Question

#Quality Question
@Matrices

If $X=  \begin{bmatrix} 4&1 \\  - 1&2 \end{bmatrix}\ \textless \ br /\ \textgreater$
then prove that

$6X - {X}^{2} - 9I = O$
Where I is unit matrix of order 2.​

Answer 1

▪ Given :-

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

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▪To Prove :-

$\bold{6 X -  {X}^{2}  - 9I  =O}$

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▪ 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!

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Answer 2

See the Answer

${X}^{2}    = \begin{bmatrix}4&1 \\  - 1&2 \end{bmatrix}\begin{bmatrix}4&1 \\  - 1&2 \end{bmatrix} \\  \\  =  \begin{bmatrix}16 - 1& 4 + 2\\  - 4 - 2& - 1 + 4 \end{bmatrix} \\  \\ \large  { {X {}^{2}  = \begin{bmatrix}15&6 \\  - 6&3 \end{bmatrix}}}$

$6X = \begin{bmatrix}4&1 \\  - 1&2 \end{bmatrix} \\  \\ \large { 6X  = \begin{bmatrix}24&6 \\  - 6&12 \end{bmatrix}}$

$9I = 9 \begin{bmatrix}1&0 \\   0&1\end{bmatrix}\\  \\ \large {  9I = \begin{bmatrix}9&0 \\  0&9\end{bmatrix}}$

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

Mark it as brainliest.