Chapter 8.2, Problem 91E

To determine

The product $AB$ for several pairs of matrices.

Answer to Problem 91E

The product matrix is given below as,

  $\left(\begin{array}{l}\begin{array}{ccc}{a}_1{b}_1& 0& \mathrm{0................0}\\ 0& a_2{b}_2& \mathrm{0................0}\\ 0& 0& a_3{b}_3\mathrm{............0}\end{array}\\ 00\mathrm{0..............}{a}_n{b}_n\end{array}\right)$

Explanation of Solution

Given information:

The matrix is given below as,

  $\begin{array}{l}A=\left(\begin{array}{cc}2& 0\\ 0& 4\end{array}\right)\\ B=\left(\begin{array}{cc}3& 0\\ 0& 5\end{array}\right)\end{array}$

Formula used:

Matrix multiplication is used.

Calculation:

To find general rule for diagonal matrix, first we take different example of multiplication of diagonal matrix.

First we take $2\times 2$ diagonal matrix of same size

  $\begin{array}{l}A=\left(\begin{array}{cc}2& 0\\ 0& 4\end{array}\right)\\ B=\left(\begin{array}{cc}3& 0\\ 0& 5\end{array}\right)\end{array}$

Now multiply AB matrix

  $\begin{array}{l}\left(\begin{array}{cc}2& 0\\ 0& 4\end{array}\right)\left(\begin{array}{cc}3& 0\\ 0& 5\end{array}\right)\\ \left(\begin{array}{cc}2\times 3+0\times & 2\times 0+0\times 5\\ 0\times 3+0\times 4& 0\times 0+4\times 5\end{array}\right)\\ \left(\begin{array}{cc}6& 0\\ 0& 20\end{array}\right)\end{array}$

Now take another diagonal matrix of different dimension $\left(3\times 3\right)$

  $\begin{array}{l}A=\left(\begin{array}{ccc}2& 0& 0\\ 0& 4& 0\\ 0& 0& 6\end{array}\right)\\ B=\left(\begin{array}{ccc}3& 0& 0\\ 0& 5& 0\\ 0& 0& 7\end{array}\right)\end{array}$

Multiply both matrixes AB

  $\begin{array}{l}\left(\begin{array}{ccc}2& 0& 0\\ 0& 4& 0\\ 0& 0& 6\end{array}\right)\left(\begin{array}{ccc}3& 0& 0\\ 0& 5& 0\\ 0& 0& 7\end{array}\right)\\ \left(\begin{array}{ccc}2\times 3+0\times 0+0\times 0& 2\times 0+0\times 5+0\times 0& 2\times 0+0\times 0+0\times 7\\ 0\times 3+4\times 0+0\times 0& 0\times 0+4\times 5+0\times 0& 0\times 0+4\times 0+0\times 7\\ 0\times 3+0\times 0+6\times 0& 0\times 0+0\times 5+6\times 0& 0\times 0+0\times 0+6\times 7\end{array}\right)\\ \left(\begin{array}{ccc}6& 0& 0\\ 0& 20& 0\\ 0& 0& 42\end{array}\right)\end{array}$

Now we take a general diagonal matrix A and B

  $\begin{array}{l}A=\left(\begin{array}{ccc}{a}_1& 0& 0\\ 0& a_2& 0\\ 0& 0& a_3\end{array}\right)\\ B=\left(\begin{array}{ccc}{b}_1& 0& 0\\ 0& b_2& 0\\ 0& 0& b_3\end{array}\right)\end{array}$

Multiply matrix A and B

  $\begin{array}{l}\left(\begin{array}{ccc}{a}_1& 0& 0\\ 0& a_2& 0\\ 0& 0& a_3\end{array}\right)\left(\begin{array}{ccc}{b}_1& 0& 0\\ 0& b_2& 0\\ 0& 0& b_3\end{array}\right)\\ \left(\begin{array}{ccc}{a}_1\times b_1+0\times 0+0\times 0& a_1\times 0+0\times b_2+0\times 0& a_1\times 0+0\times 0+0\times b_3\\ 0\times b_1+a_2\times 0+0\times 0& 0\times 0+a_2\times b_2+0\times 0& 0\times 0+a_2\times 0+0\times b_3\\ 0\times b_1+0\times 0+a_3\times 0& 0\times 0+0\times b_2+a_3\times 0& 0\times 0+0\times 0+a_3\times b_3\end{array}\right)\\ \left(\begin{array}{ccc}{a}_1{b}_1& 0& 0\\ 0& a_2{b}_2& 0\\ 0& 0& a_3{b}_3\end{array}\right)\end{array}$

Above example show that on multiplying two diagonal matrix of same size, perform multiplication of the diagonal element without using row to column multiplication. The product of that matrix in general gives the product matrix in following way.

  $\left(\begin{array}{l}\begin{array}{ccc}{a}_1{b}_1& 0& \mathrm{0................0}\\ 0& a_2{b}_2& \mathrm{0................0}\\ 0& 0& a_3{b}_3\mathrm{............0}\end{array}\\ 00\mathrm{0..............}{a}_n{b}_n\end{array}\right)$

Conclusion:

The product matrix is given below as,

  $\left(\begin{array}{l}\begin{array}{ccc}{a}_1{b}_1& 0& \mathrm{0................0}\\ 0& a_2{b}_2& \mathrm{0................0}\\ 0& 0& a_3{b}_3\mathrm{............0}\end{array}\\ 00\mathrm{0..............}{a}_n{b}_n\end{array}\right)$