Matrices and Determinants

$ \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} $

We have,

$A$ and $B$ are two square matrices

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

If $A$ is symmetric matrix and $B$ is symmetric matrix.

$ \begin{aligned} \because \quad A^T & =A \text { and } B^T=B \\ A(B) & =B A \\ A B & =B A \quad \text { (which is not possible) } \end{aligned} $

Now, if $A$ and $B$ are skew-symmetric matrix.

$ \begin{aligned} \because \quad A^T & =-A \text { and } B^T=-B \\ -A-B & =B A \\ A B & =B A \quad \text { (which is not possible) } \end{aligned} $

Thus, $A$ and $B$ are neither symmetric nor skew-symmetric matrix.

Given,

$ \operatorname{adj}\left[\begin{array}{ccc} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{array}\right]=\left[\begin{array}{ccc} 5 & m & -2 \\ 1 & 1 & 0 \\ -2 & -2 & n \end{array}\right] $

Adjoint of matrix $=$ transpose of cofactor matrix

$ \begin{aligned} & C_{11}=\left|\begin{array}{cc} 1 & -2 \\ 2 & 1 \end{array}\right|=5 \\ & C_{12}=-\left|\begin{array}{cc} -1 & -2 \\ 0 & 1 \end{array}\right|=1 \\ & C_{13}=\left|\begin{array}{cc} -1 & 1 \\ 0 & 2 \end{array}\right|=-2 \end{aligned} $

$ \begin{aligned} C_{21} & =-\left|\begin{array}{ll} 0 & 2 \\ 2 & 1 \end{array}\right|=4 \\ C_{22} & =\left|\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right|=1 \\ C_{23} & =-\left|\begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right|=-2 \\ C_{31} & =\left|\begin{array}{cc} 0 & 2 \\ 1 & -2 \end{array}\right|=-2 \\ C_{32} & =-\left|\begin{array}{cc} 1 & 2 \\ -1 & -2 \end{array}\right|=0 \\ C_{33}=\left|\begin{array}{cc} 1 & 0 \\ -1 & 1 \end{array}\right| & =1 \\ \text { So, Adj }\left[\begin{array}{ccc} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{array}\right] & =\left[\begin{array}{ccc} 5 & 1 & -2 \\ 4 & 1 & -2 \\ -2 & 0 & 1 \end{array}\right] \\ {\left[\begin{array}{ccc} 5 & m & -2 \\ 1 & 1 & 0 \\ -2 & -2 & n \end{array}\right] } & =\left[\begin{array}{ccc} 5 & 4 & -2 \\ 1 & 1 & 0 \\ -2 & -2 & 1 \end{array}\right] \end{aligned} $

Hence, $m=4, n=1$

$ \therefore m+n=4+1=5 $

We have,

$ \begin{gathered} A=\left[\begin{array}{ll} 0 & 3 \\ 0 & 0 \end{array}\right] \text { and } \\ f(x)=x+x^2+x^3+\ldots+x^{2023} \\ \text { Then, } A^2=A \cdot A=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ A^3=A A^2=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ A^4=A A^3=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ A^n=A A^{n-1}=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \end{gathered} $

So, $f(A)=A+A^2+A^3+\ldots+A^{2023}$

$ \begin{aligned} & =\left[\begin{array}{ll} 0 & 3 \\ 0 & 0 \end{array}\right]+\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]+\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]+\ldots+\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ & =\left[\begin{array}{ll} 0 & 3 \\ 0 & 0 \end{array}\right] \end{aligned} $

Hence, $f(A)+I=\left[\begin{array}{ll}0 & 3 \\ 0 & 0\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$

$ \text { } \begin{aligned} A & =\left[\begin{array}{lll} b & a & 0 \\ c & 0 & b \\ a & a & b \end{array}\right], B=\left[\begin{array}{lll} 0 & a & b \\ b & 0 & c \\ b & a & a \end{array}\right] \\ A B & =\left[\begin{array}{lll} b & a & 0 \\ c & 0 & b \\ a & a & b \end{array}\right]\left[\begin{array}{lll} 0 & a & b \\ b & 0 & c \\ b & a & a \end{array}\right] \\ & =\left[\begin{array}{ccc} 0+a b+0 & a b+0+0 & b^2+c a+0 \\ 0+0+b^2 & a c+0+a b & b c+0+a b \\ 0+a b+b^2 & a^2+0+a b & a b+a c+a b \end{array}\right] \\ & =\left[\begin{array}{ccc} a b & a b & b^2+a c \\ a b+b^2 & a c+a b & b c+a b \\ a & a b+a c+a b \end{array}\right] \\ & =\left[\begin{array}{lll} 2 & 2 & 7 \\ 1 & 8 & 5 \\ 3 & 6 & 10 \end{array}\right] \end{aligned} $

