

$\Rightarrow \quad \operatorname{Rank}([C: D])<3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) and (ii), we have

$\operatorname{Rank}([A: D]) \geq \operatorname{rank}[(C: B)]$

We have, $A$ and $B$ are two non-square matrices such that

$ P=A+B, Q=A^T B, R=A B^T $

Since, $A+B$ is defined,

So order of $A$ and $B$ are same.

Let order of $A$ and $B$ be $m \times n$

$\therefore$ Order of $A^T$ is $n \times m$ and order of $B^T$ is $n \times m$

Now, order of $P=$ order of $A+B=m \times n$

Order of $Q=$ order of $A^T B=n \times n$

And order of $R=$ order of $A B^T=m \times m$

Now, order of $P Q=m \times n$

Order of $Q R=$ not defined

Order of $R Q=$ not defined

Order of $Q P=$ not defined

Order of $R P=m \times n$

$\therefore$ Order of $A=$ order of $P Q$ and $R P$

Given equations are $x+y-z=1$,

$ 2 x+4 y-z=0 \text { and } 3 x+4 y+5 z=18 $

The augmented matrix for the system is

$ A=\left[\begin{array}{cccc} 1 & 1 & -1 & 1 \\ 2 & 4 & -1 & 0 \\ 3 & 4 & 5 & 18 \end{array}\right] $

We need to transform this matrix into

$ B=\left[\begin{array}{cccc} 1 & a & 0 & -1 \\ 0 & 2 & 1 & b \\ 0 & 0 & c & 32 \end{array}\right] $

Now, perform the row operation for matrix $A$, we get

$ \begin{gathered} R_2 \rightarrow R_2-2 R_1 \text { and } R_3 \rightarrow R_3-3 R_1 \\ A=\left[\begin{array}{cccc} 1 & 1 & -1 & 1 \\ 0 & 2 & 1 & -2 \\ 0 & 1 & 8 & 15 \end{array}\right] \end{gathered} $

Now, $R_3 \rightarrow R_3-\frac{1}{2} R_2$

$ A=\left[\begin{array}{cccc} 1 & 1 & -1 & 1 \\ 0 & 2 & 1 & -2 \\ 0 & 0 & 7.5 & 16 \end{array}\right] $

Apply $R_3 \rightarrow 2 R_3$, we get

$ A=\left[\begin{array}{cccc} 1 & 1 & -1 & 1 \\ 0 & 2 & 1 & -2 \\ 0 & 0 & 15 & 32 \end{array}\right] $

Again apply $R_1 \rightarrow R_1+R_2$, we get

$ A=\left[\begin{array}{cccc} 1 & 3 & 0 & -1 \\ 0 & 2 & 1 & -2 \\ 0 & 0 & 15 & 32 \end{array}\right] $

$ \begin{aligned} &\text { Thus, } a=3, b=-2 \text { and } c=15\\ &\begin{array}{ll} \Rightarrow & c=15 \\ \therefore & \sqrt{a+b+c}=\sqrt{3+(-2)+15} \\ & =\sqrt{1+15}=\sqrt{16}=4 \end{array} \end{aligned} $

Given, $K=\left|\begin{array}{ccc}9 & 25 & 16 \\ 16 & 36 & 25 \\ 25 & 49 & 36\end{array}\right|$

$ \begin{aligned} & =9\left|\begin{array}{ll} 36 & 25 \\ 49 & 36 \end{array}\right|-25\left|\begin{array}{ll} 16 & 25 \\ 25 & 36 \end{array}\right|+16\left|\begin{array}{ll} 16 & 36 \\ 25 & 49 \end{array}\right| \\ & =9(1296-1225)-25(576-625) +16(784-900) \\ & =9(71)-25(-49)+16(-116) \\ & =639+1225-1856=8 \\ & \therefore K=8 \end{aligned} $

So, the roots are $K$ and $K+1$, i.e., 8 and 9

$ \begin{aligned} \text { Sum of roots } & =8+9=17 \\ & =\frac{-b}{a}=\alpha+\beta \end{aligned} $

And product of roots $=8 \times 9=72$

$ =\frac{c}{a}=\alpha \beta $

So, the general form of a quadratic equation is $x^2-(\alpha+\beta) x+\alpha \beta=0$

$ \Rightarrow \quad x^2-17 x+72=0 $

Given, $A=\left[\begin{array}{ccc}1 & -3 & -5 \\ -2 & 4 & -6 \\ 7 & -11 & 13\end{array}\right]$

$ \begin{aligned} & \quad|A|=\left|\begin{array}{ccc} 1 & -3 & -5 \\ -2 & 4 & -6 \\ 7 & -11 & 13 \end{array}\right| \\ & =1(4 \times 13-(-6) \times(-11))-(-3)(-2 \times 13 -(-6) \times 7)-5(-2 \times(-11)-7 \times 4) \\ & =1(52-66)+3(-26+42)-5(22-28) \\ & =-14+48+30=64 \end{aligned} $

Since, $|\operatorname{adj} A|=|A|^{n-1}$, for $n \times n$ matrix $A$

So, for $3 \times 3$ matrix,

$ \therefore \quad \begin{aligned} |\operatorname{adj} A| & =|A|^{3-1}=|A|^2 \\ \therefore \quad|\operatorname{adj} A| & =|A|^2=(64)^2=4096 \\ \sqrt{|\operatorname{adj} A|} & =\sqrt{4096}=64 \end{aligned} $

