Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Matrices

Question:

If B = \(\begin{bmatrix}x&0\\0&y\end{bmatrix}\) then the value of B-B

Options:

\(\begin{bmatrix}x^2&0\\0&y^2\end{bmatrix}\)

\(\begin{bmatrix}2x&0\\0&2y\end{bmatrix}\)

\(\begin{bmatrix}0&0\\0&0\end{bmatrix}\)

None of these

Correct Answer:

\(\begin{bmatrix}0&0\\0&0\end{bmatrix}\)

Explanation:

Given that B = \(\begin{bmatrix}x&0\\0&y\end{bmatrix}\)

Then B-B = \(\begin{bmatrix}x&0\\0&y\end{bmatrix}\) - \(\begin{bmatrix}x&0\\0&y\end{bmatrix}\)

              = \(\begin{bmatrix}x-x&0\\0&y-y\end{bmatrix}\)

               = \(\begin{bmatrix}0&0\\0&0\end{bmatrix}\)