#### Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question 54 sub question (ii) maths textbook solution

$P Q=\left[\begin{array}{ccc}x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c\end{array}\right]=Q P$

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:

$P=\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right] and \ \ Q=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right]$

We have,

$\\\\ P Q=\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right]\\\\ P Q=\left[\begin{array}{ccc}x \times a & 0 & 0 \\ 0 & y \times b & 0 \\ 0 & 0 & z \times c\end{array}\right] \\\\ =\left[\begin{array}{ccc}x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c\end{array}\right]$

And we have,

$\begin{array}{l} Q P=\left[\begin{array}{ccc} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right]\left[\begin{array}{ccc} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right] \\\\ =\left[\begin{array}{ccc} a \times x & 0 & 0 \\ 0 & b \times y & 0 \\ 0 & 0 & c \times z \end{array}\right] \end{array}$

$=\left[\begin{array}{ccc} a x & 0 & 0 \\ 0 & b y & 0 \\ 0 & 0 & c z \end{array}\right]$

As xa=ax, yb=by, zc=cz

Therefore,

$P Q=\left[\begin{array}{ccc}x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c\end{array}\right]=Q P$

Hence proved.