$ \text {} \begin{aligned} \Delta r & =\left|\begin{array}{cc} \frac{1}{3 r-2} & \frac{2}{3 r-5} \\ 0 & \frac{3}{3 r+1} \end{array}\right| \\ & =\frac{1}{(3 r-2)} \times \frac{3}{(3 r+1)}-\frac{2}{3 r-5} \times 0 \\ & =\frac{3}{(3 r-2)(3 r+1)} \end{aligned} $

Using partial fractions

$ \begin{aligned} \Delta r & =\frac{A}{3 r-2}+\frac{B}{3 r+1} \\ \Rightarrow \quad 3 & =A(3 r+1)+B(3 r-2) \end{aligned} $

If $\quad r=\frac{2}{3}$, then

$ \begin{aligned} & \Rightarrow \quad 3=A\left(3 \times \frac{2}{3}+1\right)+B\left(3 \times \frac{2}{3}-2\right) \\ & \Rightarrow 3 A=3 \\ & \Rightarrow \quad A=1 \end{aligned} $

If $\quad r=-\frac{1}{3}$, then

$ \begin{aligned} & \quad 3=B\left(3 \times\left(-\frac{1}{3}\right)-2\right)=B(-1-2) \\ & \Rightarrow \quad B=-1 \\ & \text { So, } \Delta r=\frac{1}{3 r-2}-\frac{1}{3 r+1} \\ & \therefore \sum_{r=1}^{33} \Delta r=\sum_{r=1}^{33}\left(\frac{1}{3 r-2}-\frac{1}{3 r+1}\right) \\ & \Rightarrow\left(\frac{1}{1}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{10}\right)+ \\ & \quad \ldots+\left(\frac{1}{97}-\frac{1}{100}\right) \\ & \Rightarrow 1-\frac{1}{100}=\frac{99}{100}=0.99 \end{aligned} $

Given the matrices $A$ and $B$:

$ A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} x & y \\ 1 & 2 \end{bmatrix} $

We have the condition:

$ (A+B)(A-B) = A^2 - B^2 $

This implies:

$ AB - BA = 0 \quad \Rightarrow \quad AB = BA $

Let's compute $AB$ and $BA$:

$ AB = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x & y \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} x + 2 & y + 4 \\ 2x + 1 & 2y + 2 \end{bmatrix} $

$ BA = \begin{bmatrix} x & y \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} x + 2y & 2x + y \\ 1 + 4 & 2 + 2 \end{bmatrix} = \begin{bmatrix} x + 2y & 2x + y \\ 5 & 4 \end{bmatrix} $

Equating $AB$ and $BA$, we have:

$ \begin{bmatrix} x + 2 & y + 4 \\ 2x + 1 & 2y + 2 \end{bmatrix} = \begin{bmatrix} x + 2y & 2x + y \\ 5 & 4 \end{bmatrix} $

From this, we derive:

$ \begin{aligned} 2x + 1 &= 5 \\ 2y + 2 &= 4 \\ \end{aligned} $

Thus:

$ \begin{aligned} 2x &= 4 \quad \Rightarrow \quad x = 2 \\ 2y &= 2 \quad \Rightarrow \quad y = 1 \\ \end{aligned} $

Given matrix $C$:

$ C = \begin{bmatrix} x & 2 \\ 1 & y \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 1 & 1 \end{bmatrix} $

The trace of matrix $C$, denoted as $\operatorname{tr}(C)$, is:

$ \operatorname{tr}(C) = x + y = 2 + 1 = 3 $

To solve the given determinant equation, we start by simplifying the expression:

$\left|\begin{array}{ccc}x-2 & 3(x-1) & 5(x-1) \\ x-4 & 3(x-3) & 5(x-5) \\ x-8 & 3(x-9) & 5(x-25)\end{array}\right| = 0$

This simplifies to:

$\left|\begin{array}{ccc}x-2 & x-1 & x-1 \\ x-4 & x-3 & x-5 \\ x-8 & x-9 & x-25\end{array}\right| = 0$

Next, perform column operations to simplify the determinant. Subtract the second column from the first:

$C_1 \rightarrow C_1 - C_2 \implies \left|\begin{array}{ccc} -1 & x-1 & x-1 \\ -1 & x-3 & x-5 \\ 1 & x-9 & x-25 \end{array}\right| = 0 $

Now, apply row operations:

Add the third row to the first row: $ R_1 \rightarrow R_1 + R_3 $

Add the third row to the second row: $ R_2 \rightarrow R_2 + R_3 $

This results in:

$\left|\begin{array}{ccc} 0 & 2x-10 & 2x-26 \\ 0 & 2x-12 & 2x-30 \\ 1 & x-9 & x-25 \end{array}\right| = 0 $

To solve, expand the determinant by the first column:

$ (2x-30)(2x-10) - (2x-12)(2x-26) = 0 $

This simplifies to:

$ 2(x-15) \cdot 2(x-5) - 2(x-6) \cdot 2(x-13) = 0 $

Further simplifying, we have:

$(x-15)(x-5) - (x-6)(x-13) = 0$

Solving this, we calculate:

$x^2 - 20x + 75 - (x^2 - 19x + 78) = 0$

This leads to:

$-x - 3 = 0 \implies x = -3$

Therefore, $x = -3$ satisfies the equation $x^2 + 2x - 3 = 0$.