$ b^2=1, \quad a b=2 $

$ \begin{array}{ll} \therefore & a^2+a b=6 \\ \Rightarrow & a^2+2=6 \Rightarrow a^2=4 \end{array} $

$ \begin{aligned} \Rightarrow & & a & = \pm 2 \\ & \therefore & b^2+a c & =7 \\ \Rightarrow & & 1+a c & =7 \Rightarrow a c=6 \end{aligned} $

$ \begin{aligned} c & =\frac{6}{ \pm 2}= \pm 3 \Rightarrow c^2=9 \\ \therefore \quad a^2+b^2+c^2 & =9+1+4=14 \end{aligned} $

$A=\left[\begin{array}{lll}1 & a & 3 \\ b & 2 & c \\ 3 & d & 4\end{array}\right]$

Since, $A^T=A$

$ \begin{gathered} {\left[\begin{array}{lll} 1 & a & 3 \\ b & 2 & c \\ 3 & d & 4 \end{array}\right]=\left[\begin{array}{lll} 1 & b & 3 \\ a & 2 & d \\ 3 & c & 4 \end{array}\right]} \\ a=b \text { and } c=d \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ B=\left[\begin{array}{ccc} 0 & 5 & b \\ -5 & 0 & -7 \\ 6 & c & 0 \end{array}\right] \end{gathered} $

$ \begin{aligned} &\begin{aligned} & \text { Since, } B^T=-B \\ & \Rightarrow \quad\left[\begin{array}{ccc} 0 & -5 & 6 \\ 5 & 0 & c \\ b & -7 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & -5 & -b \\ 5 & 0 & 7 \\ -6 & -c & 0 \end{array}\right] \end{aligned}\\ &\text { So, } b=-6, a=-6, c=7 \text { and } d=7\\ &\begin{aligned} A B & =\left[\begin{array}{ccc} 1 & -6 & 3 \\ -6 & 2 & 7 \\ 3 & 7 & 4 \end{array}\right]\left[\begin{array}{ccc} 0 & 5 & -6 \\ -5 & 0 & -7 \\ 6 & 7 & 0 \end{array}\right] \\ & =\left[\begin{array}{ccc} 0+30+18 & 5+0+21 & -6+42 \\ 0-10+42 & -30+49 & 22 \\ -11 & 43 & -67 \end{array}\right] \\ & =\left[\begin{array}{ccc} 48 & 26 & 36 \\ 32 & 19 & 22 \\ -11 & 43 & -67 \end{array}\right] \end{aligned} \end{aligned} $

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

Co-factor matrix of $A$

$ \begin{aligned} & =\left[\begin{array}{ccc} -3 & +6 & -3 \\ +7 & 1 & -5 \\ -4 & 2 & -1 \end{array}\right] \\ \operatorname{adj}(A) & =\left[\begin{array}{ccc} -3 & 7 & -4 \\ 6 & 1 & 2 \\ -3 & -5 & -1 \end{array}\right] \\ |A| & =-1(5-8)+3(0-6)-2(0-3) \end{aligned} $

$ \begin{gathered} =3-18+6=-9 \\ \therefore A^{-1}=\frac{\operatorname{Adj} A}{|A|}=\left|\begin{array}{ccc} 1 / 3 & -7 / 9 & 4 / 9 \\ -2 / 3 & -1 / 9 & -2 / 9 \\ 1 / 3 & 5 / 9 & 1 / 9 \end{array}\right| \\ \therefore a_1+c_2+b_3=\frac{1}{3}+\frac{5}{9}+\left(-\frac{2}{9}\right) \\ \quad=\frac{3+5-2}{9}=\frac{6}{9}=\frac{2}{3} \end{gathered} $

Given, the system of linear equations

$ \begin{aligned} & 2 x-3 y+5 z=12 \\ & 5 x+2 y+3 z=11 \end{aligned} $

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

has the unique solution $(\alpha, \beta, \gamma)$.

