If $\begin{bmatrix}x-y&0\\x+y&1\end{bmatrix}$ is an identity matrix and $\begin{bmatrix}x&y\\z&x\end{bmatrix}$ is a singular matrix then:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $x > y = z$
Given:
$\begin{bmatrix} x - y & 0 \\ x + y & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \Rightarrow x - y = 1,\ x + y = 0$
$\Rightarrow x = \frac{1}{2},\ y = -\frac{1}{2}$
Singular matrix condition: $\begin{vmatrix} x & y \\ z & x \end{vmatrix} = 0 \Rightarrow x^2 - yz = 0$
$\Rightarrow \left(\frac{1}{2}\right)^2 - (-\frac{1}{2})z = 0 \Rightarrow \frac{1}{4} + \frac{z}{2} = 0 \Rightarrow z = -\frac{1}{2}$
Hence, $x = \frac{1}{2},\ y = -\frac{1}{2},\ z = -\frac{1}{2} \Rightarrow x > y = z$