We start with the well-known formula for any invertible matrix $ A $:

$ A \cdot \operatorname{adj}(A) = |A| I. $

Since $ A $ is non-singular ($|A| \neq 0$), we can write

$ \operatorname{adj}(A) = |A|\, A^{-1}. $

Now, replace $ A $ with $ A^{-1} $. Then, applying the same formula for $ A^{-1} $, we have:

$ A^{-1} \cdot \operatorname{adj}(A^{-1}) = |A^{-1}|\, I. $

Recall that $ |A^{-1}| = \frac{1}{|A|} $. Thus,

$ A^{-1} \cdot \operatorname{adj}(A^{-1}) = \frac{1}{|A|}\, I. $

Multiplying both sides on the left by $ A $ (which is valid since $ A $ is invertible) gives:

$ A \cdot A^{-1} \cdot \operatorname{adj}(A^{-1}) = A \cdot \frac{1}{|A|}\, I. $

Since $ A \cdot A^{-1} = I $, this simplifies to:

$ \operatorname{adj}(A^{-1}) = \frac{1}{|A|}\, A. $

On the other hand, we earlier had:

$ \operatorname{adj}(A) = |A|\, A^{-1}. $

Taking the inverse of both sides (and recalling that for any nonzero scalar $ c $ and invertible matrix $ B $, $(cB)^{-1} = \frac{1}{c}\,B^{-1}$), we get:

$ (\operatorname{adj}(A))^{-1} = (|A|\, A^{-1})^{-1} = \frac{1}{|A|}\, (A^{-1})^{-1} = \frac{1}{|A|}\, A. $

Thus, we obtain:

$ \operatorname{adj}(A^{-1}) = (\operatorname{adj}(A))^{-1}. $

Therefore, the correct option is:

Option A – $(\operatorname{adj} A)^{-1}$.

$\left|\begin{array}{rrr}1 & -2 & 3 \\\\ 2 & 4 & -5 \\\\ 3 & \lambda & \mu\end{array}\right|=0$

$ \begin{aligned} & 1(4 \mu+5 \lambda)+2(2 \mu+15)+3(2 \lambda-12)=0 \\\\ & \Rightarrow 4 \mu+5 \lambda+4 \mu+30+6 \lambda-36=0 \\\\ & 11 \lambda+8 \mu-6=0 \\\\ & \therefore \quad 8 \mu+11 \lambda=6 \end{aligned} $

$\frac{x^2}{2 x^4+7 x^2+6}=\frac{x^2}{\left(x^2+2\right)\left(2 x^2+3\right)}$

$ =\frac{A x+B}{x^2+a}+\frac{C x+D}{a x^2+3} $

which gives, $a=2$

$ \begin{aligned} & \frac{x^2}{\left(x^2+2\right)\left(2 x^2+3\right)}=\frac{A x+B}{x^2+2}+\frac{C x+D}{2 x^2+3} \\\\ & x^2=(A x+B)\left(2 x^2+3\right)+(C x+D)\left(x^2+2)\right. \\\\ & 0=2 A+C \\\\ & 1=2 B+D \end{aligned} $

On solving these equations, we get .

$ \begin{aligned} 0 & =3 A+2 C \\\\ 0 & =3 B+2 D \\\\ A & =C=0 \\\\ B & =2, D=-3 \\\\ \therefore \quad A+ & B+C-2 D=0+2+0-2(-3) \\\\ & =2+6=8 \\\\ & =4(2)=4 a \end{aligned} $

Let's consider the $3 \times 3$ matrix $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$ where each element is a positive integer and each row sums to 6, with $a_{22} = 2$.

The matrix properties provide the following conditions:

$a_{22} = a_{23} + a_{32}$ implies $a_{23} + a_{32} = 2$.

$a_{11} = a_{12} + a_{21}$.

$a_{33} = a_{31} + a_{13}$.

Substituting the conditions about row sums, we get:

$a_{11} + a_{12} + a_{13} = 6$.

$a_{21} + a_{22} + a_{23} = 6$.

$a_{31} + a_{32} + a_{33} = 6$.

From $a_{23} + a_{32} = 2$, if $a_{23} = 1$, then $a_{32} = 1$.

From the second row sum, we have: $a_{21} + 2 + 1 = 6$ which gives $a_{21} = 3$.

Given $a_{11} = a_{12} + a_{21}$ and the first row sum,

with $2a_{12} + a_{13} = 3$,

choosing $a_{12} = 1$ and $a_{13} = 1$, results in $a_{11} = 4$.

Finally, using the third row equation $2a_{31} + a_{32} + a_{13} = 6$,

substituting $a_{32} = 1$ and $a_{13} = 1$ results in $a_{31} = 2$.

Now, from $a_{31} + a_{32} + a_{33} = 6$, with $a_{31} = 2$ and $a_{32} = 1$,

results in $a_{33} = 3$.

Matrix $A$ is:

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

To find the determinant $|A|$, we calculate:

$ |A| = 4(2 \cdot 3 - 1 \cdot 1) - 1(3 \cdot 3 - 1 \cdot 2) + 1(3 \cdot 1 - 2 \cdot 2) $

$ = 4(6 - 1) - 1(9 - 2) + 1(3 - 4) $

$ = 4 \cdot 5 - 1 \cdot 7 + 1 \cdot (-1) $

$ = 20 - 7 - 1 = 12 $

Thus, the determinant $|A| = 12$.

Given that $ |\operatorname{adj} A| = x $ and $ |\operatorname{adj} B| = y $, we are tasked with finding $ \left|(\operatorname{adj}(A B))^{-1}\right| $.

