Chapter 11.4, Problem 55AYU

In Problems 45-64, use the inverses found in Problems 35-44 to solve each system of equations.

$\{\begin{array}{l}x-y+z=2\\ -2y+z=2\\ -2x-3y=\frac{1}{2}\end{array}$

To determine

To solve: System of equation using matrix inverse.

Answer to Problem 55AYU

Solution:

$x=\frac{1}{2},y=-\frac{1}{2}\text{ and }z=1$.

Explanation of Solution

Given:

$x-y+z=2$

$-2y+z=2$

$-2x-3y=\frac{1}{2}$

Calculation:

The original system of equations can be written compactly as a matrix equation.

$AX=B$

Where,

$A=\left[\begin{array}{ccc}1& -1& 1\\ 0& -2& 1\\ -2& -3& 0\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\text{ and }B=\left[\begin{array}{c}2\\ 2\\ \frac{1}{2}\end{array}\right]$

To solve the for $x\text{ and }y$, the solution of $X$ has to be found as given below:

$AX=B$

Multiply both sides by $A^{-1}$,

$A^{-1}\left(AX\right)=A^{-1}B$

By association property of matrix multiplication, the above equation can be rewritten as,

$(A^{-1}A)X=A^{-1}B$

By the definition of inverse matrix $(A^{-1}A)=I_n$.

$I_{n}X=A^{-1}B$

By the property of Identity matrix, $I_{n}X=X$.

Hence,

$X=A^{-1}B$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=A^{-1}B$

Determining $A^{-1}$.

Step 1: Construction of $\left[A|I_n\right]$.

$\left[A|I_3\right]=[\begin{array}{ccc}1& -1& 1\\ 0& -2& 1\\ -2& -3& 0\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}]$

Step 2: Transform the matrix $\left[A|I_n\right]$ into reduced row echelon form.

$=[\begin{array}{ccc}1& -1& 1\\ 0& -2& 1\\ -2& -3& 0\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}]$

$\underset{R_3=r_3+2r_1}{\to }[\begin{array}{ccc}1& -1& 1\\ 0& -2& 1\\ 0& -5& 2\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 2& 0& 1\end{array}]$

$\underset{R_2=\left(-\frac{1}{2}\right)r_2}{\to }[\begin{array}{ccc}1& -1& 1\\ 0& 1& -\frac{1}{2}\\ 0& -5& 2\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& -\frac{1}{2}& 0\\ 2& 0& 1\end{array}]$

$\underset{\begin{array}{c}{R}_1=r_1+r_2\\ R_3=r_3+5r_2\end{array}}{\to }[\begin{array}{ccc}1& 0& 1/2\\ 0& 1& -\frac{1}{2}\\ 0& 0& -\frac{1}{2}\end{array}|\begin{array}{ccc}1& -\frac{1}{2}& 0\\ 0& -\frac{1}{2}& 0\\ 2& -\frac{5}{2}& 1\end{array}]$

$\underset{\begin{array}{c}{R}_1=r_1+r_3\\ R_2=r_2-r_3\\ R_3=-2r_3\end{array}}{\to }[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{ccc}3& -3& 1\\ -2& 2& -1\\ -4& 5& -2\end{array}]$

Step 3: Separate the matrix on the right of the vertical bar which is the inverse of $A$.

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

Solving $x,y\text{ and }z$.

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=A^{-1}B$

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

$=\left[\begin{array}{c}6-6+\frac{1}{2}\\ -4+4-\frac{1}{2}\\ -8+10-1\end{array}\right]$

$=\left[\begin{array}{c}\frac{1}{2}\\ -\frac{1}{2}\\ 1\end{array}\right]$

The solution for the given system of equations is $x=\frac{1}{2},y=-\frac{1}{2}\text{ and }z=1$.