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

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

Answers (1)

Hence proved AB \neq BA

\begin{bmatrix} 1 &0 &0 \\ 3& -5 &3 \\ 0&0 &1 \end{bmatrix}\neq \begin{bmatrix} -1 &-9 &0 \\ 0& -5 & 0\\ 0& -27 & 1 \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} 1 &3 & -1\\ 2&-1 &-1 \\ 3& 0 &-1 \end{bmatrix}, B =\begin{bmatrix} -2 &3 &-1 \\ -1& 2 &-1 \\ -6& 9 &-4 \end{bmatrix}

Consider,

AB=\begin{bmatrix} 1 &3 & -1\\ 2&-1 &-1 \\ 3& 0 &-1 \end{bmatrix}\begin{bmatrix} -2 &3 &-1 \\ -1& 2 &-1 \\ -6& 9 &-4 \end{bmatrix}

A B=\left[\begin{array}{ccc} 1 \times(-2)+3 \times(-1)+(-1) \times(-6) & 1 \times 3+3 \times 2+(-1) \times 9 & 1 \times(-1)+3 \times(-1)+(-1) \times(-4) \\ 2 \times(-2)+(-1) \times(-1)+(-1) \times(-6) & 2 \times 3+(-1) \times 2+(-1) \times 9 & 2 \times(-1)+(-1) \times(-1)+(-1) \times(-4) \\ 3 \times(-2)+0 \times(-1)+(-1) \times(-6) & 3 \times 3+0 \times 2+(-1) \times 9 & 3 \times(-1)+0 \times(-1)+(-1) \times(-4) \end{array}\right]

A B=\left[\begin{array}{lll} -2-3+6 & 3+6-9 & -1-3+4 \\ -4+1+6 & 6-2-9 & -2+1+4 \\ -6+0+6 & 9+0-9 & -3+0+4 \end{array}\right]

AB=\begin{bmatrix} 1 &0 &0 \\ 3& -5 &3 \\ 0&0 &1 \end{bmatrix}                                           ...(i)

Now again consider,

BA=\begin{bmatrix} -2 &3 &-1 \\ -1& 2 &-1 \\ -6& 9 &-4 \end{bmatrix}\begin{bmatrix} 1 &3 & -1\\ 2&-1 &-1 \\ 3& 0 &-1 \end{bmatrix}

B A=\left[\begin{array}{lll} -2 \times 1+3 \times 2+(-1) \times 3 & -2 \times 3+3 \times(-1)+(-1) \times 0 & -2 \times(-1)+3 \times(-1)+(-1) \times(-1) \\ -1 \times 1+2 \times 2+(-1) \times 3 & -1 \times 3+2 \times(-1)+(-1) \times 0 & -1 \times(-1)+2 \times(-1)+(-1) \times(-1) \\ -6 \times 1+9 \times 2+(-4) \times 3 & -6 \times 3+9 \times(-1)+(-4) \times 0 & -6 \times(-1)+9 \times(-1)+(-4) \times(-1) \end{array}\right]

B A=\left[\begin{array}{ccc} -2+6-3 & -6-3+0 & 2-3+1 \\ -1+4-3 & -3-2+0 & 1-2+1 \\ -6+18-12 & -18-9+0 & 6-9+4 \end{array}\right]

BA= \begin{bmatrix} 1 &-9 &0 \\ 0& -5 &0 \\ 0& -27 &1 \end{bmatrix}                                              ...(ii)

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

Posted by

infoexpert21

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