CUET Preparation Today
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a b c d |
c |
$\begin{bmatrix}1&x\end{bmatrix}\begin{bmatrix}2&-1\\1&2\end{bmatrix}\begin{bmatrix}1\\3\end{bmatrix}=[0]$ $⇒\begin{bmatrix}1&x\end{bmatrix}\begin{bmatrix}-1\\7\end{bmatrix}=[0]$ $⇒\begin{bmatrix}-1+7x\end{bmatrix}=[0]$ $⇒x=\frac{1}{7}$ |
