Chapter 11.4, Problem 54AYU

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

$\{\begin{array}{l}x+2z=6\\ -x+2y+3z=-5\\ x-y=6\end{array}$

To determine

To solve: System of equation using matrix inverse.

Answer to Problem 54AYU

Solution:

$x=4,y=-2\text{ and }z=1$.

Explanation of Solution

Given:

$x+2z=6$

$-x+2y+3z=-5$

$x-y=6$

Calculation:

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

$AX=B$

Where,

$A=\left[\begin{array}{ccc}1& 0& 2\\ -1& 2& 3\\ 1& -1& 0\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\text{ and }B=\left[\begin{array}{c}6\\ -5\\ 6\end{array}\right]$

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,

$X=A^{-1}B$

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

Determining $A^{-1}$.

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

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

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

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

$\underset{\begin{array}{c}{R}_2=r_2+r_3\\ R_3=r_3-r_1\end{array}}{\to } [\begin{array}{ccc}1& 0& 2\\ 0& 1& 3\\ 0& -1& -2\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ -1& 0& 1\end{array}]$

$\underset{R_3=r_3+r_2}{\to } [\begin{array}{ccc}1& 0& 2\\ 0& 1& 3\\ 0& 0& 1\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ -1& 1& 2\end{array}]$

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

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

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

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

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

$=\left[\begin{array}{ccc}3& -2& -4\\ 3& -2& -5\\ -1& 1& 2\end{array}\right]\left[\begin{array}{c}6\\ -5\\ 6\end{array}\right]$

$=\left[\begin{array}{c}18+10-24\\ 18+10-30\\ -6-5+12\end{array}\right]$

$=\left[\begin{array}{c}4\\ -2\\ 1\end{array}\right]$

The solution for the given system of equations is $x=4,y=-2\text{ and }z=1$.