Question

help me to solve this matrix problem ​

Answer 1

Step-by-step explanation:

I EXPLAINED IT IN IMAGE

MARK IT BRAINLEAST ANSWER BRO

PLEASEEEEEEEW

CLICK THE IMAGE

AND VERIFY IT

Answer 2

Given :-

A = $\left[\begin{array}{c c } 2&0 \\-1&7 \end{array}\right]$  

and $I = \left[\begin{array}{c c} 1 & 0 \\0 & 1\end{array}\right]$

$\sf{A^{2} = 9A + mI }$

To Find :-

Value of m .

SOLUTION :-

Firstly we will find out the value of $A^{2}$

$A \times A = \left[\begin{array}{c c } 2&-0 \\-1&7 \end{array}\right] \left[\begin{array}{c c } 2&-0 \\-1&7 \end{array}\right]$

Now the matrix will be like this :-

$\implies \left[\begin{array}{c c } a_{11}&a_{12} \\a_{21}&a_{22} \end{array}\right]$

$\implies a_{11} = [2 \times 2] + [0 \times -1] = 4$

$\implies a_{12} = [2 \times 0] + [0 \times 7] = 0$

$\implies a_{21} = [-1 \times 2] + [ 7 \times -1] = -9$

$\implies a_{22} = [-1 \times 0] + [ 7 \times 7] = 49$

So our Matrix is :-

$\implies A^{2} = \left[\begin{array}{c c } 4&0 \\-9&49 \end{array}\right]$

Now $A^{2} = 9A + mI$

$\implies \left[\begin{array}{c c } 4&0 \\-9&49 \end{array}\right] = 9 \left[\begin{array}{c c } 2&0 \\-1&7 \end{array}\right] + m \left[\begin{array}{c c } 1&0 \\0&1 \end{array}\right]$

$\implies \left[\begin{array}{c c } 4&0 \\-9&49 \end{array}\right] = \left[\begin{array}{c c } 18&0 \\-9&63 \end{array}\right] + m\left[\begin{array}{c c } 1&0 \\0&1 \end{array}\right]$

Taking value of a left hand side

$\implies \left[\begin{array}{c c } 4&0 \\-9&49 \end{array}\right] - \left[\begin{array}{c c } 18&0 \\-9&63 \end{array}\right] = m \left[\begin{array}{c c } 1&0 \\0&1 \end{array}\right]$

$\implies \left[\begin{array}{c c } -14&0 \\0&-14 \end{array}\right] = m \left[\begin{array}{c c } 1&0 \\0&1 \end{array}\right]$

Taking -14 common

$\implies - 14 \left[\begin{array}{c c } 1&0 \\0&1 \end{array}\right]= m \left[\begin{array}{c c } 1&0 \\0&1 \end{array}\right]$

Now by comparing both sides we got

$\boxed{\sf{\implies\; m = -14}}$