Matrices and Determinants

(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} $

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

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

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

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

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

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

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

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

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

$\begin{aligned} & \left|\operatorname{adj}\left(A-2 A^T\right) \cdot \operatorname{adj}\left(2 A-A^T\right)\right|=2^8 \\ & P=A-2 A^{\top} \\ & Q=2 A^T-A \Rightarrow Q^T=2 A^T-A=-P \\ & |\operatorname{adj}(P) \operatorname{adj}(Q)| \Rightarrow|P Q|=-2^4 \\ & \Rightarrow|P|(-|P|)=-2^4 \Rightarrow|P|=4 \text { and }|Q|=-4 \\ & \left|A-2 A^T\right|=4 \\ & A-2 A^T=\left[\begin{array}{lll} 1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{array}\right]-2\left[\begin{array}{lll} 1 & 1 & 0 \\ 2 & 0 & 1 \\ \alpha & 1 & 2 \end{array}\right]=\left[\begin{array}{ccc} -1 & 0 & \alpha \\ -3 & 0 & -1 \\ -2 \alpha & -1 & -2 \end{array}\right] \\ & \Rightarrow\left|A-2 A^T\right|=1+3 \alpha=4 \Rightarrow \alpha=1 \Rightarrow|A|=-4 \\ & \Rightarrow|A|^2=16 \end{aligned}$

$\begin{aligned} & x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\ & x+(\cos \alpha) y+(\sin \alpha) z=0 \\ & x+(\sin \alpha) y-(\cos \alpha) z=0 \end{aligned}$

$\because$ Non-trivial solution

$\Rightarrow D=0$

$\begin{aligned} & \left|\begin{array}{ccc} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha \end{array}\right|=0 \\ & 1{\left[ { - {{\cos }^2}\alpha - \sin \alpha } \right]^{ - 1}}\left[ { - \sqrt 2 \sin \alpha \cos \alpha - \sqrt 2 \sin \alpha \cos \alpha } \right] + 1\left[ {\sqrt 2 {{\sin }^2}\alpha - \sqrt 2 {{\cos }^2}\alpha } \right] = 0 \\ \end{aligned}$

$\begin{aligned} & -1+2 \sqrt{2} \sin \alpha \cos \alpha+\sqrt{2}\left(\sin ^2 \alpha-\cos ^2 \alpha\right)=0 \\ & \sqrt{2} \sin 2 \alpha-\sqrt{2} \cos 2 \alpha=1 \\ & \frac{\sin 2 \alpha}{\sqrt{2}}-\frac{\cos 2 \alpha}{\sqrt{2}}=\frac{1}{2} \\ & \sin \left(2 \alpha-\frac{\pi}{4}\right)=\sin \frac{\pi}{6} \\ & \Rightarrow 2 \alpha-\frac{\pi}{4}=n \pi+(-1)^n \frac{\pi}{6} \text { for } n=0 \\ & \Rightarrow \alpha=\frac{5 \pi}{24} \end{aligned}$

The solution involves understanding matrix operations and properties such as multiplication, transpose, and determinant. Given $\mathrm{A}$, $\mathrm{B}$, and that $\mathrm{C} = \mathrm{ABA}^{\mathrm{T}}$, and $\mathrm{X} = \mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{A}$, we find $\operatorname{det} \mathrm{X}$ as follows:

First, we express $|\mathrm{C}|$, the determinant of $\mathrm{C}$, in terms of the determinants of $\mathrm{A}$ and $\mathrm{B}$, using the property that $|\mathrm{ABC}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{C}|$:

$\begin{aligned} |\mathrm{C}| &= |\mathrm{ABA}^{\mathrm{T}}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{A}^{\mathrm{T}}| \\\\ &= |\mathrm{A}|^2 \cdot |\mathrm{B}| \quad \text{since } |\mathrm{A}^{\mathrm{T}}| = |\mathrm{A}|. \end{aligned}$

Next, we find $|\mathrm{X}|$, using the property $|\mathrm{ABC}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{C}|$ and substituting $|\mathrm{C}|$:

$\begin{aligned} |\mathrm{X}| &= |\mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{A}| = |\mathrm{A}^{\mathrm{T}}| \cdot |\mathrm{C}|^2 \cdot |\mathrm{A}| \\\\ &= |\mathrm{A}|^2 \cdot |\mathrm{C}|^2. \end{aligned}$

Substituting the expression for $|\mathrm{C}|$ obtained earlier:

$\begin{aligned} |\mathrm{X}| &= |\mathrm{A}|^2 \cdot (|\mathrm{A}|^2 \cdot |\mathrm{B}|)^2 \\\\ &= (|\mathrm{A}|^3 \cdot |\mathrm{B}|)^2. \end{aligned}$

The determinants of $\mathrm{A}$ and $\mathrm{B}$ are calculated as:

Finally, substituting these values into our expression for $|\mathrm{X}|$:

$\begin{aligned} |\mathrm{X}| &= (3^3 \cdot 1)^2 \\\\ &= 729. \end{aligned}$

Therefore, $\operatorname{det} \mathrm{X} = 729$.

