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Question:

If $A=\left[\begin{array}{rr}2 & 3 \\ -1 & 0\end{array}\right]$, show that $A^{2}-2 A+3 / 2=0$

Solution:

Given : $A=\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right]$

Now,

$A^{2}=A A$

$\Rightarrow A^{2}=\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{cc}4-3 & 6+0 \\ -2+0 & -3+0\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{cc}1 & 6 \\ -2 & -3\end{array}\right]$

$A^{2}-2 A+3 I_{2}$

$\Rightarrow A^{2}-2 A+3 I_{2}=\left[\begin{array}{cc}1 & 6 \\ -2 & -3\end{array}\right]-2\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right]+3\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow A^{2}-2 A+3 I_{2}=\left[\begin{array}{cc}1 & 6 \\ -2 & -3\end{array}\right]-\left[\begin{array}{cc}4 & 6 \\ -2 & 0\end{array}\right]+\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]$

$\Rightarrow A^{2}-2 A+3 I_{2}=\left[\begin{array}{cc}1-4+3 & 6-6+0 \\ -2+2+0 & -3+0+3\end{array}\right]$

$\Rightarrow A^{2}-2 A+3 I_{2}=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0$

Hence proved.

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