First, recognize that the determinant of the adjugate of a matrix can be expressed with respect to the determinants of the matrix itself. Using this relation, we have:

$ \left|\operatorname{adj}(AB)\right| = \left|\operatorname{adj}(\operatorname{adj}(A))\right| \cdot \left|\operatorname{adj}(\operatorname{adj}(B))\right| $

The inverse of the adjugate adjoint product can be decomposed as:

$ \left|(\operatorname{adj}(A B))^{-1}\right| = \left|\left(\operatorname{adj} B)^{-1} (\operatorname{adj} A)^{-1}\right)\right| $

Because $ |\operatorname{adj} A| = x $ and $ |\operatorname{adj} B| = y $, the properties of determinants give us:

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

$ \left|(\operatorname{adj} B)^{-1}\right| = \frac{1}{|\operatorname{adj} B|} = \frac{1}{y} $

Thus, the determinant of the inverse of the product is:

$ \left|(\operatorname{adj}(A B))^{-1}\right| = \frac{1}{x} \times \frac{1}{y} = \frac{1}{xy} $

Therefore, the result is $ \frac{1}{xy} $.

Given, $ x+3 b y+b z=0 $

$ x+2 a y+a z=0 $

$ x+4 c y+c z=0 $

$ \Rightarrow \Delta=\left|\begin{array}{lll}1 & 3 b & b \\ 1 & 2 a & a \\ 1 & 4 c & c\end{array}\right| $

$ \Delta=2 a c+3 a b-2 a b+4 b c-4 a c-3 b c $

$ \Delta=-2 a c+a b+b c $

For non-zero solution $ \Delta=0 $

$ \Rightarrow 2 a c=a b+b c $

To evaluate the determinant of the matrix:

$ \left|\begin{array}{ccc}\frac{-bc}{a^{2}} & \frac{c}{a} & \frac{b}{a} \\ \frac{c}{b} & -\frac{ac}{b^{2}} & \frac{a}{b} \\ \frac{b}{c} & \frac{a}{c} & -\frac{ab}{c^{2}}\end{array}\right| $

we can simplify the process by factoring out $ \frac{1}{a^{2}}, \frac{1}{b^{2}}, $ and $ \frac{1}{c^{2}} $ from rows $ R_1, R_2, $ and $ R_3 $ respectively. This allows us to focus on a simpler determinant:

$ \Rightarrow \quad \frac{1}{a^{2} b^{2} c^{2}}\left|\begin{array}{ccc} -bc & ac & ab \\ bc & -ac & ab \\ bc & ac & -ab \end{array}\right| $

Next, apply row operations to simplify further: replace $ R_2 $ with $ R_2 + R_1 $ and $ R_3 $ with $ R_3 + R_1 $:

$ \Rightarrow \quad \frac{1}{a^{2} b^{2} c^{2}}\left|\begin{array}{ccc} -bc & ac & ab \\ 0 & 0 & 2ab \\ 0 & 2ac & 0 \end{array}\right| $

This matrix is now triangular, allowing us to compute the determinant by multiplying the diagonal elements:

$ \Rightarrow \quad \frac{1}{a^{2} b^{2} c^{2}} \times (4a^{2}b^{2}c^{2}) = 4 $

Thus, the value of the determinant is 4.

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

and

$ A=\left[\begin{array}{lll} x & y & y \\\\ y & x & y \\\\ y & y & x \end{array}\right] $

$ \Rightarrow \quad 5 A^{-1} A=5 I $

$ \left[\begin{array}{ccc} -3 & 2 & 2 \\\\ 2 & -3 & 2 \\\\ 2 & 2 & -3 \end{array}\right]\left[\begin{array}{lll} x & y & y \\\\ y & x & y \\\\ y & y & x \end{array}\right]=\left[\begin{array}{lll} 5 & 0 & 0 \\\\ 0 & 5 & 0 \\\\ 0 & 0 & 5 \end{array}\right] $

$ \left[\begin{array}{ccc} -3 x+4 y & -y+2 x & -y+2 x \\\\ -y+2 x & -3 x+4 y & -y+2 x \\\\ -y+2 x & -y+2 x & -3 x+4 y \end{array}\right] $

$ =\left[\begin{array}{lll} 5 & 0 & 0 \\\\ 0 & 5 & 0 \\\\ 0 & 0 & 5 \end{array}\right] $

$ \begin{aligned} & \Rightarrow \quad y=2 x \text { and }-3 x+4 y=5 \\\\ & \Rightarrow \quad 5 x=5 \\\\ & \Rightarrow \quad x=1 \end{aligned} $

Put $x=1$, then $y=2$

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

$ \begin{aligned} A^2 & =\left[\begin{array}{lll} 1 & 2 & 2 \\\\ 2 & 1 & 2 \\\\ 2 & 2 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 2 \\\\ 2 & 1 & 2 \\\\ 2 & 2 & 1 \end{array}\right] \\\\ & =\left[\begin{array}{lll} 9 & 8 & 8 \\\\ 8 & 9 & 8 \\\\ 8 & 8 & 9 \end{array}\right] \\\\ A^2-4 A & =\left[\begin{array}{lll} 9 & 8 & 8 \\\\ 8 & 9 & 8 \\\\ 8 & 8 & 9 \end{array}\right]-4\left[\begin{array}{lll} 1 & 2 & 2 \\\\ 2 & 1 & 2 \\\\ 2 & 2 & 1 \end{array}\right] \\\\ & =\left[\begin{array}{lll} 5 & 0 & 0 \\\\ 0 & 5 & 0 \\\\ 0 & 0 & 5 \end{array}\right]=5 I \end{aligned} $

