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) → \(x \gt y = z\)
Comparing the first matrix with the identity matrix \(\begin{bmatrix}1&0\\0&1\end{bmatrix}\):
\(x - y = 1,\quad x + y = 0\)
\( \displaystyle x = \frac{1}{2},\quad y = -\frac{1}{2} \)
A singular matrix has determinant zero:
\( \displaystyle \begin{vmatrix}x&y\\z&x\end{vmatrix} = x^2 - yz = 0 \)
\( \displaystyle \frac{1}{4} + \frac{z}{2} = 0,\quad z = -\frac{1}{2} \)
So \(x = \frac{1}{2}\) and \(y = z = -\frac{1}{2}\), which gives \(x \gt y = z\).