Now, by using the Cramer's rule, we get

$ \begin{aligned} \alpha & =\frac{\left|\begin{array}{ccc} 12 & -3 & 5 \\ 11 & 2 & 3 \\ -3 & 2 & -3 \end{array}\right|}{\left|\begin{array}{ccc} 2 & -3 & 5 \\ 5 & 2 & 3 \\ 1 & 2 & -3 \end{array}\right|}=\frac{-76}{-38}=2 \\ \beta & =\frac{\left|\begin{array}{ccc} 2 & 12 & 5 \\ 5 & 11 & 3 \\ 1 & -3 & -3 \end{array}\right|}{\left|\begin{array}{ccc} 2 & -3 & 5 \\ 5 & 2 & 3 \\ 1 & 2 & -3 \end{array}\right|}=\frac{38}{-38}=-1 \\ \text { and } \gamma & =\frac{\left|\begin{array}{ccc} 2 & -3 & 12 \\ 5 & 2 & 11 \\ 1 & 2 & -3 \end{array}\right|}{\left|\begin{array}{ccc} 2 & -3 & 5 \\ 5 & 2 & 3 \\ 1 & 2 & -3 \end{array}\right|}=\frac{-38}{-38}=1 \end{aligned} $

$ \begin{array}{r} \therefore 2 \alpha+5 \beta+3 \gamma=2(2)+5(-1)+3(1) \\ =4-5+3=7-5=2 \end{array} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} A & =\left[\begin{array}{ccc} 1 & 2 & -1 \\ -1 & 0 & 2 \\ 1 & 2 & 0 \end{array}\right], B=\left[\begin{array}{ccc} -3 & -2 & 4 \\ 2 & 2 & -1 \\ -2 & 0 & 3 \end{array}\right] \\ A^2 & =\left[\begin{array}{ccc} 1 & 2 & -1 \\ -1 & 0 & 2 \\ 1 & 2 & 0 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & -1 \\ -1 & 0 & 2 \\ 1 & 2 & 0 \end{array}\right] \\ & =\left[\begin{array}{ccc} -2 & 0 & 3 \\ 1 & 2 & 1 \\ -1 & 2 & 3 \end{array}\right]=A+B \end{aligned} \end{aligned} $

Let $\Delta_1=\left|\begin{array}{lll}2 & 3 & 5 \\ 3 & 5 & 2 \\ 5 & 2 & 3\end{array}\right|=-70$

$ \begin{aligned}and\,\, \Delta_2 & =\left|\begin{array}{ccc} 1 & 1 & 1 \\ 7 & 11 & 13 \\ 49 & 121 & 169 \end{array}\right| \\ & C_2 \rightarrow C_2-C_1 \\ & C_3 \rightarrow C_3-C_1 \\ & =\left|\begin{array}{ccc} 1 & 0 & 0 \\ 7 & 4 & 6 \\ 49 & 72 & 120 \end{array}\right|=48 \end{aligned} $

Now, $\Delta_1+\Delta_2=-70+48=-22$

$\operatorname{det} A=190$

$ \begin{aligned} & {\left[\begin{array}{ccc} k & 5 & 2 \\ 2 & -k & 5 \\ 5 & 2 & -k \end{array}\right]=190 } \\ \Rightarrow & (k-3)\left(k^2+3 k+19\right)=0 \Rightarrow k=3 \end{aligned} $

$\operatorname{adj} A=$ Transpose of co-factor matrix of $A$

$ =\left[\begin{array}{ccc} -1 & 19 & 31 \\ 31 & -19 & -11 \\ 19 & 19 & -19 \end{array}\right] $

System of equations

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

It has unique solution.

If $\Delta \neq 0$ and $x=\frac{\Delta_1}{\Delta}, y=\frac{\Delta_2}{\Delta}, z=\frac{\Delta_3}{\Delta}$

$ \begin{array}{ll} \text { Now, } & \Delta=\left|\begin{array}{ccc} 3 & -2 & 1 \\ 2 & 3 & -2 \\ 1 & 4 & 3 \end{array}\right|=72 \neq 0 \\ \text { and } & \Delta_3=\left|\begin{array}{ccc} 3 & -2 & 5 k \\ 2 & 3 & -5 k \\ 1 & 4 & k \end{array}\right| \\ & =k[69+14+25]=108 k \\ \therefore & z=3\,\,\,\, (given) \\ \Rightarrow & \frac{\Delta_3}{\Delta}=3 \Rightarrow \frac{108 k}{72}=3 \\ \Rightarrow & k=2 \end{array} $

