Chapter 12.1, Problem 11PPSE

Solution

To find: The solution of the matrix equation.

Required solution is, $X=\left[\begin{array}{cc}4& -8\\ -1& -1\\ 11& 1\end{array}\right]$

Given information: Matrix equation is, $\left[\begin{array}{cc}1& 2\\ 2& 1\\ -3& 4\end{array}\right]+X=\left[\begin{array}{cc}5& -6\\ 1& 0\\ 8& 5\end{array}\right]$

Formula used: The difference of the matrices is,

  $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]-\left[\begin{array}{cc}e& f\\ g& h\end{array}\right]=\left[\begin{array}{cc}a-e& b-f\\ c-g& d-h\end{array}\right]$

Calculation:

Consider the matrix equation as,

  $\left[\begin{array}{cc}1& 2\\ 2& 1\\ -3& 4\end{array}\right]+X=\left[\begin{array}{cc}5& -6\\ 1& 0\\ 8& 5\end{array}\right]$

This can be rewritten as,

  $X=\left[\begin{array}{cc}5& -6\\ 1& 0\\ 8& 5\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 2& 1\\ -3& 4\end{array}\right]$

Then the difference of the two matrices is,

  $\begin{array}{c}X=\left[\begin{array}{cc}5& -6\\ 1& 0\\ 8& 5\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 2& 1\\ -3& 4\end{array}\right]\\ =\left[\begin{array}{cc}5-1& -6-2\\ 1-2& 0-1\\ 8-\left(-3\right)& 5-4\end{array}\right]\\ =\left[\begin{array}{cc}4& -8\\ -1& -1\\ 11& 1\end{array}\right]\end{array}$

Therefore, required solution is, $X=\left[\begin{array}{cc}4& -8\\ -1& -1\\ 11& 1\end{array}\right]$