Chapter 4.8, Problem 58E

To determine

To solve: The given system of linear equations using the matrix capabilities of a graphing utility.

Answer to Problem 58E

The solution of the system of linear equation is $x_1=-32$ , $x_2=-13$ , $x_3=-37$ and $x_4=15$ .

Explanation of Solution

Given information :

The system of linear equations:

  $\begin{array}{l}{x}_1-2x_2-x_3-2x_4=\text{ 1}\\ 3x_1-5x_2-2x_3-3x_4=-2\\ 2x_1-5x_2-2x_3-5x_4=0\\ -x_1+4x_2+4x_3+11x_4=-3\end{array}$

The inverse matrix: (as obtained in EX 28)

  $A^{-1}=\left[\begin{array}{cccc}-24& 7& 1& -2\\ -10& 3& 0& -1\\ -29& 7& 3& -2\\ 12& -3& -1& 1\end{array}\right]$

Formula used:

To solve a system of linear equations:

If $A$ is an invertible matrix, the system of linear equations represented by $AX=B$ has a unique solution given by $X=A^{-1}B$ .

Calculation :

As per problem,

The given system of linear equations:

  $\begin{array}{l}{x}_1-2x_2-x_3-2x_4=\text{ 1}\\ 3x_1-5x_2-2x_3-3x_4=-2\\ 2x_1-5x_2-2x_3-5x_4=0\\ -x_1+4x_2+4x_3+11x_4=-3\end{array}$

Write the system in matrix form, $AX=B$ .

  $\left[\begin{array}{cccc}1& -2& -1& -2\\ 3& -5& -2& -3\\ 2& -5& -2& -5\\ -1& 4& 4& 11\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]=\left[\begin{array}{c}1\\ -2\\ 0\\ -3\end{array}\right]$

To obtain, $X=A^{-1}B$

Find the product of the given inverse matrix ( $A^{-1}$ ) with $B$ using graphing utility.

The solution of the system of linear equation using graphing utility is obtained as:

1. Enter the first matrix by using the combination of keystrokes $\boxed{\text{2nd}}+\boxed{x^{-1}}$
2. Use cursor key to select the edit option and then select row 1 of matrix A and press $\boxed{\text{Enter}}$ .
3. Enter the dimension of the matrix A (4 rows by 4 columns).
4. Enter the values of the matrix A (row by row).
5. Enter the second matrix by using the combination of keystrokes $\boxed{\text{2nd}}+\boxed{x^{-1}}$
6. Use cursor key to select the edit option and press [2] or highlight 2. [B] and press $\boxed{\text{Enter}}$ .
7. Enter the dimension of the matrix B (4 rows by 1 column).
8. Enter the values of the matrix B (row by row).
9. Press $\boxed{\text{2nd}}+\boxed{\text{MODE}}$ to quit out of the matrix screen.
10. Press $\boxed{\text{2nd}}+\boxed{x^{-1}}$ and stay at the NAMES menu
11. Press $\boxed{\text{Enter}}$ to select marix A
12. To select matrix B, go back intot the matrix menu by pressing $\boxed{\text{2nd}}+\boxed{x^{-1}}$ .
13. Press [2] or highlight 2. [B] and press $\boxed{\text{Enter}}$ .
14. Press $\boxed{\text{Enter}}$ to multiply the matrices.

The resultant matrix is displayed as:

  $\left[\begin{array}{c}-32\\ -13\\ -37\\ 15\end{array}\right]$

The corresponding system is,

  $\{\begin{array}{c}{x}_1=-32\\ \begin{array}{l}{x}_2=-13\\ x_3=-37\\ x_4=15\end{array}\end{array}$

Hence,

The solution of the system of linear equation is $x_1=-32$ , $x_2=-13$ , $x_3=-37$ and $x_4=15$ .