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Question:

If $A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, then find $\lambda$ so that $A^{2}=5 A+N$.

Solution:

Given : $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$

Now,

$A^{2}=A A$

$\Rightarrow A^{2}=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]$

$A^{2}=5 A+\lambda I$

$\Rightarrow\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]=5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]+\lambda\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]=\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]+\left[\begin{array}{ll}\lambda & 0 \\ 0 & \lambda\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]=\left[\begin{array}{cc}15+\lambda & 5+0 \\ -5+0 & 10+\lambda\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]=\left[\begin{array}{cc}15+\lambda & 5 \\ -5 & 10+\lambda\end{array}\right]$

The corresponding elements of two equal matrices are equal.

$\therefore 8=15+\lambda$

$\Rightarrow 8-15=\lambda$

$\Rightarrow-7=\lambda$

$\therefore \lambda=-7$

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