Chapter 7.5, Problem 41E

To determine

To find: The product of two given matrices.

Answer to Problem 41E

The required product of the given matrices is $AB=\left[\begin{array}{ccc}-2& 0& 0\\ 0& -1& 0\\ 0& 0& \frac{-1}{3}\end{array}\right]$ .

Explanation of Solution

Given: The two matrices are $A=\left[\begin{array}{ccc}6& 0& 0\\ 0& 4& 0\\ 0& 0& -2\end{array}\right]$ and $B=\left[\begin{array}{ccc}\frac{-1}{3}& 0& 0\\ 0& \frac{-1}{4}& 0\\ 0& 0& \frac{1}{6}\end{array}\right]$ .

Concept Used: Matrix multiplication is possible if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. Thus, two matrices $A={\left[a_{ij}\right]}_{m\times n}$ of order $m\times n$ and $B={\left[b_{ij}\right]}_{n\times p}$ of order $n\times p$ can be multiplied which results in a matrix of order $m\times p$ given by $AB={\left[a_{ij}{b}_{ji}\right]}_{m\times p}$ , i.e.- the respective terms in the rows of the first matrix is multiplied to the respective terms in the columns of the second matrix, or vice versa.

In case of diagonal matrix since all other entries except the diagonal terms are zero, the product of two diagonal matrices will also be a diagonal matrix where the diagonal terms are obtained by multiplying the numbers in the diagonals of the two given matrices.

Calculations: The given matrices are $A=\left[\begin{array}{ccc}6& 0& 0\\ 0& 4& 0\\ 0& 0& -2\end{array}\right]$ and $B=\left[\begin{array}{ccc}\frac{-1}{3}& 0& 0\\ 0& \frac{-1}{4}& 0\\ 0& 0& \frac{1}{6}\end{array}\right]$ .

Since number of columns in A is equal to the number of rows in B , hence AB exists which will be a matrix of order $3\times 3$ . Also, as both the matrices are diagonal matrices, the product will be a diagonal matrix whose diagonal terms are obtained by multiplying the numbers in the diagonal of both matrices.

Thus the required product is-

  $AB$

  $=\left[\begin{array}{ccc}6& 0& 0\\ 0& 4& 0\\ 0& 0& -2\end{array}\right]\left[\begin{array}{ccc}\frac{-1}{3}& 0& 0\\ 0& \frac{-1}{4}& 0\\ 0& 0& \frac{1}{6}\end{array}\right]$

  $=\left[\begin{array}{ccc}6\times \left(\frac{-1}{3}\right)& 0& 0\\ 0& 4\times \left(\frac{-1}{4}\right)& 0\\ 0& 0& -2\times \left(\frac{1}{6}\right)\end{array}\right]$

  $=\left[\begin{array}{ccc}-2& 0& 0\\ 0& -1& 0\\ 0& 0& \frac{-1}{3}\end{array}\right]$

Thus, the required product is $AB=\left[\begin{array}{ccc}-2& 0& 0\\ 0& -1& 0\\ 0& 0& \frac{-1}{3}\end{array}\right]$ .