Chapter 7.5, Problem 29E

To determine

To find:The solution of the given matrix equation.

Answer to Problem 29E

The required matrix is $X=\left[\begin{array}{ccc}-4& 6& -2\\ 4& 0& 10\end{array}\right]$ .

Explanation of Solution

Given: The given equation is $X=2A+2B$ 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-

  $X=2A+2B$

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

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

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

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

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

  $=\left[\begin{array}{ccc}-4& 6& -2\\ 4& 0& 10\end{array}\right]$

Thus, the required solution is $X=\left[\begin{array}{ccc}-4& 6& -2\\ 4& 0& 10\end{array}\right]$ .