Question

answers for this step by step explanation​

Answer 1

༒ Given ➽

$\bf X= \begin{bmatrix}4&1 \\ - 1&2 \end{bmatrix}$

___________________________

༒ To Prove ➽

$\bold{6 X - {X}^{2} - 9I =O}$

___________________________

༒ Proof ➽

${X}^{2} = X.X \\ \\ = \sf\begin{bmatrix}4&1 \\ - 1&2 \end{bmatrix}\begin{bmatrix}4&1 \\ - 1&2 \end{bmatrix} \\ \\ = \begin{bmatrix}16 - 1& 4 + 2\\ - 4 - 2& - 1 + 4 \end{bmatrix} \\ \\ \Large \purple{ \bold {X {}^{2} = \begin{bmatrix}15&6 \\ - 6&3 \end{bmatrix}}}$

Now,

$6X = 6 \bigg(\begin{bmatrix}4&1 \\ - 1&2 \end{bmatrix} \bigg) \\ \\ \Large \purple{ \bold 6X = \begin{bmatrix}24&6 \\ - 6&12 \end{bmatrix}}$

Also,

$9I = 9 \bigg(\begin{bmatrix}1&0 \\ 0&1\end{bmatrix}\bigg) \\ \\ \Large \purple{ \bold 9I = \begin{bmatrix}9&0 \\ 0&9\end{bmatrix}}$

Hence,

$6 X - {X}^{2} - 9I \\ \\ = \begin{bmatrix}24&6 \\ - 6&12 \end{bmatrix} - \begin{bmatrix}15&6\\ - 6&3 \end{bmatrix} - \begin{bmatrix}9&0 \\ 0&9 \end{bmatrix} \\ \\ = \begin{bmatrix}24 - 15 - 9 \: \: \: &6 - 6 - 0\\ - 6 + 6 - 0 \: \: \: &12 - 3 - 9 \end{bmatrix} \\ \\ =\sf \begin{bmatrix}0&0\\ 0&0 \end{bmatrix} = O$

Hence Proved

___________________________