$ \begin{aligned} & \text { Given, } A=\left[\begin{array}{lll} 9 & 3 & 0 \\\\ 1 & 5 & 8 \\\\ 7 & 6 & 2 \end{array}\right] \\\\ & \text { and } A A^T-A^2=\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \end{array}\right] \end{aligned} $

Now,

$ A^2=\left[\begin{array}{lll} 9 & 3 & 0 \\\\ 1 & 5 & 8 \\\\ 7 & 6 & 2 \end{array}\right]\left[\begin{array}{lll} 9 & 3 & 0 \\\\ 1 & 5 & 8 \\\\ 7 & 6 & 2 \end{array}\right]=\left[\begin{array}{lll} 84 & 42 & 24 \\\\ 70 & 76 & 56 \\\\ 83 & 63 & 52 \end{array}\right] $

$ \begin{aligned} & A A^T-A^2=\left[\begin{array}{lll} 9 & 3 & 0 \\\\ 1 & 5 & 8 \\\\ 7 & 6 & 2 \end{array}\right]\left[\begin{array}{lll} 9 & 1 & 7 \\\\ 3 & 5 & 6 \\\\ 0 & 8 & 2 \end{array}\right] \\\\ & -\left[\begin{array}{lll} 84 & 42 & 24 \\\\ 70 & 76 & 56 \\\\ 83 & 63 & 52 \end{array}\right] \\\\ & =\left[\begin{array}{lll} 90 & 24 & 81 \\\\ 24 & 90 & 53 \\\\ 81 & 53 & 89 \end{array}\right]-\left[\begin{array}{lll} 84 & 42 & 24 \\\\ 70 & 76 & 56 \\\\ 83 & 63 & 52 \end{array}\right] \\\\ & =\left[\begin{array}{ccc} 6 & -18 & 57 \\\\ -46 & 14 & -3 \\\\ -2 & -10 & 37 \end{array}\right] \\\\ & \sum a_{i j}=6-18+57-46+14 \\ & 1 \leq i \leq 3 \\ & 1 \leq j \leq 3 \end{aligned} $

Given,

$ \begin{array}{l} \Delta_{1}=\left|\begin{array}{lll} 1 & a^{2} & b c \\\\ 1 & b^{2} & c a \\\\ 1 & c^{2} & a b \end{array}\right| \\\\ R_{2} \rightarrow R_{2}-R_{1} \text { and } R_{3} \rightarrow R_{3}-R_{1} \\\\ \Delta_{1}=\left|\begin{array}{ccc} 1 & a^{2} & b c \\\\ 0 & b^{2}-a^{2} & c(a-b) \\\\ 0 & c^{2}-a^{2} & b(a-c) \end{array}\right| \\\\ =(a-b)(c-a)\left|\begin{array}{ccc} 1 & a^{2} & b c \\\\ 0 & -(a+b) & c \\\\ 0 & c+a & -b \end{array}\right| \\\\ =(a-b)(c-a)\left(a b+b^{2}-c^{2}-a c\right) \\\\ =(a-b)(c-a)(a(b-c)+(b-c)(b+c) \\\\ \Delta_{1}=(a-b)(b-c)(c-a)(a+b+c) \ldots \\\\ \text { and } \Delta_{2}=\left|\begin{array}{ccc} 1 & 1 & 1 \\\\ a^{2} & b^{2} & c^{2} \\\\ a^{3} & b^{3} & c^{3} \end{array}\right| \\\\ C_{2} \rightarrow C_{2}-C_{1} \text { and } C_{3} \rightarrow C_{3}-C_{1} \\\\ =\left|\begin{array}{ccc} 1 & 0 & 0 \\\\ a^{2} & b^{2}-a^{2} & c^{2}-a^{2} \\\\ a^{3} & b^{3}-a^{3} & c^{3}-a^{3} \end{array}\right| \\\\ =-(a-b)(c-a) \\\\ \left|\begin{array}{ccc} 1 & 0 & 0 \\\\ a^{2} & a+b & c+a \\\\ a^{3} & \left(a^{2}+b^{2}+a b\right) & c^{2}+a^{2}+a c \end{array}\right| \\\\ C_{2} \rightarrow C_{2}-C_{3} \\\\ =-(a-b)(c-a) \\\\ \left|\begin{array}{ccc} 1 & 0 & 0 \\\\ a^{2} & b-c & c+a \\\\ a^{3} & \left(b^{2}-c^{2}\right)+a(b-c) & c^{2}+a^{2}+a c \end{array}\right| \\\\ =-(a-b)(b-c)(c-a) \\\\ \left|\begin{array}{ccc} 1 & 0 & 0 \\\\ a^{2} & 1 & c+a \\\\ a^{3} & b+c+a & c^{2}+a^{2}+a c \end{array}\right| \end{array} $

$ \begin{array}{l} =-(a-b)(b-c)(c-a) \\\\ {\left[c^{2}+a^{2}+a c-b c-c^{2}-a c-a b-a c-a^{2}\right]} \\\\ =-(a-b)(b-c)(c-a)[-b c-a b-a c] \\\\ \Delta_{2}=(a-b)(b-c)(c-a)[a b+b c+c a] \end{array} $

From Ist and 2nd equation replace $ \Delta_{1} $ and $ \Delta_{2} $

$ \begin{array}{l} \frac{\Delta_{1}}{\Delta_{2}}=\frac{6}{11} \Rightarrow \frac{a+b+c}{a b+b c+c a}=\frac{6}{11} \\\\ \therefore 11(a+b+c)=6(a b+b c+c a) \end{array} $