$\left|\begin{array}{ccc}\sqrt{3} & 2 \sqrt{5} & \sqrt{5} \\ \sqrt{15} & 5 & \sqrt{10} \\ 3 & \sqrt{15} & 5\end{array}\right|$

Taking $\sqrt{3}, \sqrt{5}$ and $\sqrt{5}$ from $C_1, C_2$ and $C_3$, we get

$ \begin{aligned} & \sqrt{3} \times \sqrt{5} \times \sqrt{5}\left|\begin{array}{ccc} 1 & 2 & 1 \\ \sqrt{5} & \sqrt{5} & \sqrt{2} \\ \sqrt{3} & \sqrt{3} & \sqrt{5} \end{array}\right| \\ & =5 \sqrt{3}\left|\begin{array}{ccc} 1 & 1 & 1 \\ \sqrt{5} & 0 & \sqrt{2} \\ \sqrt{3} & 0 & \sqrt{5} \end{array}\right| \end{aligned} $

$ \begin{aligned} & \text { (Applying } C_2 \rightarrow C_2-C_1, \\ & \quad-5 \sqrt{3}(5-\sqrt{6}) \\ & =-25 \sqrt{3}+15 \sqrt{2}=15 \sqrt{2}-25 \sqrt{3} \end{aligned} $

Given that $ A $ is a non-singular matrix, and it satisfies the equation:

$ (A-2I)(A-3I) = 0 $

Since $ A $ is non-singular, it implies that $ |A| \neq 0 $, which means $ A^{-1} $ exists.

Expanding the given equation:

$ A^2 - 2A - 3A + 6I = 0 $

Simplifies to:

$ A^2 - 5A + 6I = 0 $

Multiplying both sides of this equation by $ A^{-1} $ yields:

$ A - 5I + 6A^{-1} = 0 $

Dividing through by 5 gives:

$ \frac{A}{5} + \frac{6}{5} A^{-1} = I $

Given that $A B=$ Scalar matrix

$ =\left[\begin{array}{ll} \lambda & 0 \\ 0 & \lambda \end{array}\right] $

where $\lambda=$ scalar

Also, given that $B=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]$ and

$ \begin{aligned} & \operatorname{det}(3 A)=27 \\ & \Rightarrow \quad \operatorname{det} A=\frac{27}{9}=3 \end{aligned} $

Now, let $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

$ \therefore \quad A B=\left[\begin{array}{ll} \lambda & 0 \\ 0 & \lambda \end{array}\right] $

$ \begin{aligned} &\begin{aligned} & \Rightarrow\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right]=\left[\begin{array}{ll} \lambda & 0 \\ 0 & \lambda \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ll} a & 2 a+3 b \\ c & 2 c+3 d \end{array}\right]=\left[\begin{array}{ll} \lambda & 0 \\ 0 & \lambda \end{array}\right] \end{aligned}\\ &\text { Comparing on both sides, we get }\\ &\begin{aligned} & a=\lambda, c=0,2 a+3 b=0,2 c+3 d=\lambda \\ & \therefore \quad b=-\frac{2 \lambda}{3} \text { and } d=\frac{\lambda}{3} \\ & \text { Thus, } A=\left[\begin{array}{cc} \lambda & -\frac{2 \lambda}{3} \\ 0 & \frac{\lambda^3}{3} \end{array}\right] \\ & \qquad|A|=\left|\begin{array}{cc} \lambda & -2 \lambda \\ 0 & \frac{\lambda}{3} \end{array}\right|=\frac{\lambda^2}{3}=3 \quad \text { (given) } \\ & \Rightarrow \quad \lambda^2=9 \Rightarrow \lambda= \pm 3 \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \therefore \quad A^2=\left[\begin{array}{cc} \lambda & -\frac{2 \lambda}{3} \\ 0 & \frac{\lambda^2}{3} \end{array}\right]\left[\begin{array}{cc} \lambda & -\frac{2 \lambda}{3} \\ 0 & \frac{\lambda^2}{3} \end{array}\right] \\ & \quad=\left[\begin{array}{cc} \lambda^2 & -\frac{2 \lambda^2}{3}-\frac{2 \lambda^2}{9} \\ 0 & \frac{\lambda^2}{9} \end{array}\right]=\left[\begin{array}{cr} \lambda^2 & -\frac{8 \lambda^2}{9} \\ 0 & \frac{\lambda^2}{9} \end{array}\right] \\ & \quad A^2=\left[\begin{array}{cc} 9 & -8 \\ 0 & 1 \end{array}\right] \\ & \text { Clearly, } A^{-1}=\frac{1}{3}\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right] \\ & \Rightarrow \quad 3 A^{-1}=\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right] \end{aligned}\\ &\text { Thus, }\\ &3 A^{-1}+A^2=\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right]+\left[\begin{array}{cc} 9 & -8 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 10 & -6 \\ 0 & 4 \end{array}\right] \end{aligned} $

