Question

Choose the correct answer in the following questions: If \(\begin{bmatrix} α & β  \\ \gamma & -\alpha \end{bmatrix}\) is such that A²=I then  

• A. 1+α²+βγ=0
• B. 1-α²+βγ=0
• C. 1-α²-βγ=0
• D. 1+α²-βγ=0

Answer 1

The Correct Option is C

A=\(\begin{bmatrix} α & β  \\ \gamma & -\alpha \end{bmatrix}\)

A²=A.A =A=\(\begin{bmatrix} α & β  \\ \gamma & -\alpha \end{bmatrix}\)A=\(\begin{bmatrix} α & β  \\ \gamma & -\alpha \end{bmatrix}\)

= \(\begin{bmatrix} α^2+\beta\gamma & \alphaβ-\alpha\beta  \\ \alpha\gamma-\alpha\gamma & \beta\gamma+\alpha^2 \end{bmatrix}\) 

=\(\begin{bmatrix} α^2+\beta\gamma & 0  \\ 0& \beta\gamma+\alpha^2 \end{bmatrix}\)

Now A²=I ⇒ \(\begin{bmatrix} α^2+\beta\gamma & 0  \\ 0& \beta\gamma+\alpha^2 \end{bmatrix}\)= \(\begin{bmatrix} 1 & 0  \\ 0 & 1 \end{bmatrix}\)

On comparing the corresponding elements, we have: 

α²+βγ=1 

⇒α²+βγ-1=0 

⇒1-α²-βγ=0