Chapter 12.1, Problem 4LC

Solution

To find: The solution of the matrix equation.

Required solution is $X=\left[\begin{array}{cc}6& 10\\ 13& -4\end{array}\right]$

Given information: Matrix equation is $X-\left[\begin{array}{cc}2& 0\\ 5& -1\end{array}\right]=\left[\begin{array}{cc}4& 10\\ 8& -3\end{array}\right]$

Formula used: The sum 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,

  $X-\left[\begin{array}{cc}2& 0\\ 5& -1\end{array}\right]=\left[\begin{array}{cc}4& 10\\ 8& -3\end{array}\right]$

This can be rewritten as,

  $X=\left[\begin{array}{cc}4& 10\\ 8& -3\end{array}\right]+\left[\begin{array}{cc}2& 0\\ 5& -1\end{array}\right]$

Then the sum of the two matrices is,

  $\begin{array}{c}X=\left[\begin{array}{cc}4& 10\\ 8& -3\end{array}\right]+\left[\begin{array}{cc}2& 0\\ 5& -1\end{array}\right]\\ =\left[\begin{array}{cc}4+2& 10+0\\ 8+5& -3+\left(-1\right)\end{array}\right]\\ =\left[\begin{array}{cc}6& 10\\ 13& -3-1\end{array}\right]\\ =\left[\begin{array}{cc}6& 10\\ 13& -4\end{array}\right]\end{array}$

Therefore, required solution is $X=\left[\begin{array}{cc}6& 10\\ 13& -4\end{array}\right]$