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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 4 Algebra of Matrices Exercise Multiple Choice Questions Question 13 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 4 Algebra of Matrices Exercise Multiple Choice Questions  Question 13 Maths Textbook Solution.

Answers (1)

Answer: The correct option is (B).

Given:A=\left[\begin{array}{rr} 1 & -1 \\ 2 & -1 \end{array}\right], B=\left[\begin{array}{cc} a & 1 \\ b & -1 \end{array}\right] \text { and }(A+B)^{2}=A^{2}+B^{2}

Solution: \begin{aligned} A+B &=\left[\begin{array}{ll} 1 & -1 \\ 2 & -1 \end{array}\right]+\left[\begin{array}{cc} a & 1 \\ b & -1 \end{array}\right] \\ &=\left[\begin{array}{cc} a+1 & 0 \\ b+2 & -2 \end{array}\right] \end{aligned}

                    (A+B)^{2}=\left[\begin{array}{cc} a+1 & 0 \\ b+2 & -2 \end{array}\right] \times\left[\begin{array}{cc} a+1 & 0 \\ b+2 & -2 \end{array}\right]

\begin{aligned} &=\left[\begin{array}{cc} (a+1)^{2}+0 & 0+0 \\ (b+2)(a+1)-4-b & 0+4 \end{array}\right] \\ &=\left[\begin{array}{cc} a^{2}+2 a+1 & 0 \\ 2+2 a+b+a b-4-b & 4 \end{array}\right] \\ &=\left[\begin{array}{cc} a^{2}+2 a+1 & 0 \\ 2 a+a b-2 & 4 \end{array}\right] \end{aligned}

\begin{aligned} A^{2} &=\left[\begin{array}{rr} 1 & -1 \\ 2 & -1 \end{array}\right] \times\left[\begin{array}{rr} 1 & -1 \\ 2 & -1 \end{array}\right] \\ &=\left[\begin{array}{rr} 1-2 & -1+1 \\ 2-2 & -2+1 \end{array}\right] \\ A^{2} &=\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right] \end{aligned}

\begin{aligned} &B^{2}=\left[\begin{array}{cc} a & 1 \\ b & -1 \end{array}\right] \times\left[\begin{array}{cc} a & 1 \\ b & -1 \end{array}\right] \\ &B^{2}=\left[\begin{array}{ll} a^{2}+b & a-1 \\ a b-b & b+1 \end{array}\right] \\ &A^{2}+B^{2}=\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right]+\left[\begin{array}{cc} a^{2}+b & a-1 \\ a b-b & b+1 \end{array}\right] \\ &A^{2}+B^{2}=\left[\begin{array}{cc} a^{2}+b-1 & a-1 \\ a b-b & b \end{array}\right] \end{aligned}

As, A^{2}+B^{2}=(A+B)^{2}

\begin{aligned} &{\left[\begin{array}{cc} a^{2}+b-1 & a-1 \\ a b-b & b \end{array}\right]=\left[\begin{array}{cc} a^{2}+2 a+1 & 0 \\ 2 a+a b-2 & 4 \end{array}\right]} \\ &a-1=0 \\ &a=1 \\ &b=4 \end{aligned}

So, the correct option is (B).

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