$\text { Let } A=\left[\begin{array}{lll} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{array}\right]$

$\text { Given } A\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \\ 2 \end{array}\right] \quad \text{ ..... (1)}$

$\therefore\left[\begin{array}{l} \mathrm{x}_1+\mathrm{z}_1 \\ \mathrm{x}_2+\mathrm{z}_2 \\ \mathrm{x}_3+\mathrm{z}_3 \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \\ 2 \end{array}\right]$

$\begin{aligned} \therefore \mathrm{x}_1+\mathrm{z}_1 & =2 \quad \text{.... (2)}\\ \mathrm{x}_2+\mathrm{z}_2 & =0 \quad \text{.... (3)}\\ \mathrm{x}_3+\mathrm{z}_3 & =0 \quad \text{.... (4)} \end{aligned}$

$\begin{aligned} & \text { Given } A\left[\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} -4 \\ 0 \\ 4 \end{array}\right] \\ & \therefore\left[\begin{array}{l} -\mathrm{x}_1+\mathrm{z}_1 \\ -\mathrm{x}_2+\mathrm{z}_2 \\ -\mathrm{x}_3+\mathrm{z}_3 \end{array}\right]=\left[\begin{array}{l} 4 \\ 0 \\ 4 \end{array}\right] \end{aligned}$

$\begin{array}{r} \Rightarrow-\mathrm{x}_1+\mathrm{z}_1=-4 \quad \text{... (5)}\\ -\mathrm{x}_2+\mathrm{x}_2=0 \quad \text{... (6)}\\ -\mathrm{x}_3+\mathrm{z}_3=4 \end{array}$

$\begin{aligned} & \text { Given } A\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]=\left[\begin{array}{l} 0 \\ 2 \\ 0 \end{array}\right] \\ & \therefore\left[\begin{array}{l} \mathrm{y}_1 \\ \mathrm{y}_2 \\ \mathrm{y}_3 \end{array}\right]=\left[\begin{array}{l} 0 \\ 2 \\ 0 \end{array}\right] \end{aligned}$

$\begin{aligned} & \therefore \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 \\ & \therefore \text { from }(2),(3),(4),(5),(6) \text { and }(7) \\ & \mathrm{x}_1=3 \mathrm{x}, \mathrm{x}_2=0, \mathrm{x}_3=-1 \\ & \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 \\ & \mathrm{z}_1=-1, \mathrm{z}_2=0, \mathrm{z}_3=3 \end{aligned}$

$\begin{gathered} \therefore A=\left[\begin{array}{ccc} 3 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 3 \end{array}\right] \\ \therefore \text { Now }(A-31)\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} -1 \\ 2 \\ 3 \end{array}\right] \\ \therefore\left[\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} -1 \\ 2 \\ 3 \end{array}\right] \\ {\left[\begin{array}{c} -z \\ -y \\ -x \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]} \\ {[z=-1],[y=-2],[x=-3]} \end{gathered}$

$\begin{aligned} & D=\left|\begin{array}{ccc} 1 & -2 & 1 \\ 2 & \alpha & 3 \\ 3 & -1 & \beta \end{array}\right| \\ & =1(\alpha \beta+3)+2(2 \beta-9)+1(-2-3 \alpha) \\ & =\alpha \beta+3+4 \beta-18-2-3 \alpha \end{aligned}$

For infinite solutions $\mathrm{D}=0, \mathrm{D}_1=0, \mathrm{D}_2=0$ and

$\begin{aligned} & \mathrm{D}_3=0 \\ & \mathrm{D}=0 \end{aligned}$

$\alpha \beta-3 \alpha+4 \beta=17$ ..... (1)

$\begin{aligned} & \mathrm{D}_1=\left|\begin{array}{ccc} -4 & -2 & 1 \\ 5 & \alpha & 3 \\ 3 & -1 & \beta \end{array}\right|=0 \\ & \mathrm{D}_2=\left|\begin{array}{ccc} 1 & -4 & 1 \\ 2 & 5 & 3 \\ 3 & 3 & \beta \end{array}\right|=0 \\ & \Rightarrow 1(5 \beta-9)+4(2 \beta-9)+1(6-15)=0 \\ & 13 \beta-9-36-9=0 \end{aligned}$

$13 \beta=54, \beta=\frac{54}{13}$ put in (1)

