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  • Provide solution for RD Sharma maths class12 Chapter Algebra of Matrices exercise 4.3 question 4 sub question (ii)

Provide solution for RD Sharma maths class12 Chapter Algebra of Matrices exercise 4.3 question 4 sub question (ii)

Answers (1)

Hence proved AB\neq BA

\begin{bmatrix} -3 &1 &14 \\ 4& -2 &-17 \\ -5& 1 & 15 \end{bmatrix}\neq \begin{bmatrix} 6 & -4 &1 \\ -5& 3 & -2\\ -5& 1 &1 \end{bmatrix}

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

Given: =A=\begin{bmatrix} 10 &-4 & -1\\ -11 & 5 & 0\\ 9&-5 &1 \end{bmatrix}, B = \begin{bmatrix} 1 &2 &2 \\ 3& 4 & 1\\ 1& 3 & 2 \end{bmatrix}

Consider,

AB=\begin{bmatrix} 10 &-4 & -1\\ -11 & 5 & 0\\ 9&-5 &1 \end{bmatrix} \begin{bmatrix} 1 &2 &2 \\ 3& 4 & 1\\ 1& 3 & 2 \end{bmatrix}

\begin{aligned} &A B=\left[\begin{array}{ccc} 10 \times 1+(-4) \times 3+(-1) \times 1 & 10 \times 2+(-4) \times 4+(-1) \times 3 & 10 \times 2+(-4) \times 1+(-1) \times 2 \\ -11 \times 1+5 \times 3+0 \times 1 & -11 \times 2+5 \times 4+0 \times 3 & -11 \times 2+5 \times 1+0 \times 2 \\ 9 \times 1+(-5) \times 3+1 \times 1 & 9 \times 2+(-5) \times 4+1 \times 3 & 9 \times 2+(-5) \times 1+1 \times 2 \end{array}\right] \\ &A B=\left[\begin{array}{ccc} 10-12-1 & 20-16-3 & 20-4-2 \\ -11+15+0 & -22+20+0 & -22+5+0 \\ 9-15+1 & 18-20+3 & 18-5+2 \end{array}\right] \end{aligned}

AB =\begin{bmatrix} -3 &1 &14 \\ 4& -2& -17\\ -5 &1 &15 \end{bmatrix}                                           ...(i)

Now again consider,

BA=\begin{bmatrix} 1 &2 &2 \\ 3& 4 & 1\\ 1& 3 & 2 \end{bmatrix}\begin{bmatrix} 10 &-4 & -1\\ -11 & 5 & 0\\ 9&-5 &1 \end{bmatrix}

B A=\left[\begin{array}{lll} 1 \times 10+2 \times(-11)+2 \times 9 & 1 \times(-4)+2 \times 5+2 \times(-5) & 1 \times(-1)+2 \times 0+2 \times 1 \\ 3 \times 10+4 \times(-11)+1 \times 9 & 3 \times(-4)+4 \times 5+1 \times(-5) & 3 \times(-1)+4 \times 0+1 \times 1 \\ 1 \times 10+3 \times(-11)+2 \times 9 & 1 \times(-4)+3 \times 5+2 \times(-5) & 1 \times(-1)+3 \times 0+2 \times 1 \end{array}\right]

B A=\left[\begin{array}{ccc} 10-22+18 & -4+10-10 & -1+0+2 \\ 30-44+9 & -12+20-5 & -3+0+1 \\ 10-33+18 & -4+15-10 & -1+0+2 \end{array}\right]

B A=\begin{bmatrix} 6 &-4 &1 \\ -5 &3 &-2 \\ -5& 1 & 1 \end{bmatrix}                                                  ...(ii)

From equation (i) and (ii), it is clear that AB\neq BA

Posted by

infoexpert21

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