Question

If $A = \begin{bmatrix} 1 & - 1 \\ 2 & - 1 \end{bmatrix},B = \begin{bmatrix} a & 1 \\ b & - 1 \end{bmatrix}$and $(A + B)^{2} = A^{2} + B^{2}$then value of a and b are

• A. $a = 4,b = 1$
• B. $a = 1,b = 4$
• C. $a = 0,b = 4$
• D. $a = 2,b = 4$

Answer 1

Answer: (B)

$a = 1,b = 4$

Answer 2

We have $(A + B)^{2} = A^{2} + B^{2} + A.B + BA$

∵ $AB + BA = 0$

∴ $\left[ \begin{array} { c c } a - b & 2 \\ 2 a - b & 3 \end{array} \right] + \left[ \begin{array} { l l } a + 2 & - a - 1 \\ b - 2 & - b + 1 \end{array} \right] = 0$ $\begin{bmatrix} 2a + 2 - b & - a + 1 \\ 2a - 2 & 4 - b \end{bmatrix} = 0$. On comparing, we get, $- a + 1 = 0$⇒ $a = 1$ and $4 - b = 0$

⇒ $b = 4$