If A=1−12−1,B=a−1b−1 and (A+B)2=A2+B2, then (a,b)=

A=1-12-1,B=a-1b-1(A+B)2=A2+B2⇒1+a−22+b−22=1−12−12+a−1b−12L.H.S1+a−22+b−21+a−22+b−2=a2+1+2a−4−2b−2−2a+42a+2+ab+b−4−2b−2(2+b)+4=a2+2a−2b−3−2a+22a+ab−b−2−2bR.H.S=1−12−11−12−1+a−1b−1a−1b−1=−100−1+a2−b−a+1ab−b−b+1=a2−b−11−aab−b−bL.H.S=R.H.Sa12⇒−2a+2=1−a⇒−a=−1⇒a=1a 22⇒−2b=−b⇒b=0∴(a,b)=(1,0)

If A=1−12−1,B=a−1b−1 and (A+B)2=A2+B2, then (a,b)=

A=1-12-1,B=a-1b-1(A+B)2=A2+B2⇒1+a−22+b−22=1−12−12+a−1b−12L.H.S1+a−22+b−21+a−22+b−2=a2+1+2a−4−2b−2−2a+42a+2+ab+b−4−2b−2(2+b)+4=a2+2a−2b−3−2a+22a+ab−b−2−2bR.H.S=1−12−11−12−1+a−1b−1a−1b−1=−100−1+a2−b−a+1ab−b−b+1=a2−b−11−aab−b−bL.H.S=R.H.Sa12⇒−2a+2=1−a⇒−a=−1⇒a=1a 22⇒−2b=−b⇒b=0∴(a,b)=(1,0)

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If $A = [\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}], B = [\begin{matrix} a & - 1 \\ b & - 1 \end{matrix}]$ and ${(A + B)}^{2} = A^{2} + B^{2}$ , then $(a, b) =$

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$(2, 0)$

$(1, 0)$

$(0, 1)$

$(0, 2)$

answer is B.

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Detailed Solution

$A = [\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}], B = [\begin{matrix} a & - 1 \\ b & - 1 \end{matrix}]$

${(A + B)}^{2} = A^{2} + B^{2}$

$\Rightarrow {[\begin{matrix} 1 + a & - 2 \\ 2 + b & - 2 \end{matrix}]}^{2} = {[\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}]}^{2} + {[\begin{matrix} a & - 1 \\ b & - 1 \end{matrix}]}^{2}$

$L.H.S$

$[\begin{matrix} 1 + a & - 2 \\ 2 + b & - 2 \end{matrix}] [\begin{matrix} 1 + a & - 2 \\ 2 + b & - 2 \end{matrix}] = [\begin{matrix} a^{2} + 1 + 2 a - 4 - 2 b & - 2 - 2 a + 4 \\ 2 a + 2 + a b + b - 4 - 2 b & - 2 (2 + b) + 4 \end{matrix}]$

$= [\begin{matrix} a^{2} + 2 a - 2 b - 3 & - 2 a + 2 \\ 2 a + a b - b - 2 & - 2 b \end{matrix}]$

$R.H.S$

$= [\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}] [\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}] + [\begin{matrix} a & - 1 \\ b & - 1 \end{matrix}] [\begin{matrix} a & - 1 \\ b & - 1 \end{matrix}]$

$= [\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix}] + [\begin{matrix} a^{2} - b & - a + 1 \\ a b - b & - b + 1 \end{matrix}] = [\begin{matrix} a^{2} - b - 1 & 1 - a \\ a b - b & - b \end{matrix}]$