If $ A $ is a symmetric matrix with real entries, it means $ A^T = A $.

Now, suppose the inverse $ A^{-1} $ exists.

For symmetric matrices, the transpose of the inverse is equal to the inverse of the transpose, so:

$ (A^{-1})^T = (A^T)^{-1} = A^{-1} $

This equation shows that $ A^{-1} $ is also symmetric. Therefore, if the inverse of a symmetric matrix exists, it is necessarily symmetric as well.

$ \begin{aligned} &\text { }\left|\begin{array}{ccc} \omega+\omega^2 & \omega^2+\omega^9 & \omega^9+\omega \\ \omega^{27}+\omega^{31} & \omega^{31}+\omega^{17} & \omega^{17}+\omega^{27} \\ \omega^{30}+\omega^{41} & \omega^{41}+\omega^{19} & \omega^{19}+\omega^{30} \end{array}\right|\\ &\begin{aligned} = & \left|\begin{array}{ccc} \omega & \omega^2 & \omega^9 \\ \omega^{27} & \omega^{31} & \omega^{17} \\ \omega^{30} & \omega^{41} & \omega^{19} \end{array}\right|+\left|\begin{array}{ccc} \omega^2 & \omega^9 & \omega \\ \omega^{31} & \omega^{17} & \omega^{27} \\ \omega^{41} & \omega^{19} & \omega^{30} \end{array}\right| \\ = & \omega \times \omega^{17} \times \omega^{19}\left|\begin{array}{ccc} 1 & \omega & \omega^8 \\ \omega^{10} & \omega^{14} & 1 \\ \omega^{11} & \omega^{22} & 1 \end{array}\right| +\omega \times \omega^{17} \times \omega^{19}\left|\begin{array}{ccc} \omega & \omega^8 & 1 \\ \omega^{14} & 1 & \omega^{10} \\ \omega^{22} & 1 & \omega^{11} \end{array}\right| \\ = & \omega^{37}\left|\begin{array}{ccc} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & \omega & 1 \end{array}\right|+\omega^{37}\left|\begin{array}{ccc} \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \\ \omega & 1 & \omega^2 \end{array}\right| \\ = & \omega^{37} \times 0+\omega^{37} \times 0 \\ = & 0+0=0 \end{aligned} \end{aligned} $

Given a non-singular matrix $ P $ such that:

$ I + P + P^2 + \ldots + P^n = 0 $

we aim to determine $ P^{-1} $.

First, recognize the expression as a series whose sum is zero:

$ I + P + P^2 + \ldots + P^n = 0 $

This expression can be seen as a finite geometric series. Now, apply the formula for the sum of a geometric series:

$ S = I \frac{P^{n+1} - I}{P - I} $

Given that the entire series equals zero:

$ \frac{P^{n+1} - I}{P - I} = 0 $

This implies:

$ P^{n+1} - I = 0 $

Thus, we have:

$ P^{n+1} = I $

This equation indicates that $ P $ raised to the power $ n+1 $ results in the identity matrix $ I $. Therefore, multiplying both sides by $ P^{-1} $, we deduce:

$ P^{n+1} \cdot P^{-1} = I \cdot P^{-1} $

Hence:

$ P^n = P^{-1} $

Therefore, the inverse of $ P $ is $ P^n $.

To solve for $|\alpha|$, we begin with the matrix $A$:

$ A = \begin{bmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{bmatrix} $

Given that $\operatorname{det}(A^2) = 25$, we can derive that $(\operatorname{det}(A))^2 = 25$. Hence, $|A| = \pm 5$.

