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If a square matrix $A$ is such that $\left(A^T-\frac{1}{2} I\right)\left(A-\frac{1}{2} I\right)$ $=\left(A^T+\frac{1}{2} I\right)\left(A+\frac{1}{2} I\right)=I$, where $I$ is a unit matrix, then $A$ is
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Verified Answer
The correct answer is:
skew-symmetric matrix
$$
\begin{aligned}
\text { Given, } & \left(A^T-\frac{1}{2} I\right)\left(A-\frac{1}{2} I\right) \\
= & \left(A^T+\frac{1}{2} I\right)\left(A+\frac{1}{2} I\right)=I
\end{aligned}
$$
Taking starting two, we get
$$
\begin{aligned}
& A^T A-\frac{1}{2} A^T I-\frac{1}{2} I A+\frac{1}{4} I^2=A^T A+\frac{1}{2} A^T I+\frac{1}{2} I A \\
& +\frac{1}{4} I^2 \\
& \Rightarrow-\frac{1}{2} A^T-\frac{1}{2} A=\frac{1}{2} A^T+\frac{1}{2} A
\end{aligned}
$$
[U sing cancellation law) and $I A=I$ ]
$$
\begin{aligned}
\Rightarrow & & 0 & =\frac{1}{2} A^{\mathrm{T}}+\frac{1}{2} A^T+\frac{1}{2} A+\frac{1}{2} A \\
& & & =A^T+A \\
\Rightarrow & & A^T & =-A
\end{aligned}
$$
$\Rightarrow A$ is skew-symmetric matrix.
\begin{aligned}
\text { Given, } & \left(A^T-\frac{1}{2} I\right)\left(A-\frac{1}{2} I\right) \\
= & \left(A^T+\frac{1}{2} I\right)\left(A+\frac{1}{2} I\right)=I
\end{aligned}
$$
Taking starting two, we get
$$
\begin{aligned}
& A^T A-\frac{1}{2} A^T I-\frac{1}{2} I A+\frac{1}{4} I^2=A^T A+\frac{1}{2} A^T I+\frac{1}{2} I A \\
& +\frac{1}{4} I^2 \\
& \Rightarrow-\frac{1}{2} A^T-\frac{1}{2} A=\frac{1}{2} A^T+\frac{1}{2} A
\end{aligned}
$$
[U sing cancellation law) and $I A=I$ ]
$$
\begin{aligned}
\Rightarrow & & 0 & =\frac{1}{2} A^{\mathrm{T}}+\frac{1}{2} A^T+\frac{1}{2} A+\frac{1}{2} A \\
& & & =A^T+A \\
\Rightarrow & & A^T & =-A
\end{aligned}
$$
$\Rightarrow A$ is skew-symmetric matrix.
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