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Provide solution for RD Sharma maths class12 Chapter Algebra of Matrices exercise 4.3 question 3 sub question (ii)

Answers (1)

AB= \begin{bmatrix} 12 &17 &22 \\ -4 & -5 & -6\\ -4& -4 &-4 \end{bmatrix} and BA= \begin{bmatrix} 1 &14 \\ -3 &2 \end{bmatrix} 

Hint: matrix multiplication is only possible, when number of columns of first matrix is equal to the number of rows of second matrix.

Given: A=\begin{bmatrix} 3 &2 \\ -1 &0 \\ -1 & 1 \end{bmatrix}  andB=\begin{bmatrix} 4 &5 &6 \\ 0& 1 &2 \end{bmatrix}

Consider,

AB=\begin{bmatrix} 3 &2 \\ -1 &0 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 4 &5 &6 \\ 0& 1 &2 \end{bmatrix} 

\begin{aligned} &A B=\left[\begin{array}{ccc} 3 \times 4+2 \times 0 & 3 \times 5+2 \times 1 & 3 \times 6+2 \times 2 \\ -1 \times 4+0 \times 0 & -1 \times 5+0 \times 1 & -1 \times 6+0 \times 2 \\ -1 \times 4+1 \times 0 & -1 \times 5+1 \times 1 & -1 \times 6+1 \times 2 \end{array}\right] \\ &A B=\left[\begin{array}{lll} 12+0 & 15+2 & 18+4 \\ -4+0 & -5+0 & -6+0 \\ -4+0 & -5+1 & -6+2 \end{array}\right] \end{aligned}

AB= \begin{bmatrix} 12 &17 &22 \\ -4 & -5 & -6\\ -4& -4 &-4 \end{bmatrix}

Again consider

BA=\begin{bmatrix} 4 &5 &6 \\ 0& 1 &2 \end{bmatrix}\begin{bmatrix} 3 &2 \\ -1 &0 \\ -1 & 1 \end{bmatrix}

\begin{aligned} &B A=\left[\begin{array}{ll} 4 \times 3+5 \times(-1)+6 \times(-1) & 4 \times 2+5 \times 0+6 \times 1 \\ 0 \times 3+1 \times(-1)+2 \times(-1) & 0 \times 2+1 \times 0+2 \times 1 \end{array}\right] \\ B A & =\left[\begin{array}{cc} 12-5-6 & 8+0+6 \\ 0-1-2 & 0+0+2 \end{array}\right] \end{aligned}

BA= \begin{bmatrix} 1 &14 \\ -3 &2 \end{bmatrix}

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