Math, asked by AnbalaganKAnbalaganK, 1 day ago

answers for this step by step explanation​

Attachments:

Answers

Answered by SparklingBoy
15

༒ 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

___________________________

Similar questions