Question

Use Exercises 18 and 24 to solve the system

$\begin{array}{l}7x-8x_2+5x_3=5\\ -4x+5x_2-3x_5=-3\\ x-x_2+x_4=0\end{array}$

Short answer

The value of$x_1=1,x_2=-1,x_3=-2$

Step-by-step solution

Step 1

Given:

$\begin{array}{l}7x-8x_2+5x_3=5\\ -4x+5x_2-3x_5=-3\\ x-x_2+x_4=0\end{array}$

The system is rewritten as follows:

$\mathbf{AX}\mathbf{=}\mathit{B}$

A is the matrix containing the coefficients of:

$A=\left[\begin{array}{ccc}7& 8& 5\\ -4& 5& -3\\ 1& -2& 1\end{array}\right]$

A is the $3\times 1$-matrix containing all $x_i$-variables.

role="math" localid="1668487654224" $X=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]$

Step 2

B is the $3\times 1$-matrix containing all constants of the given equations:

$B=\left[\begin{array}{c}5\\ -3\\ 0\end{array}\right]$

We know that we can solve the system using the equation because of exercise 24. $\mathit{X}\mathbf{=}{\mathit{A}}^{\mathbf{-}\mathbf{1}}\mathit{B}$.

The inverse$A^{-1}$ of A was determine in exercise 18:

$A^{-1}=\left[\begin{array}{ccc}2& 3& -1\\ 1& 2& 1\\ -1& -1& 3\end{array}\right]$

Step 3

The solution to the system of equations can then be found:

$\begin{array}{r}\mathbf{X}={\mathbf{A}}^{-1}\mathbf{B}=\left[\begin{array}{ccc}2& 3& -1\\ 1& 2& 1\\ -1& -1& 3\end{array}\right]\left[\begin{array}{c}5\\ -3\\ 0\end{array}\right]\\ =\left[\begin{array}{c}2\left(5\right)+3(-3)+(-1)\left(0\right)\\ 1\left(5\right)+2(-3)+1\left(0\right)\\ (-1)\left(5\right)+(-1)(-3)+3\left(0\right)\end{array}\right]\\ =\left[\begin{array}{c}10-9+0\\ 5-6+0\\ -5+3+0\end{array}\right]\\ =\left[\begin{array}{c}1\\ -1\\ -2\end{array}\right]\end{array}$

Step 4

We then obtain

$\mathit{X}=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}1\\ -1\\ -2\end{array}\right]$

Thus,$x_1=1,x_2=-1,x_3=-2$