Given, $ x+3 y+7=0 $

$ \begin{array}{c} 3 x+10 y-3 z+18=0 \\\\ 3 y-9 z+2=0 \\\\ D=\left|\begin{array}{ccc} 1 & 3 & 0 \\\\ 3 & 10 & -3 \\\\ 0 & 3 & -9 \end{array}\right|=0 \\\\ D_{x}=\left|\begin{array}{ccc} -7 & 3 & 0 \\\\ -18 & 10 & -3 \\\\ -2 & 3 & -9 \end{array}\right|=99 \\\\ D_{y}=\left|\begin{array}{ccc} 1 & -7 & 0 \\\\ 3 & -18 & -3 \\\\ 0 & -2 & -9 \end{array}\right|=-33 \\\\ D_{z}=\left|\begin{array}{ccc} 1 & 3 & -7 \\\\ 3 & 10 & -18 \\\\ 0 & 3 & -2 \end{array}\right|=-11 \end{array} $

As $ D=0 $, and none $ D_{x}, D_{y} $ and $ D_{z} $ is zero, the given system of equation has no solution.

We have, $\alpha, \beta, \gamma$ are the roots of the equation $\left|\begin{array}{ccc}x & 2 & 2 \\\\ 2 & x & 2 \\\\ 2 & 2 & x\end{array}\right|=0$ and

$ \min (\alpha, \beta, \gamma)=\alpha $

Now, $\quad\left|\begin{array}{lll}x & 2 & 2 \\\\ 2 & x & 2 \\\\ 2 & 2 & x\end{array}\right|=0$

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

$\begin{gathered}{\left[C_1 \rightarrow C_1-C_3 \text { and } C_2 \rightarrow C_2-C_3\right]} \\\\ \Rightarrow \quad(x-2)^2\left|\begin{array}{ccc}1 & 0 & 2 \\\\ 0 & 1 & 2 \\\\ -1 & -1 & x\end{array}\right|=0 \\\\ {\left[C_1 \rightarrow \frac{1}{x-2} C_1 \text { and } C_2 \rightarrow \frac{1}{x-2} C_2\right]} \\\\ \Rightarrow(x-2)^2[1(x+2)-0+(-1)(0-2)]=0 \\\\ \Rightarrow(x-2)^2(x+4)=0 \Rightarrow x=-4,2,2 \\\\ \therefore \quad \alpha=-4, \beta=2 \text { and } \gamma=2 \\\\ \text { Now, } 2 \alpha+3 \beta+4 \gamma \\\\ =2(-4)+3(2)+4(2) \\\\ =-8+6+8=6\end{gathered}$

$ \begin{aligned} &\text { We have, }\\\\ &\begin{aligned} & \begin{array}{l} A=\left[\begin{array}{lll} 1 & 2 & 2 \\\\ 3 & 2 & 3 \\\\ 1 & 1 & 2 \end{array}\right] \text { and } A^{-1}=\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \end{array}\right] \\\\ |A|=1(4-3)-2(6-3)+2(3-2)=1-6+2=-3\begin{array}{l} \end{array} \\\\ C_{11}=\left|\begin{array}{ll} 2 & 3 \\\\ 1 & 2 \end{array}\right|=1 \\\\ C_{12}=-\left|\begin{array}{ll} 3 & 3 \\\\ 1 & 2 \end{array}\right|=-3 \\\\ C_{13}=\left|\begin{array}{ll} 3 & 2 \\\\ 1 & 1 \end{array}\right|=1 \\\\ C_{21}=-\left|\begin{array}{ll} 2 & 2 \\\\ 1 & 2 \end{array}\right|=-2 \\\\ C_{22}=\left|\begin{array}{ll} 1 & 2 \\\\ 1 & 2 \end{array}\right|=0 \\\\ C_{23}=-\left|\begin{array}{ll} 1 & 2 \\\\ 1 & 1 \end{array}\right|=1 \end{array} \end{aligned} \end{aligned} $

$ \begin{aligned} & C_{31}=\left|\begin{array}{ll} 2 & 2 \\\\ 2 & 3 \end{array}\right|=2, C_{32}=-\left|\begin{array}{ll} 1 & 2 \\\\ 3 & 3 \end{array}\right|=3 \\\\ & C_{33}=\left|\begin{array}{ll} 1 & 2 \\\\ 3 & 2 \end{array}\right|=-4 \\\\ & \operatorname{adj} A=\left[\begin{array}{ccc} 1 & -3 & 1 \\\\ -2 & 0 & 1 \\\\ 2 & 3 & -4 \end{array}\right]^T=\left[\begin{array}{ccc} 1 & -2 & 2 \\\\ -3 & 0 & 3 \\\\ 1 & 1 & -4 \end{array}\right] \\\\ & \therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{-3}\left[\begin{array}{ccc} 1 & -2 & 2 \\\\ -3 & 0 & 3 \\\\ 1 & 1 & -4 \end{array}\right] \end{aligned} $

Now,

$ \sum_{\substack{1 \leq i \leq 3 \\\\ 1 \leq j \leq 3}} a_{i j}=\left(-\frac{1}{3}\right) $

$ \begin{aligned} & {[1+(-2)+2+(-3)+0+3+1+1+(-4)]} \\\\ & \quad=\left(-\frac{1}{3}\right)(-1)=\frac{1}{3} \end{aligned} $