Next, we compute the determinant of $A$:

The matrix $A$ is upper triangular. Thus, the determinant $|A|$ is the product of its diagonal elements:

$ |A| = 5 \times \alpha \times 5 = 25\alpha $

Given that $|A| = \pm 5$, we equate and solve for $\alpha$:

$ 25\alpha = \pm 5 $

This implies that $\alpha = \pm \frac{1}{5}$.

Therefore, the absolute value of $\alpha$ is:

$ |\alpha| = \frac{1}{5} $

$ \begin{aligned} P & =\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right] \operatorname{Tr}(P)=a+e+i \\ P^T & =\left[\begin{array}{lll} a & d & g \\ b & e & h \\ c & f & i \end{array}\right] \operatorname{Tr}\left(P^T\right)=a+e+i \end{aligned} $

$ \Rightarrow \operatorname{Tr}(P)=\operatorname{Tr}\left(P^T\right) $

$ \operatorname{Tr}\left(P-P^T\right)+\operatorname{Tr}\left(P+P^\tau\right)+\frac{\operatorname{Tr}(P)}{\operatorname{Tr}\left(P^T\right)}+\operatorname{Tr}(P) $

$ \times \operatorname{Tr}\left(P^T\right)=0 $

$ \begin{aligned} \Rightarrow \operatorname{Tr}(P)-\operatorname{Tr}\left(P^T\right)+\operatorname{Tr}(P) & +\operatorname{Tr}\left(P^T\right)+\frac{\operatorname{Tr}(P)}{\operatorname{Tr}(P)} \\ & +\operatorname{Tr}(P) \cdot \operatorname{Tr}(P)=0 \end{aligned} $

$ \Rightarrow(\operatorname{Tr}(P))^2+2 \operatorname{Tr}(P)+1=0 $

$(\operatorname{Tr}(P)+1)^2=0$

$\operatorname{Tr}(P)=-1$

To solve the given system of equations using the matrix inversion method, we need to ensure that the coefficient matrix is invertible. This requires that the determinant of the matrix is not equal to zero.

Consider the system of equations:

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

$ 4x + 3y + kz = 14 $

$ 2x + y + 2z = 3 $

The coefficient matrix for this system is:

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

To check the invertibility, we calculate the determinant of this matrix:

$ \text{Determinant} = 1(6 - k) - 4(2k - 3) + 2(k^2 - 9) $

Simplifying the expression:

$ 1(6 - k) = 6 - k $

$ -4(2k - 3) = -8k + 12 $

$ 2(k^2 - 9) = 2k^2 - 18 $

Combining these results gives:

$ 6 - k - 8k + 12 + 2k^2 - 18 = 2k^2 - 9k $

For the matrix to be invertible, we need the determinant to be non-zero:

$ 2k^2 - 9k \neq 0 $

This expression can be factored as:

$ k(2k - 9) \neq 0 $

Thus, the values of $ k $ for which the system is solvable by matrix inversion are:

$ k \neq 0 $

$ k \neq \frac{9}{2} $

Therefore, $ k $ should not be 0 and not $ \frac{9}{2} $.

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

For option A :

${R_1} \to {R_1} + {R_2}$

$A = \left[ {\matrix{ 0 & 1 \cr 1 & { - 1} \cr } } \right]$

$\therefore$ Option A can be obtained.

For option B :

${R_1} \leftrightarrow {R_2}$

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

$\therefore$ Option B can be obtained.

Option C :

Not possible by a single elementary row operation.

Option D :

${R_2} \to {R_2} + 2{R_1}$

$A = \left[ {\matrix{ { - 1} & 2 \cr { - 1} & 3 \cr } } \right]$

$\therefore$ Option D can be obtained.

Given,

$x + y + z = 6$ ...... (1)

$2x + 5y + \alpha z = \beta $ ..... (2)

$x + 2y + 3z = 14$ ...... (3)

System of equation have infinite many solutions.

