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  • Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.3 question 68  math

Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.3 question 68  math

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Answer: A and B are two square matrices with A B \neq B A then (A B)^{2} \neq A^{2} B^{2}

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 and B be square matrices of order 3 \times 3

Solution: Let

A=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \ \ and \ \ B=\left[\begin{array}{lll}0 & 1 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]

Here

A B =\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}0 & 1 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]              

\\=\left[\begin{array}{lll}1 \times 0+0 \times 2+0 \times 0 & 1 \times 1+0 \times 1+0 \times 0 & 1 \times 0+0 \times 0+0 \times 1 \\ 1 \times 0+1 \times 2+0 \times 0 & 1 \times 1+1 \times 1+0 \times 0 & 1 \times 0+1 \times 0+0 \times 1 \\ 0 \times 0+0 \times 2+1 \times 0 & 0 \times 1+0 \times 1+1 \times 0 & 0 \times 0+0 \times 0+1 \times 1\end{array}\right] \\ \\\\=\left[\begin{array}{lll}0 & 1 & 0 \\ 2 & 2 & 0 \\ 0 & 0 & 1\end{array}\right]

And B A =\left[\begin{array}{lll}0 & 1 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]
              

\\=\left[\begin{array}{lll} 0 \times 1+1 \times 1+0 \times 0 & 0 \times 0+1 \times 1+0 \times 0 & 0 \times 0+1 \times 0+0 \times 1 \\ 2 \times 1+1 \times 1+0 \times 0 & 2 \times 0+1 \times 1+0 \times 0 & 2 \times 0+1 \times 0+0 \times 1 \\ 0 \times 1+0 \times 1+1 \times 0 & 0 \times 0+0 \times 1+1 \times 0 & 0 \times 0+0 \times 0+1 \times 1 \end{array}\right] \\\\ =\left[\begin{array}{lll} 1 & 1 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]

Here, AB \neq BANow,

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

=\left[\begin{array}{lll} 2 & 2 & 0 \\ 4 & 6 & 0 \\ 0 & 0 & 1 \end{array}\right]

 

\\ A^2=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\\\\\ A^{2}=\left[\begin{array}{lll} 1+0+0 & 0+0+0 & 0+0+0 \\ 1+1+0 & 0+1+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+1 \end{array}\right] \\\\ =\left[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]
\begin{array}{l} B^{2}=\left[\begin{array}{lll} 0 & 1 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{lll} 0 & 1 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\\\ =\left[\begin{array}{lll} 0+2+0 & 0+1+0 & 0+0+0 \\ 0+2+0 & 2+1+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+1 \end{array}\right] \\\\ =\left[\begin{array}{lll} 2 & 1 & 0 \\ 2 & 3 & 0 \\ 0 & 0 & 1 \end{array}\right] \\\\ A^{2} B^{2}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 0 \\ 2 & 3 & 0 \\ 0 & 0 & 1 \end{array}\right] \\\\ \end{array}

\\ A^{2} B^{2}=\left[\begin{array}{lll} 2+0+0 & 1+0+0 & 0+0+0 \\ 4+2+0 & 2+3+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+1 \end{array}\right] \\\\\\ A^{2} B^{2}=\left[\begin{array}{lll} 2 & 1 & 0 \\ 6 & 5 & 0 \\ 0 & 0 & 1 \end{array}\right]

We can see that if we have A and B two square matrices with AB \neq BA  then (A B)^{2} \neq A^{2} B^{2}

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