Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Matrices

Question:

If A = \(\begin{bmatrix}\alpha &\beta \\-\beta &\alpha \end{bmatrix}\) and \(A+ { A }^{ T } \) = \(\begin{bmatrix}0&1\\1&0\end{bmatrix}\), then the value of \(\alpha and \beta \) is

Options:

\(\beta = \frac{1}{2}\) and \(\alpha = 0\)

\(\beta = \frac{-1}{2}\) and \(\alpha = 0\)

Cannot determine values of \(\alpha \)

None of the above

Correct Answer:

Cannot determine values of \(\alpha \)

Explanation:

A = \(\begin{bmatrix}\alpha &\beta \\-\beta &\alpha \end{bmatrix}\)

\( { A }^{ T } \) = \(\begin{bmatrix}\alpha &-\beta \\\beta &\alpha \end{bmatrix}\) 

\(A+ { A }^{ T } \) = \(\begin{bmatrix}\alpha &\beta \\-\beta &\alpha \end{bmatrix}\) + \(\begin{bmatrix}\alpha &-\beta \\\beta &\alpha \end{bmatrix}\) = \(\begin{bmatrix}0&1\\1&0\end{bmatrix}\)

\(A+ { A }^{ T } \) = \(\begin{bmatrix}0 &2\beta\\-2\beta &0 \end{bmatrix}\) = \(\begin{bmatrix}0&1\\1&0\end{bmatrix}\)

Equating all the corresponding terms of both the matrices we get

2\(\beta \) = 1     and so \(\beta \) = \(\frac{1}{2}\)

and also

\(\alpha - \alpha = 0\)

and so we cannot determine the values of the first variable.