First slide

Operations of matrix

Question

If $A = [\begin{array}{rr} 1 - 1 \\ 2 - 1 \end{array}], B = [\begin{array}{rr} a 1 \\ b - 1 \end{array}]$
and $(A + B)^{2} = A^{2} + B^{2}$ then the values of $a$ and $b$ are

Moderate

A

$a = 4, b = 1$
B

$a = 1, b = 4$
C

$a = 0, b = 4$
D

$a = 2, b = 4$

Solution

We have,
$A + B = [\begin{array}{rr} a + 1 0 \\ b + 2 - 2 \end{array}]$
$\begin{array}{l} A^{2} = [\begin{array}{lr} 1 & - 1 \\ 2 & - 1 \end{array}] [\begin{array}{lr} 1 & - 1 \\ 2 & - 1 \end{array}] = [\begin{array}{rr} - 1 & 0 \\ 0 & - 1 \end{array}] \\ B^{2} = [\begin{array}{rr} a & 1 \\ b & - 1 \end{array}] [\begin{array}{rr} a & 1 \\ b & - 1 \end{array}] = [\begin{array}{rr} a^{2} + b & a - 1 \\ a b - b & b + 1 \end{array}] \\ (A + B)^{2} = [\begin{array}{rr} a + 1 & 0 \\ b + 2 & - 2 \end{array}] [\begin{array}{rr} a + 1 & 0 \\ b + 2 & - 2 \end{array}] = [\begin{array}{c} (a + 1)^{2} & 0 \\ (a - 1) (b + 2) & 4 \end{array}] \\ ∴ (A + B)^{2} = A^{2} + B^{2} \\ \Rightarrow [\begin{array}{cc} (a + 1)^{2} & 0 \\ (a - 1) (b + 2) & 4 \end{array}] = [\begin{array}{rr} - 1 & 0 \\ 0 & - 1 \end{array}] + [\begin{array}{cc} a^{2} + b & a - 1 \\ a b - b & b + 1 \end{array}] \\ \Rightarrow [\begin{array}{cc} (a + 1)^{2} & 0 \\ (a - 1) (b + 2) & 4 \end{array}] = [\begin{array}{cc} a^{2} + b - 1 & a - 1 \\ a b - b & b \end{array}] \\ \Rightarrow a - 1 = 0, b = 4, (a + 1)^{2} = a^{2} + b - 1, (a - 1) (b + 2) = a b - b \end{array}$
$\Rightarrow a = 1$ and $b = 4$

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