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Question : 15 of 69

Marks: +1, -0

Let

A=\begin{bmatrix}0&2\\-2&0\end{bmatrix}

(mI+nA)^{2}=A

m, n

are positive real numbers and

I

is the identity matrix. What is

(m+n)

0
$\frac{1}{2}$
1
$\frac{3}{2}$

Solution: 👈: Video Solution

(d) Let

A=\begin{bmatrix}0&2\\-2&0\end{bmatrix}

and

(ml+nA)^{2}=A

where, / is Identify matrix

\because m l+n A = m\begin{bmatrix}1&0\\0&1\end{bmatrix}+n\begin{bmatrix}0&2\\-2&0\end{bmatrix}

=\begin{bmatrix}m&0\\0&m\end{bmatrix}+\begin{bmatrix}0&2n\\-2n&0\end{bmatrix}

=\begin{bmatrix}m&2n\\-2n&m\end{bmatrix}

\therefore (ml+nA)^{2}=A

\begin{bmatrix}m&2n\\-2n&m\end{bmatrix}\begin{bmatrix}m&2n\\-2n&m\end{bmatrix}=\begin{bmatrix}0&2\\-2&0\end{bmatrix}

\begin{bmatrix}m^{2}-4n^{2}&4mn\\-4mn&m^{2}-4n^{2}\end{bmatrix}=\begin{bmatrix}0&2\\-2&0\end{bmatrix}

\Rightarrow 4mn=2 \text{ and } m^{2}-4n^{2}=0

mn=\frac{1}{2} \text{ and } m=\pm 2n

\text{When, } m=2n

(2n)(n)=\frac{1}{2}

n=\pm \frac{1}{2} \Rightarrow m=\pm 1

\therefore m+n=1+\frac{1}{2}=\frac{3}{2}

Hence, option (d) is correct.

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