Chapter 7.5, Problem 34E

To determine

To find: The solution of the given matrix equation.

Answer to Problem 34E

The required matrix is $X=\left[\begin{array}{ccc}\frac{-8}{3}& \frac{8}{3}& \frac{4}{3}\\ \frac{2}{3}& 0& 6\end{array}\right]$ .

Explanation of Solution

Given: The given equation is $3X-4A=2B$ where $A=\left[\begin{array}{ccc}-2& 1& 3\\ -1& 0& 4\end{array}\right]$ and $B=\left[\begin{array}{ccc}0& 2& -4\\ 3& 0& 1\end{array}\right]$ .

Concept Used: A matrix when multiplied with a scalar, all entries of the matrix gets multiplied by that scalar. Hence, if $A={\left[a_{ij}\right]}_{m\times n}$ is a matrix which is multiplied by a scalar k , then $kA={\left[k(a_{ij})\right]}_{m\times n}$ .

Matrix addition is possible only if the order of the matrices are equal. Thus, two matrices $A={\left[a_{ij}\right]}_{m\times n}$ and $B={\left[b_{ij}\right]}_{m\times n}$ can be added which results in $A+B={\left[(a_{ij}+b_{ij})\right]}_{m\times n}$ , i.e.- the respective terms in order get added.

Calculation: The given matrix equation is-

  $3X-4A=2B$

where $A=\left[\begin{array}{ccc}-2& 1& 3\\ -1& 0& 4\end{array}\right]$ and $B=\left[\begin{array}{ccc}0& 2& -4\\ 3& 0& 1\end{array}\right]$ .

The given equation can be simplified as-

  $3X=4A+2B$

Putting in the values of A and B in the given equation gives-

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

  $=\left[\begin{array}{ccc}-8& 4& 12\\ -4& 0& 16\end{array}\right]+\left[\begin{array}{ccc}0& 4& -8\\ 6& 0& 2\end{array}\right]$

  $=\left[\begin{array}{ccc}-8+0& 4+4& 12-8\\ -4+6& 0& 16+2\end{array}\right]$

  $=\left[\begin{array}{ccc}-8& 8& 4\\ 2& 0& 18\end{array}\right]$

  $\Rightarrow 3X=\left[\begin{array}{ccc}-8& 8& 4\\ 2& 0& 18\end{array}\right]$

  $\Rightarrow X=\frac{1}{3}\left[\begin{array}{ccc}-8& 8& 4\\ 2& 0& 18\end{array}\right]$

  $\Rightarrow X=\left[\begin{array}{ccc}\frac{-8}{3}& \frac{8}{3}& \frac{4}{3}\\ \frac{2}{3}& 0& 6\end{array}\right]$

Thus, the required solution is $X=\left[\begin{array}{ccc}\frac{-8}{3}& \frac{8}{3}& \frac{4}{3}\\ \frac{2}{3}& 0& 6\end{array}\right]$ .