$\therefore$ ${\Delta _x} = {\Delta _y} = {\Delta _z} = 0$ and $\Delta = 0$

Now, $\Delta = \left| {\matrix{ 1 & 1 & 1 \cr 2 & 5 & \alpha \cr 1 & 2 & 3 \cr } } \right| = 0$

${C_1} \to {C_1} - {C_3}$

${C_2} \to {C_2} - {C_3}$

$ \Rightarrow \left| {\matrix{ 0 & 0 & 1 \cr {2 - \alpha } & {5 - \alpha } & \alpha \cr { - 2} & { - 1} & 3 \cr } } \right| = 0$

$ \Rightarrow - 2 + \alpha + 10 - 2\alpha = 0$

$ \Rightarrow 8 - \alpha = 0$

$ \Rightarrow \alpha = 8$

Now, $x + y + z = 6$

$2x + 5y + 8z = \beta $

$x + 2y + 3z = 14$

$\therefore$ ${\Delta _x} = \left| {\matrix{ 6 & 1 & 1 \cr \beta & 5 & 8 \cr {14} & 2 & 3 \cr } } \right| = 0$

${C_1} \to {C_1} - 6{C_3}$

${C_2} \to {C_2} - {C_3}$

$ \Rightarrow \left| {\matrix{ 0 & 0 & 1 \cr {\beta - 48} & { - 3} & 8 \cr { - 4} & { - 1} & 3 \cr } } \right| = 0$

$ \Rightarrow - \beta + 48 - 12 = 0$

$ \Rightarrow \beta = 36$

$\therefore$ $\alpha + \beta = 8 + 36 = 44$

AB is zero matrix

$ \Rightarrow |A| = |B| = 0$

So neither A nor B is invertible

If $|A| = 0$

$ \Rightarrow |\mathrm{adj}\,A| = 0$ so $\mathrm{adj}\,A$

$AX = 0$ is homogeneous system and $|A| = 0$

So, it is having infinitely many solutions

(A) $M = {A^4} - {B^4}$

${M^T} = {({A^4} - {B^4})^T} = {({A^T})^4} - {({B^T})^4}$

$ = {A^4} - {( - B)^4} = {A^4} - {B^4} = M$

(B) $M = AB - BA$

${M^T} = {(AB - BA)^T} = {(AB)^T} - {(BA)^T}$

$ = {B^T}{A^T} - {A^T}{B^T}$

$ = - BA - A( - B)$

$ = AB - BA = M$

(C) $M = {B^5} - {A^5}$

${M^T} = {({B^T})^5} - {({A^T})^5} = - ({B^5} + {A^5}) \ne - M$

(D) $M = AB + BA$

${M^T} = {(AB)^T} + {(BA)^T}$

$ = {B^T}{A^T} + {A^T}{B^T} = - BA - AB = - M$

$A = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right]$

$ \Rightarrow {A^2} = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right] \times \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right] = \left[ {\matrix{ 0 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr } } \right]$

$ \Rightarrow {A^3} = \left[ {\matrix{ 0 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr } } \right]\left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] = l$

Now ${B_0} = {A^{49}} + 2{A^{98}} = {({A^3})^{16}}\,.\,A + 2{({A^3})^{32}}\,.\,{A^2}$

${B_0} = A + 2{A^2} = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right] + \left[ {\matrix{ 0 & 0 & 2 \cr 2 & 0 & 0 \cr 0 & 2 & 0 \cr } } \right] = \left[ {\matrix{ 0 & 1 & 2 \cr 2 & 0 & 1 \cr 1 & 2 & 0 \cr } } \right]$

$|{B_0}| = 9$

Since, ${B_n} = Adj\,|{B_{n - 1}}| \Rightarrow |{B_n}| = |{B_{n - 1}}{|^2}$

Hence $|{B_4}| = |{B_3}{|^2} = |{B_2}{|^4} = |{B_1}{|^8} = |{B_0}{|^{16}}$

$ = |{3^2}{|^{16}} = {3^{32}}$

Characteristic equation of A is given by

$\left| {A - \lambda I} \right| = 0$

$\left| {\matrix{ {4 - \lambda } & { - 2} \cr \alpha & {\beta - \lambda } \cr } } \right| = 0$

$ \Rightarrow {\lambda ^2} - (4 + \beta )\lambda + (4\beta + 2\alpha ) = 0$

So, ${A^2} - (4 + \beta )A + (4\beta + 2\alpha )I = 0$

$|A| = 4\beta + 2\alpha = 18$

${A^2} = \left[ {\matrix{ 1 & 2 \cr { - 2} & { - 5} \cr } } \right]\left[ {\matrix{ 1 & 2 \cr { - 2} & { - 5} \cr } } \right] = \left[ {\matrix{ { - 3} & { - 8} \cr 8 & {21} \cr } } \right]$

