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NCERT Exemplar · Q42

Q.If A=[1241]A = \begin{bmatrix}1 & 2\\ 4 & 1\end{bmatrix}, find A2+2A+7IA^2 + 2A + 7I.

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Compute A2=[9489]A^2=\begin{bmatrix}9 & 4\\8 & 9\end{bmatrix}, then add 2A2A and 7I7I entry-wise: A2+2A+7I=[1881618]A^2+2A+7I=\begin{bmatrix}18 & 8\\16 & 18\end{bmatrix}.

Step 1 — A2=A⋅AA^2=A\cdot A.

A2=[1241][1241]=[1+82+24+48+1]=[9489].A^2=\begin{bmatrix}1 & 2\\4 & 1\end{bmatrix}\begin{bmatrix}1 & 2\\4 & 1\end{bmatrix} =\begin{bmatrix}1+8 & 2+2\\4+4 & 8+1\end{bmatrix} =\begin{bmatrix}9 & 4\\8 & 9\end{bmatrix}.

Step 2 — 2A2A and 7I7I.

2A=[2482],7I=[7007].2A=\begin{bmatrix}2 & 4\\8 & 2\end{bmatrix},\qquad 7I=\begin{bmatrix}7 & 0\\0 & 7\end{bmatrix}. …

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