$\frac{54}{13} \alpha-3 \alpha+4\left(\frac{54}{13}\right)=17$

$54 \alpha-39 \alpha+216=221$

$15 \alpha=5 \quad \alpha=\frac{1}{3}$

Now, $12 \alpha+13 \beta=12 \cdot \frac{1}{3}+13 \cdot \frac{54}{13}$

$=4+54=58$

$\begin{aligned} & x \sin \theta=y \sin \left(\theta+\frac{2 \pi}{3}\right)=z \sin \left(\theta+\frac{4 \pi}{3}\right) \neq 0 \\ & \Rightarrow x, y, z \neq 0 \end{aligned}$

Also,

$\begin{aligned} & \sin \theta+\sin \left(\theta+\frac{2 \pi}{3}\right)+\sin \left(\theta+\frac{4 \pi}{3}\right)=0 \forall \theta \in \mathrm{R} \\ & \Rightarrow \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}=0 \\ & \Rightarrow \mathrm{xy}+\mathrm{yz}+\mathrm{zx}=0 \end{aligned}$

(i) $\quad \operatorname{Trace}(\mathrm{R})=\mathrm{x}+\mathrm{y}+\mathrm{z}$

If $x+y+z=0$ and $x y+y z+z x=0$

$\Rightarrow \mathrm{x}=\mathrm{y}=\mathrm{z}=0$

Statement (i) is False

(ii) $\quad \operatorname{Adj}(\operatorname{Adj}(\mathrm{R}))=|\mathrm{R}| \mathrm{R}$

Trace $(\operatorname{Adj}(\operatorname{Adj}(\mathrm{R})))$

$=x y z(x+y+z) \neq 0$

Statement (ii) is also False

$\begin{aligned} & \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & \lambda^2 \\ 1 & 3 & \lambda \end{array}\right|=0 \\ & \Rightarrow 2 \lambda^2-\lambda-1=0 \\ & \lambda=1,-\frac{1}{2} \\ & \left|\begin{array}{ccc} 1 & 1 & 5 \\ 2 & \lambda^2 & 9 \\ 3 & \lambda & \mu \end{array}\right|=0 \Rightarrow \mu=13 \end{aligned}$

Infinite solution $\lambda=1 \& \mu=13$

For unique $\operatorname{sol}^{\mathrm{n}} \lambda \neq 1$

For no $\operatorname{sol}^{\mathrm{n}} \lambda=1 \& \mu \neq 13$

If $\lambda \neq 1$ and $\mu \neq 13$

Considering the case when $\lambda=-\frac{1}{2}$ and $\mu \neq 13$ this will generate no solution case

$x+y+z=4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda{ }^2 z=\mu^2+15$,

$\Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 2 \lambda \\ 1 & 3 & 4 \lambda^2 \end{array}\right|=(2 \lambda-1)^2$

For unique solution $\Delta \neq 0,2 \lambda-1 \neq 0,\left(\lambda \neq \frac{1}{2}\right)$

Let $\Delta=0, \lambda=\frac{1}{2}$

$\begin{aligned} & \Delta_y=0, \Delta_x=\Delta_z=\left|\begin{array}{ccc} 4 \mu & 1 & 1 \\ 10 \mu & 2 & 1 \\ \mu^2+15 & 3 & 1 \end{array}\right| \\ & =(\mu-15)(\mu-1) \end{aligned}$

For infinite solution $\lambda=\frac{1}{2}, \mu=1$ or 15

$\begin{aligned} \left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{I}\right| & =\left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{P}^{-1} \mathrm{P}\right| \\ & =\left|\mathrm{P}^{-1}(\mathrm{~A}-2 \mathrm{I}) \mathrm{P}\right| \\ & =\left|\mathrm{P}^{-1}\right||\mathrm{A}-2 \mathrm{I}||\mathrm{P}| \\ & =|\mathrm{A}-2 \mathrm{I}| \\ & =\left|\begin{array}{ccc} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{array}\right|=69 \end{aligned}$

So, Prime factor of 69 is 3 & 23

So, sum = 26

$\begin{aligned} & |\mathrm{A}|=\alpha^2-\beta^2 \\ & |2 \mathrm{~A}|^3=2^{21} \Rightarrow|\mathrm{A}|=2^4 \\ & \alpha^2-\beta^2=16 \\ & (\alpha+\beta)(\alpha-\beta)=16 \Rightarrow \alpha=4 \text { or } 5 \end{aligned}$

