• Problem:

• Let $P$ be an invertible matrix with $I+p+p^2+\ldots+p^n=0$ (i.e, the zero matrix).

• Find $P^{-1}$ in terms of $P$.

• Solution:

• $I+P+p^2+\cdots+P^n=0$.

• Multiply this equation by

• $p^{-1}$.

• $p^{-1}+ I +p+\cdots+p^{n-1}=0$

• $\Rightarrow p^{-1} =-I-p-\cdots+p^{n-1} =-\sum_0^{n-1} p^i .$

• Problem:

• If $A=\left[\begin{array}{ccc}1 & 2 & 1 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right]$, find $A^3$.

• Solution:

• $A=\left[\begin{array}{ccc}1 & 2 & 1 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right]$ $\begin{aligned} & A^2=\left[\begin{array}{ccc} 1 & 2 & 1 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & 1 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{array}\right] \\ & \end{aligned}$

• $\begin{aligned} & =\left[\begin{array}{llll} 1+10-2 & 2+4-1 & 1+12 -3 \\ 5+10-12 & 10+4-6 & 5+12-18 \\ -2-5+6 & -4-2+3 & -2-6+9 \end{array}\right] \end{aligned}$
• $\begin{aligned} =\left[\begin{array}{ccc} 9 & 5 & 10 \\ 3 & 10 & -1 \\ -1 & -3 & 1 \end{array}\right] \\ & \end{aligned}$

$\begin{aligned} & A^3=A^2 \cdot A=\left[\begin{array}{ccc} 9 & 5 & 10 \\ 3 & 10 & -1 \\ -1 & -3 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & 1 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{array}\right] \\ & =\left[\begin{array}{ccc} 9+25-20 & 18+10-20 & 9+30-30 \\ 3+50+2 & 6+20+1 & 3+60+3 \\ -1+15-2 & -2-6-1 & -1+18-3 \end{array}\right] \ & \end{aligned}$ $\begin{aligned} & =\left[\begin{array}{ccc} 14 & 8 & 9 \\ 55 & 29 & 66 \\ -18 & -9 & 14 \end{array}\right] \ & \end{aligned}$

• Problem:

• The set of natural numbers is partitioned into arrays of rows & column in the form of matrices an $M_1=[1], M_2=\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right],\\ M_3=\left[\begin{array}{lll}6 & 1 & 11 \\ 9 & 10 & 11 \\ 13 & \text { is } & 14\end{array}\right]$ and so on. Find the trace of $M_n$.

• Solution:

• $\{{1,2,6,15,…, t_n…}\}$

• $\therefore \text { Let } S_n= 1+2+6+15+\cdots+t_n$

• $\therefore\quad 0=S_ n-S_ n\\$

• $=\left(1+2+6+15+\cdots+t_ n\right)-\left(1+2+6+15+\cdots+t_n\right)$

• $= 1+(2-1)+(6-2)+(15-6)+\cdots+\left(t_{n-1}-t_{n-2}\right)$

• $+\left(t_n-t_{n-1}\right)+t_n$

• $\begin{aligned} & 0=1+1^2+2^2+3^2+\cdots+\left(t_n-t_{n-1}\right)-t_n \\ & \therefore \quad t_n=1+1^2+2^2+3^2+\cdots+\left(t_n-t_{n-1}\right) \\ &=1+\sum_{i=1}^{n-1} i^2 =1+\frac{n(n-1)(2 n-1)}{6} . \end{aligned}$

• Let " $a_{11}$ denote the $1^{\text {st }}$ element of $M_n$.

• Then $a_{11}=1+\frac{n(n-1)(2 n-1)}{6}$

• $a_{i i}=1+\frac{n(n-1)(2 n-1)}{6}+(i-1)(n+1), 1 \leqslant i \leqslant n .$

• $\begin{aligned} \therefore & \operatorname{trace}\left(M_n\right) \end{aligned}$

• $\begin{aligned} & =\sum_{i=1}^n a_{i i} \\ & =\sum_{i=1}^n\left[\frac{1+n(n-1)(2 n-1)}{6}+(i-1)(n+1)\right] \\ & =\sum_{i=1}^n \left(1+\frac{n(n-1)(2 n-1)}{6}\right)+\sum_{i=1}^n(i-1)(n+1) \end{aligned}$

• $=n\left[1+\frac{n(n-1)(2 n-1)}{6}\right]+(n+1) \sum_{i=1}^n i=1$

• $A=n\left[1+\frac{n(n-1)(2 n-1)}{6}\right]+(n+1) \sum_{i=1}^n i$

• $\begin{aligned} & =n\left[1+\frac{n(n-1)(2 n-1)}{6}\right]+\frac{(n+1) \frac{(n-1) n}{2}}{6} \\ & =\frac{n}{6}\left[6+2 n^3-3 n^2+n+3 n^2-30\right] \\ & =\frac{n}{6}\left[2 n^3+2 n+6\right] \\ & \therefore \operatorname{tr}(M n)=\frac{n}{6}\left(2 n^3+n+3\right) .\end{aligned}$

• Problem:

• If $A=\left[\begin{array}{cc}\frac{-1+i \sqrt{3}}{2 i} & \frac{-1-i \sqrt{5}}{2 i} \\\ \frac{1+i \sqrt{3}}{2 i} & \frac{1-i \sqrt{3}}{2 i}\end{array}\right]$ & $f(x)=x^2+1$,

• then, find $f(A)$.

