Practicing Success
If B = \(\begin{bmatrix}x&0\\0&y\end{bmatrix}\) then the value of B-B |
\(\begin{bmatrix}x^2&0\\0&y^2\end{bmatrix}\) \(\begin{bmatrix}2x&0\\0&2y\end{bmatrix}\) \(\begin{bmatrix}0&0\\0&0\end{bmatrix}\) None of these |
\(\begin{bmatrix}0&0\\0&0\end{bmatrix}\) |
Given that B = \(\begin{bmatrix}x&0\\0&y\end{bmatrix}\) Then B-B = \(\begin{bmatrix}x&0\\0&y\end{bmatrix}\) - \(\begin{bmatrix}x&0\\0&y\end{bmatrix}\) = \(\begin{bmatrix}x-x&0\\0&y-y\end{bmatrix}\) = \(\begin{bmatrix}0&0\\0&0\end{bmatrix}\) |