CUET Preparation Today
$\begin{bmatrix}1-x^2 & 2y\\x+5 & 0\end{bmatrix}=\begin{bmatrix} -3 & y-1\\7 & 0\end{bmatrix}$, then value of $(x-y)^2$ is : |
9 4 1 16 |
9 |
The correct answer is Option (1) → 9 $1-x^2=-3,x+5=7$ $x^2=4$ $x=±2$ $∴x=2$ $y-1=2y$ $y=-1$ $∴(x-y)^2=(2-(1))^2=3^2=9$ |