• Solution:

• $f(x)=x^2+1$

• $\therefore \quad f(A)=A^2+I$

• $A^2=\left[\begin{array}{cc} \frac{-1+i \sqrt{3}}{2 i} & \frac{-1-i \sqrt{3}}{2 i} \\ \frac{1+i \sqrt{3}}{2 i} & \frac{1-i \sqrt{5}}{2 i} \end{array}\right]\left[\begin{array}{cc} \frac{-1+i \sqrt{3}}{2 i} & \frac{-1-i \sqrt{5}}{2 i} \\ \frac{1+i \sqrt{3}}{2 i} & \frac{1-i \sqrt{3}}{2 i} \end{array}\right]$

• $\begin{aligned} & =\frac{-1}{4}\left[\begin{array}{cc}-1+i \sqrt{5} & -1-i \sqrt{5} \\ 1+i \sqrt{5} & 1-i \sqrt{5}\end{array}\right]\left[\begin{array}{cc}-1+i \sqrt{5} & -1-i \sqrt{5} \\ 1+i \sqrt{3} & 1-i \sqrt{3}\end{array}\right] \\ & =\frac{-1}{4}\left[\begin{array}{lc}(-1+i \sqrt{3})^2-(1+i \sqrt{3})^2 & 1+3+1+3 \\ -3-1+1+3 & -(1+i \sqrt{3})^2+(1-i \sqrt{3})^2\end{array}\right] \\ &\end{aligned}$

• $=-\frac{1}{4}\left[\begin{array}{cc} 1-2 i \sqrt{3}-3-1-2 i \sqrt{3}+3 & 0 \\ 0 & -1-2 i \sqrt{3}-3+1-2 i \sqrt{3}-3 \\ \end{array}\right]$

• $=-\frac{1}{4}\left[\begin{array}{cc}-4 i \sqrt{3} & 0 \\ 0 & -4 i \sqrt{3}\end{array}\right]=\left[\begin{array}{cc}i \sqrt{3} & 0 \\ 0 & i \sqrt{3}\end{array}\right]$

• $\begin{aligned} \therefore\quad f(A) & =A^2+I \\ & =\left[\begin{array}{cc} i \sqrt{5} & 0 \\ 0 & i n \end{array}\right]+\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & =\left[\begin{array}{cc} 1+i \sqrt{6} & 0 \\ 0 & 1+i \sqrt{3} \end{array}\right] . \end{aligned}$

• Problem:

• If $A=\left[\begin{array}{ccc} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{array}\right]$ is an idempotent matrix, then find the value of $x$.

• A square matrix $A$ which satifies $A^2=A$ is called an idempotent matrix.

• $\begin{aligned} A^2 & =\left[\begin{array}{ccc} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{array}\right]\left[\begin{array}{ccc} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{array}\right] \\ & =\left[\begin{array}{cccc} 4+2-4 & -4-6+2 & -8-8-4 x \\ -2 -3+4 & 2+9-8 & 4+12+4 x \\ 2+2+x & -2-6-2 x & -4-8+x^2 \end{array}\right] \end{aligned}$

• $=\left[\begin{array}{ccc} 2 & -2 & -16-4 x \\ -1 & 3 & 16+4 x \\ 4+x & -8-2 x & -12+x^2 \end{array}\right]$

• $\begin{aligned} & A^2=A ~ \text { i.e., }\left[\begin{array}{ccc} 2 & -2 & -16-4 x \\ -1 & 3 & 16+4 x \\ 4+x & -8-2 x & -12+x^2 \end{array}\right] \\ & \end{aligned}$

• $\begin{aligned} =\left[\begin{array}{ccc} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{array}\right] \\ & \end{aligned}$

• Suppose a matrix $A$ satisfies $A^2-5 A+7 I=0$. If $A^5=a A+b I$, then find the value of $a$ and $b$.

