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Question:

If $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2\end{array}\right]$, then $a I+b A+2 A^{2}$ equals

(a) $A$

(b) $-A$

(c) $a b A$

(d) none of these

Solution:

(d) none of these

$A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{ccc}1+a & b & 3 \\ a & b & 2 \\ 3 a & 2 b & a+b+4\end{array}\right]$

Now,

$a I+b A+2 A^{2}=\left[\begin{array}{ccc}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right]+\left[\begin{array}{ccc}b & 0 & b \\ 0 & 0 & b \\ a b & b^{2} & 2 b\end{array}\right]+\left[\begin{array}{ccc}2+2 a & 2 b & 6 \\ 2 a & 2 b & 4 \\ 6 a & 6 b & 2 a+2 b+8\end{array}\right]$

$=\left[\begin{array}{ccc}3 a+b+2 & 2 b & b+6 \\ 2 a & a+2 b & b+4 \\ a b+6 a & b^{2}+6 b & 3 a+4 b+8\end{array}\right]$

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