We have,

$A X=D$, where

$ \begin{aligned} A & =\left[\begin{array}{ccc} 3 & -4 & 7 \\\\ 5 & 2 & -4 \\\\ 8 & -6 & -1 \end{array}\right], X=\left[\begin{array}{l} x \\\\ y \\\\ z \end{array}\right] \text { and } D=\left[\begin{array}{c} -6 \\\\ -9 \\\\ -5 \end{array}\right] \\\\ |A| & =3(-2-24)+4(-5+32)+7(-30-16) \\\\ & =-78+108-322 \\\\ & =-292 \neq 0 \end{aligned} $

$\therefore$ Rank $(A)=$ Order of $A=3$

$ A D=\left[\begin{array}{ccc} 3 & -4 & 7 \\\\ 5 & 2 & -4 \\\\ 8 & -6 & -1 \end{array}\right]\left[\begin{array}{l} -6 \\\\ -9 \\\\ -5 \end{array}\right]=\left[\begin{array}{c} -17 \\\\ -28 \\\\ 11 \end{array}\right] $

Clearly, Rank $([A D])=3$

Hence, Rank $(A)=\operatorname{Rank}([A D])=3$

We have,

$(x, y, z)=(\alpha, \beta, \gamma)$ is a unique solution of the system of simultaneous linear equations

$ \begin{aligned} 3 x-4 y+z & =-7 \\\\ 2 x+3 y-z & =10 \\\\ x-2 y-3 z & =3 \end{aligned} $

so, by cramer's rule,

$ \begin{aligned} & x=\alpha=\frac{\left|\begin{array}{ccc} -7 & -4 & 1 \\\\ 10 & 3 & -1 \\\\ 3 & -2 & -3 \end{array}\right|}{\left|\begin{array}{ccc} 3 & -4 & 1 \\\\ 2 & 3 & -1 \\\\ 1 & -2 & -3 \end{array}\right|} \\\\ & =\frac{-7(-9-2)+4(-30+3)+(-20-9)}{3(-9-2)+4(-6+1)+(-4-3)} \\\\ & =\frac{77-108-29}{-33-20-7} \\\\ & =\frac{-60}{-60}=1 \end{aligned} $

Given,

$\alpha, \beta, \gamma$ are the roots of the equation

$ \begin{gathered} 2 x^3-5 x^2+4 x-3=0 \\\\ \therefore \alpha+\beta+\gamma=\frac{-(-5)}{2}=\frac{5}{2} \quad \ldots \text { (i) } \\\\ \alpha \beta+\beta \gamma+\gamma \alpha=\frac{4}{2}=2 \quad \ldots \text { (ii) } \\\\ \alpha \beta \gamma=\frac{-(-3)}{2}=\frac{3}{2} \quad \ldots \text { (iii) } \end{gathered} $

Now, $\Sigma \alpha \beta(\alpha+\beta)$

$ \begin{aligned} & =\alpha \beta(\alpha+\beta)+\beta \gamma(\beta+\gamma)+\gamma \alpha(\alpha+\gamma) \\\\ & =\alpha^2 \beta+\alpha \beta^2+\beta^2 \gamma+\beta \gamma^2+\alpha^2 \gamma+\alpha \gamma^2 \\\\ & =(\alpha+\beta+\gamma)(\alpha \beta+\beta \gamma+\gamma \alpha)-3 \alpha \beta \gamma \\\\ & =\left(\frac{5}{2}\right)(2)-3\left(\frac{3}{2}\right) \end{aligned} $

[using Eqs. (i), (ii) and (iii)]

$ =\frac{10}{2}-\frac{9}{2}=\frac{1}{2} $

$\because A+B$ is symmecric

$ \begin{aligned} & \Rightarrow \quad(A+B)^T=A+B \\\\ & \Rightarrow \quad A^T+B^T=A+B \quad \ldots \text { (i) } \end{aligned} $

and $A-B$ is skew-symmetric

$ \begin{aligned} \Rightarrow \quad\left(A-A^T\right. & =-(A-B) \\\\ A^T-B^T & =-A+B \quad \ldots \text { (ii) } \end{aligned} $

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

$ B=A^T $

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

Given $D$ is the transpose of $C$

$ \begin{aligned} & \Rightarrow \quad=C^r=\left[\begin{array}{ccc} 0 & 1 & -2 \\\\ 2 & -1 & 0 \\\\ 0 & 2 & 1 \end{array}\right]^T \\\\ & D=\left[\begin{array}{ccc} 0 & 2 & 0 \\\\ 1 & -1 & 2 \\\\ -2 & 0 & 1 \end{array}\right] \\\\ & 50, B+D= {\left[\begin{array}{ccc} -1 & 4 & 3 \\\\ 2 & 3 & -4 \\\\ 3 & -2 & 5 \end{array}\right]+\left[\begin{array}{ccc} 0 & 2 & 0 \\\\ 1 & -1 & 2 \\\\ -2 & 0 & 1 \end{array}\right] } \\\\ &= {\left[\begin{array}{ccc} -1 & 6 & 3 \\\\ 3 & 2 & -2 \\\\ 1 & -2 & 6 \end{array}\right] } \end{aligned} $