• Solution:

• Given $A^2-5 A+7 I=0$

• $\Rightarrow A^2=5 A-7 I \text {. }$

• $A^3=A^2 \cdot A$

• $=(5 A-7 I), A$

• $=5 A^2-7 A=5(5 A-7 x)-7 A$

• $=18 A-7 I .$

• $\begin{aligned} A^5 & =A^3 \cdot A^2 \\ & =(18 A-7 I)(5 A-7 I) \\ & =90 A^2-126 A-35 A+49 I \\ & =90(5 A-7 I)-161 A+49 I \\ & =450 A-630 I-161 A+49 I \\ & =28 9 A-54 I \longrightarrow \text { (1) } \end{aligned}$

• Also, we are given that $A^5=a A+b I \rightarrow$ (2)

• Comparing (1) & (2), we get

• $a=289, ~ ~ b=-581$

• Problem:

• $A=\left[\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right]$ & $A^2 = I$ then find $\alpha$

• Solution:

• $\begin{aligned} A^2 & =\left[\begin{array}{ll} \alpha & 0 \\ 1 & 1 \end{array}\right]\left[\begin{array}{ll} \alpha & 0 \\ 1 & 1 \end{array}\right] \\ & =\left[\begin{array}{ll} \alpha^2 & 0 \\ \alpha+1 & 1 \end{array}\right] \\ A^2 & =I \text { i.e., }\left[\begin{array}{ll} \alpha^2 & 0 \\ \alpha+1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \alpha^2 & =1 \quad \alpha+1=0 \\ \Rightarrow \alpha & =-1 . \end{aligned}$

• If $A=\left[\begin{array}{ll}\alpha & 0 \\ 2 & 3\end{array}\right]$ & $A^2=9 I$, then find $\alpha$.

• Solution.

• $\begin{aligned} A^2 & =\left[\begin{array}{ll} \alpha & 0 \\ 2 & 3 \end{array}\right]\left[\begin{array}{ll} \alpha & 0 \\ 2 & 3 \end{array}\right] \\ & =\left[\begin{array}{ll} \alpha^2 & 0 \\ 2 \alpha+6 & 9 \end{array}\right] \end{aligned}$

• $\begin{gathered} A^2=9 I \text { i.e., }\left[\begin{array}{ll} \alpha^2 & 0 \\ 2 \alpha+6 & 9 \end{array}\right]=\left[\begin{array}{ll} 9 & 0 \\ 0 & 9 \end{array}\right] \end{gathered}$

• $\Rightarrow \alpha^2=9 \quad \Rightarrow 2 \alpha+6=0 \Rightarrow \alpha=-3$

• $\begin{gathered} \text { If }\left[\begin{array}{lll} x & x & x \\ x & x & x \\ x & x & x \end{array}\right]\left[\begin{array}{lll} y & y & y \\ y & y & y \\ y & y & y \end{array}\right] =\frac{1}{3}\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right] \end{gathered}$

• then show that $x=1 / 9 y$.

• $\left[\begin{array}{lll}x & x & x \\ x & x & x \\ x & x & x\end{array}\right]\left[\begin{array}{lll}y & y & y \\ y & y & y \\ y & y & y\end{array}\right]=\left[\begin{array}{lll}1 / 3 & 1 / 3 & 1 / 3 \\ 1 / 3 & 1 / 3 & 1 / 6 \\ 1 / 3 & 1 / 3 & 1 / 3\end{array}\right]$

• $\begin{aligned} & {\left[\begin{array}{lll}x & x & x \\ x & x & x \\ x & x & x\end{array}\right]\left[\begin{array}{lll}y & y & y \\ y & y & y \\ y & y & y\end{array}\right]} =\left[\begin{array}{lll}3 x y & 3 x y & 3 x y \\ 3 x y & 3 x y & 3 x y \\ 3 x y & 3 x y & 3 x y\end{array}\right] \\ & 3 x y=1 / 3 \\ & \Rightarrow x y=1 / 9 \\ & \text { or } x=1 / 9 y .\end{aligned}$

If the equations, $\begin{aligned} x - 2y + 3z &= 0 \\ -2x + 2y + 2z &= 0 \\ -8x + \lambda y &= 0 \end{aligned}$

• have non-trivial solution, then find $\lambda$.

$\begin{aligned} & \text { Solution: }\left[\begin{array}{ccc} 1 & -2 & 3 \\ -2 & 3 & 2 \\ -8 & \lambda & 0 \end{array}\right] \\ & R_2 \rightarrow R_2+2 R_1\left[\begin{array}{ccc} 1 & -2 & 3 \\ 0 & -1 & 1 \\ 0 & \lambda-16 & 24 \end{array}\right] \end{aligned}$

• $\begin{aligned} & R_2 \rightarrow-R_2\left[\begin{array}{ccc} 1 & -2 & 3 \\ 0 & 1 & -1 \\ 0 & \lambda-16 & 24 \end{array}\right] \\ & R_1 \rightarrow R_1+2 R_2\left[\begin{array}{ccc} 1 & 0 & -13 \\ 0 & 1 & -8 \\ 0 & 0 & 24+(16-\lambda)(-2) \end{array}\right] \\ & R_3 \rightarrow R_3+(16-\lambda) R_2\left[\begin{array}{ccc} 1 & 0 & -13 \\ 0 & 1 & -8 \\ 0 & 0 & \lambda-104 \end{array}\right] \end{aligned}$

• For the existence of a non-trivial solution, $8 \lambda-104$ should be equal to 0.

• $\begin{aligned} \therefore & 8\lambda-104=0 \\ & \Rightarrow \lambda=104 / 8=13 . \end{aligned}$

• Thus $\lambda=13$, is the only value for which the given system will have a non-trivial solution.