If for the matrix, A = [ \beginmatrix 1 & - \alpha \alpha & \beta \endmatrix ] , AA^T = I_2 , then the value of \alpha ^4 + \beta ^4 is :

Matrices and Determinants PYQ — JEE Main 2021 · 25 Feb · Shift 2 · Q37 with Solution

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Correct answer: C

We are given $A=\begin{bmatrix}1&-\alpha\\ \alpha&\beta\end{bmatrix}$ and $AA^T=I_2.$

This means the rows of $A$ are orthonormal.

So,

\begin{bmatrix}1&-\alpha\\ \alpha&\beta\end{bmatrix} \begin{bmatrix}1&\alpha\\ -\alpha&\beta\end{bmatrix}.$$ 3. Multiply: $$AA^T= \begin{bmatrix} 1+\alpha^2 & \alpha-\alpha\beta\\[4pt] \alpha-\alpha\beta & \alpha^2+\beta^2 \end{bmatrix}.$$ Since $AA^T=I_2$, we compare entries with $$I_2=\begin{bmatrix}1&0\\0&1\end{bmatrix}.$$ Thus, - From $(1,1)$ entry: $$1+\alpha^2=1 \implies \alpha^2=0 \implies \alpha=0.$$ - From $(2,2)$ entry: $$\alpha^2+\beta^2=1 \implies 0+\beta^2=1 \implies \beta^2=1.$$ So, $$\beta=\pm 1.$$ 4. Now calculate $$\alpha^4+\beta^4=0^4+(\pm1)^4=1.$$ 5. Check options: - A: $3$ ❌ - B: $2$ ❌ - C: $1$ ✅ - D: $4$ ❌ Therefore, the correct answer is **C**.

View written solutionFree

Correct answer: C

We are given $A=\begin{bmatrix}1&-\alpha\\ \alpha&\beta\end{bmatrix}$ and $AA^T=I_2.$

This means the rows of $A$ are orthonormal.

So,

\begin{bmatrix}1&-\alpha\\ \alpha&\beta\end{bmatrix} \begin{bmatrix}1&\alpha\\ -\alpha&\beta\end{bmatrix}.$$ 3. Multiply: $$AA^T= \begin{bmatrix} 1+\alpha^2 & \alpha-\alpha\beta\\[4pt] \alpha-\alpha\beta & \alpha^2+\beta^2 \end{bmatrix}.$$ Since $AA^T=I_2$, we compare entries with $$I_2=\begin{bmatrix}1&0\\0&1\end{bmatrix}.$$ Thus, - From $(1,1)$ entry: $$1+\alpha^2=1 \implies \alpha^2=0 \implies \alpha=0.$$ - From $(2,2)$ entry: $$\alpha^2+\beta^2=1 \implies 0+\beta^2=1 \implies \beta^2=1.$$ So, $$\beta=\pm 1.$$ 4. Now calculate $$\alpha^4+\beta^4=0^4+(\pm1)^4=1.$$ 5. Check options: - A: $3$ ❌ - B: $2$ ❌ - C: $1$ ✅ - D: $4$ ❌ Therefore, the correct answer is **C**.