$\alpha {A^2} + \beta A = \left[ {\matrix{ { - 3\alpha } & { - 8\alpha } \cr {8\alpha } & {21\alpha } \cr } } \right] + \left[ {\matrix{ \beta & {2\beta } \cr { - 2\beta } & { - 5\beta } \cr } } \right]$

$ = \left[ {\matrix{ { - 3\alpha + \beta } & { - 8\alpha + 2\beta } \cr {8\alpha - 2\beta } & {21\alpha - 5\beta } \cr } } \right] = \left[ {\matrix{ 2 & 0 \cr 0 & 2 \cr } } \right]$

On Comparing

$8\alpha = 2\beta ,\, - 3\alpha + \beta = 2,\,21\alpha - 5\beta = 2$

$ \Rightarrow \alpha = 2,\,\beta = 8$

So, $\alpha + \beta = 10$

$A'BA = \left[ {\matrix{ 1 & 1 & 1 \cr } } \right]\left[ {\matrix{ {{9^2}} & { - {{10}^2}} & {{{11}^2}} \cr {{{12}^2}} & {{{13}^2}} & { - {{14}^2}} \cr { - {{15}^2}} & {{{16}^2}} & {{{17}^2}} \cr } } \right]A$

$ = \left[ {\matrix{ {{9^2} + {{12}^2} - {{15}^2}} & { - {{10}^2} + {{13}^2} + {{16}^2}} & {{{11}^2} - {{14}^2} + {{17}^2}} \cr } } \right]\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$

$ = \left[ {{9^2} + {{12}^2} - {{15}^2} - {{10}^2} + {{13}^2} + {{16}^2} + {{11}^2} - {{14}^2} + {{17}^2}} \right]$

$ = [({9^2} - {10^2}) + ({11^2} + {12^2}) + ({13^2} - {14^2}) + ({16^2} - {15^2}) + {17^2}]$

$ = [ - 19 + 265 + ( - 27) + 31 + 289]$

$ = [585 - 46] = [539]$

$\Delta = \left| {\matrix{ 8 & 1 & 4 \cr 1 & 1 & 1 \cr \lambda & { - 3} & 0 \cr } } \right|$

$ = 8(3) - 1( - \lambda ) + 4( - 3 - \lambda )$

$ = 24 + \lambda - 12 - 4\lambda $

$ = 12 - 3\lambda $

So for $\lambda = 4$, it is having infinitely many solutions.

${\Delta _x} = \left| {\matrix{ { - 2} & 1 & 4 \cr 0 & 1 & 1 \cr \mu & { - 3} & 0 \cr } } \right|$

$ = - 2(3) - 1( - \mu )+4( - \mu )$

$ = - 6 - 3\mu = 0$

For $\mu = - 2$

Distance of $(4, - 2,{{ - 1} \over 2})$ from $8x + y + 4z + 2 = 0$

$ = {{32 - 2 - 2 + 2} \over {\sqrt {64 + 1 + 16} }} = {{10} \over 3}$ units

$|(A + I)(adj\,A + I)| = 4$

$ \Rightarrow |A\,adj\,A + A + adj\,A + I| = 4$

$ \Rightarrow |(A)I + A + adj\,A + I| = 4$

$|A| = - 1 \Rightarrow |A + adj\,A| = 4$

$A = \left[ {\matrix{ a & b \cr c & d \cr } } \right]\,adj\,A = \left[ {\matrix{ a & { - b} \cr { - c} & d \cr } } \right]$

$ \Rightarrow \left| {\matrix{ {(a + d)} & 0 \cr 0 & {(a + d)} \cr } } \right| = 4$

$ \Rightarrow a + d = \, \pm \,2$

$\Delta = \left| {\matrix{ 2 & { - 3} & 5 \cr 1 & 3 & { - 1} \cr 3 & { - 1} & {{\lambda ^2} - |\lambda |} \cr } } \right| = 2\left( {3{\lambda ^2} - 3|\lambda | - 1} \right) + 3\left( {{\lambda ^2} - |\lambda | + 3} \right) + 5( - 1 - 9)$

$ = 9{\lambda ^2} - 9|\lambda | - 43$

$ = 9|\lambda {|^2} - 9|\lambda | - 43$

$\Delta = 0$ for 2 values of $|\lambda |$ out of which one is $-\mathrm{ve}$ and other is $+\mathrm{ve}$

So, 2 values of $\lambda$ satisfy the system of equations to obtain no solution.