Question

Let \(A=\) [\(a_{ij}\)]\(_{2\times2}\) be a matrix and \(A^2 = I\) where \(a_{ij} \neq0\). If a sum of diagonal elements and b=det(A), then \(3a^2+4b^2\) is

• A. 10
• B. 12
• C. 4
• D. 8

Answer 1

The Correct Option is C

The correct answer is (C) : 4
Let A \(=\begin{bmatrix} p & q \\r & s \end{bmatrix}\)
\(A^2=\begin{bmatrix} p^2+qr & pq+qs \\ pr+rs & qs+s^2 \end{bmatrix}\)
⇒ p² +qr=1 (1) pq + qs = 0
⇒ q(p+s) = 0 (3)
⇒ s² + qr =1 (2) pr + rs = 0
⇒ r(p+s) = 0 (4)
From , eqn (1) - eqn (2)
p² = s² ⇒ p+s=0
Now 3a² + 4b²
= 3(p+s)² + 4(ps-qr)
= 3.0 + 4(-p²-qr)²
= 4(p² + qr )²
= 4

Answer 2

Matrix Properties Calculation 

Given that $A^2 = I$, this implies that $|A^2| = |I|$. Since $|A^2| = |A|^2$ and $|I| = 1$, we have $|A|^2 = 1$, so $|A| = \pm 1 = b$.

Let $A = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}$. Then

$A^2 = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix} = \begin{bmatrix} \alpha^2 + \beta\gamma & \alpha\beta + \beta\delta \\ \alpha\gamma + \gamma\delta & \gamma\beta + \delta^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

From this, we get the following equations:

1. $\alpha^2 + \beta\gamma = 1$
2. $\alpha\beta + \beta\delta = 0 \implies \beta(\alpha + \delta) = 0$
3. $\alpha\gamma + \gamma\delta = 0 \implies \gamma(\alpha + \delta) = 0$
4. $\gamma\beta + \delta^2 = 1$

Since $a_{ij} \neq 0$, $\beta \neq 0$ and $\gamma \neq 0$. Therefore, from equations (2) and (3), we must have $\alpha + \delta = 0$, which means $\delta = -\alpha$. Substituting this into equations (1) and (4), we get:

1. $\alpha^2 + \beta\gamma = 1$
2. $\gamma\beta + \alpha^2 = 1$

These equations are consistent. The sum of the diagonal elements is $a = \alpha + \delta = \alpha - \alpha = 0$.

Then $3a^2 + 4b^2 = 3(0)^2 + 4(\pm 1)^2 = 4$.

Conclusion: $3a^2 + 4b^2 = 4$.