If A=≤ft[{array}{ll}a & b \\ b & a{array}] and (A I)²=≤ft[{array}{ll}α & β \\ β & α{array}] where I is the identity matrix then

Correct Answer is : α=a2+b2,β=2abα=a²+b², β=2 a bα=a2+b2,β=2ab

Correct Answer is : α=a2+b2,β=2abα=a²+b², β=2 a bα=a2+b2,β=2ab

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COMEDK 2025 Evening Solved Paper

Question : 130 of 180

Marks: +1, -0

A=\left[\begin{array}{ll}a & b \\ b & a\end{array}\right]

(A I)^2=\left[\begin{array}{ll}\alpha & \beta \\ \beta & \alpha\end{array}\right]

where I is the identity matrix then

Solution:

Concept:

Matrix multiplication and the fact that

AI = A

, so

(AI)^2 = A^2

Explanation:

Given

A = \begin{pmatrix} a & b \\ b & a \end{pmatrix}

and

I

is the identity matrix.

Since

AI = A

, we have

(AI)^2 = A^2

Compute

A^2

A^2 = \begin{pmatrix} a & b \\ b & a \end{pmatrix} \begin{pmatrix} a & b \\ b & a \end{pmatrix}

Entry

(1,1): a \cdot a + b \cdot b = a^2 + b^2

Entry

(1,2): a \cdot b + b \cdot a = ab + ba = 2ab

Entry

(2,1): b \cdot a + a \cdot b = ba + ab = 2ab

Entry

(2,2): b \cdot b + a \cdot a = b^2 + a^2 = a^2 + b^2

Thus

A^2 = \begin{pmatrix} a^2+b^2 & 2ab \\ 2ab & a^2+b^2 \end{pmatrix}

Comparing with

\begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix}

, we get

\alpha = a^2+b^2

and

\beta = 2ab

Answer:

Option A:

\alpha = a^2 + b^2

\beta = 2ab

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