We have, $A^2+I=2 A \Rightarrow A^2=2 A-I$

$ \begin{aligned} & \because A^9=A^2 \cdot A^2 A^2 A^2 A \\\\ & =(2 A-I)(2 A-I)(2 A-I)(2 A-I) A \\\\ & \qquad \begin{aligned} \Rightarrow A^9= & \left(4 A^2+I-4 A\right)\left(4 A^2+I-4 A\right) A \\\\ & =(4 A-3 I)(4 A-3 I) A \\\\ & =\left(16 A^2+9 I-24 A\right) A \\\\ & =(8 A-7 I) A \\\\ & =8 A^2-7 A \\\\ & =16 A-8 I-7 A \\\\ & =9 A-8 I \end{aligned} \end{aligned} $

Let $\Delta=\left[\begin{array}{ccc}\frac{a^2+b^2}{c} & c & c \\\\ a & \frac{b^2+c^2}{a} & a \\\\ b & b & \frac{c^2+a^2}{b}\end{array}\right]$

On multiply $R_1, R_2$ and $R_3$ by $c, a$ and $b$, respectively

$ \Rightarrow \Delta=\frac{1}{a b c}\left[\begin{array}{ccc} a^2+b^2 & c^2 & c^2 \\\\ a^2 & b^2+c^2 & a^2 \\\\ b^2 & b^2 & c^2+a^2 \end{array}\right] $

Apply $R_1 \rightarrow R_1-\left(R_2+R_3\right)$

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

Apply $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$, we get

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

We have,

$ \begin{aligned} & x-2 y+3 z=4 \\\\ & 3 x+y-2 z=7 \\\\ & 2 x+3 y+z=6 \end{aligned} $

Here,

$ \begin{aligned} & D=\left|\begin{array}{ccc} 1 & -2 & 3 \\\\ 3 & 1 & -2 \\\\ 2 & 3 & 1 \end{array}\right| \\\\ &=1(1+6)+2(3+4)+3(9-2)=42 \\\\ & D_x=\left|\begin{array}{ccc} 4 & -2 & 3 \\\\ 7 & 1 & -2 \\\\ 6 & 3 & 1 \end{array}\right| \\\\ &=4(1+6)+2(7+12)+3(21-6)=111 \\\\ & D_y=\left|\begin{array}{ccc} 1 & 4 & 3 \\\\ 3 & 7 & -2 \\\\ 2 & 6 & 1 \end{array}\right|=3 \\\\ & D_z=\left|\begin{array}{ccc} 1 & -2 & 4 \\\\ 3 & 1 & 7 \\\\ 2 & 3 & 6 \end{array}\right|=21 \end{aligned} $

So, system of linear equations has unique solution and $z=\frac{D_z}{D}=\frac{21}{42}=\frac{1}{2}$

Given, $X_{4 \times 3}, Y_{4 \times 3}$ and $P_{2 \times 3}$ are the matrices.

$ \text { } \begin{aligned} Since, & {\left[P_{2 \times 3}\left(X_{3 \times 4}^T Y_{4 \times 3}\right)^{-1} P_{3 \times 2}^T\right]^T } \\ &=\left[P_{2 \times 3}\left(X^T Y\right)_{3 \times 3}^{-1} P_{3 \times 2}^T\right]^T \\ &=\left(P\left(X^T Y\right)^{-1} P^T\right)_{2 \times 2}^T \end{aligned} $

$ \begin{aligned} & \text { Here, } A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 5 \end{array}\right] \\ & A^2=\left[\begin{array}{ll} 1 & 2 \\ 3 & 5 \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 3 & 5 \end{array}\right]=\left[\begin{array}{ll} 7 & 12 \\ 18 & 31 \end{array}\right] \\ & \alpha A^2-\beta A=2 I \\ & \Rightarrow \alpha\left[\begin{array}{ll} 7 & 12 \\ 18 & 31 \end{array}\right]-\beta\left[\begin{array}{ll} 1 & 2 \\ 3 & 5 \end{array}\right]=2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & \Rightarrow \quad 7 \alpha-\beta=2 \\ & \,\,\,\,\,\,\, \quad 18 \alpha-3 \beta=0 \\ & \Rightarrow \quad \beta=6 \alpha \\\Rightarrow & \alpha=2, \beta=12 \\ & \alpha^2+\beta=4+12=16 \end{aligned} $

At $\alpha=-2$, determinant becomes

$ \left|\begin{array}{ccc} 1 & 9 & 25 \\ 0 & 4 & 16 \\ 1 & 1 & 9 \end{array}\right|=k $

On expanding with $C_1$,

$ \begin{aligned} (36-16)+0+1(144-100) & =k \\\Rightarrow && k =64 \end{aligned} $

Given, the system of equations,

$ x+y+z=5, x+2 y+2 z=6 $

and  $ x+3 y+\lambda z=\mu $

Since, equation is solvable, then $D \neq 0$

$ \begin{aligned} & \Rightarrow \quad\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 3 & \lambda \end{array}\right| \neq 0 \\ & \Rightarrow(2 \lambda-6)-1(\lambda-2)+1(3-2) \neq 0 \\ & \Rightarrow \lambda \neq 3, \mu \in R \end{aligned} $

As we know,

If $n$ is the order of a square matrix.

Then, $|\operatorname{adj}(\operatorname{adj} A)|=|A|^{(n-1)^2}$

So, $\left|\operatorname{adj}\left(\operatorname{adj} A^2\right)\right|=\left|A^2\right|^{(n-1)^2}$

∵ Order of $A=n=3$

$ \text { } \begin{aligned}Then, \left|\operatorname{adj}\left(\operatorname{adj} A^2\right)\right| & =\left|A^2\right|^{(3-1)^2} \\ & =\left|A^2\right|^4=|A|^8 \end{aligned} $