Choose the correct answer in the following questions: If \(\begin{bmatrix} α & β \\ \gamma & -\alpha \end{bmatrix}\) is such that A2=I then
1+α2+βγ=0
1-α2+βγ=0
1-α2-βγ=0
1+α2-βγ=0
The Correct Option is C
Solution and Explanation
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
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