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

Answer: (C)

4

Answer 2

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