$\mathrm{AA}^{\mathrm{T}}=\mathrm{I}=\mathrm{A}^{\mathrm{T}} \mathrm{A}$

On solving given expression, we get

$\begin{aligned} & \frac{1}{2} \mathrm{~A}\left[\mathrm{~A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2+2 \mathrm{~A} \mathrm{~A}^{\mathrm{T}}+\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2-2 \mathrm{~A} \mathrm{~A}^{\mathrm{T}}\right] \\ & =\mathrm{A}\left[\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2\right]=\mathrm{A}^3+\mathrm{A}^{\mathrm{T}} \end{aligned}$

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

$\begin{aligned} & \Rightarrow(2 \alpha+3)\left\{\frac{7 \alpha}{6}\right\}-(3 \alpha+1)\left\{\frac{-7}{6}\right\}=0 \\ & \Rightarrow(2 \alpha+3) \cdot \frac{7 \alpha}{6}+(3 \alpha+1) \cdot \frac{7}{6}=0 \\ & \Rightarrow 2 \alpha^2+3 \alpha+3 \alpha+1=0 \\ & \Rightarrow 2 \alpha^2+6 \alpha+1=0 \\ & \Rightarrow \alpha=\frac{-3+\sqrt{7}}{2}, \frac{-3-\sqrt{7}}{2} \end{aligned}$

Hence option (C) is correct.

$\begin{aligned} & f(-x)=\left[\begin{array}{ccc} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & f(x) \cdot f(-x)=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I \end{aligned}$

Hence statement- I is correct

Now, checking statement II

$\begin{aligned} & f(y)=\left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & f(x) \cdot f(y)=\left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \Rightarrow f(x) \cdot f(y)=f(x+y) \end{aligned}$

Hence statement-II is also correct.

$\begin{aligned} \text{(A)}\quad & \mathrm{ax}^2+2 \mathrm{bxy}+\mathrm{cy}^2>0 \forall(\mathrm{x}, \mathrm{y}) \in \mathbb{R}^2-\{(0,0)\} \\ & \Rightarrow \mathrm{ax}+2 \mathrm{bxy}+\mathrm{cy}^2 \text { must represent pair of imaginary lines and } \mathrm{a}, \mathrm{c}>0 . \\ & \Rightarrow \mathrm{b}^2<\mathrm{ac} \end{aligned}$

$\mathrm{(B)}\quad\mathrm{b}^2<3 \times \frac{1}{12} \Rightarrow|2 \mathrm{~b}|<1$

$\mathrm{(C)}\quad$ since $\mathrm{b}^2 \neq \mathrm{ac}$

$\Rightarrow \mathrm{ax}+\mathrm{by}=1 \text { and } \mathrm{bx}+\mathrm{cy}=-1$

are not parallel lines.

$\mathrm{(D)}\quad\mathrm{ac}+\mathrm{a}+\mathrm{c}>\mathrm{b}^2 \Rightarrow$ lines are not parallel.

$\Rightarrow$ Options (B), (C), (D) are Correct.

$\alpha, \beta$ are roots of $x^2+x-1=0$

$\begin{aligned} & \therefore \alpha+\beta=-1 \Rightarrow 1+\alpha+\beta=0 \\ & M=\left[\begin{array}{lll} \mathrm{a}_{11} & \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{21} & \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{31} & \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right] \end{aligned}$

(P) $\quad M=\left[\begin{array}{lll}1 & \alpha & \beta \\ \alpha & \beta & 1 \\ \beta & 1 & \alpha\end{array}\right] \Rightarrow 3 ! \times 2=12$

For one arrangement of row 1 we can arrange other two rows exactly in two ways and row 1 can be arranged in 3! ways

$\therefore 3!\times 2=12 \text { ways }$

(Q) $\quad M=\left[\begin{array}{lll}x & a & b \\ a & y & c \\ b & c & z\end{array}\right] \Rightarrow$ Consider one such arrangement with $a=\alpha, b=\beta, c=1$

a, b, c can be arranged in 3! ways and corresponding entries can be arranged in 1 way.

(R)

$\begin{aligned} & {\left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} a \\ 0 \\ -c \end{array}\right]} \\ & a y+b z=a \\ & -a x+c z=0 \\ & -b x-c y=-c \end{aligned}$

It is observed that $\mathrm{D}=\mathrm{D}_{\mathrm{x}}=\mathrm{D}_{\mathrm{y}}=\mathrm{D}_{\mathrm{z}}=0$

$\therefore$ infinite solution

(S)

$\begin{aligned} & {\left[\begin{array}{lll} 1 & \alpha & \beta \\ \beta & \alpha & 1 \\ \alpha & 1 & \beta \end{array}\right] } \\ & \Rightarrow \alpha \beta-1-\alpha \beta^2+\alpha^2+\beta^2-\alpha^2 \beta=0 \quad(\text { since } \alpha \beta=\alpha+\beta=-1) \end{aligned}$