Chapter 11.4, Problem 57AYU

To determine

To solve: System of equation using matrix inverse.

Answer to Problem 57AYU

$x=-\frac{34}{7},y=\frac{85}{7}$ and $z=\frac{12}{7}$.

Explanation of Solution

Given:

$x+y+z=9$

$3x+2y-z=8$

$3x+y+2z=1$

Calculation:

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

$AX=B$

Where,

$A=\left[\begin{array}{ccc}1& 1& 1\\ 3& 2& -1\\ 3& 1& 2\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ and $B=\left[\begin{array}{c}9\\ 8\\ 1\end{array}\right]$.

To solve the for $x$ 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 $\left[A|I_n\right]$.

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

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

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

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

$\underset{\begin{array}{c}{R}_1=r_1-r_2\\ R_3=r_3+2r_2\end{array}}{\to }[\begin{array}{ccc}1& 0& 4\\ 0& 1& -3\\ 0& 0& -7\end{array}|\begin{array}{ccc}1& -1& 1\\ 0& 1& -1\\ -3& 2& -1\end{array}]$

$\underset{R_3=\left(- \frac{1}{7}\right)r_3}{\to }[\begin{array}{ccc}1& 0& 4\\ 0& 1& -3\\ 0& 0& 1\end{array}|\begin{array}{ccc}1& -1& 1\\ 0& 1& -1\\ \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}]$

$\underset{\begin{array}{c}{R}_1=r_1-4r_3\\ R_2=r_2+3r_3\end{array}}{\to }[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{ccc}-\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ \frac{9}{7}& \frac{1}{7}& -\frac{4}{7}\\ \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}]$

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

$A^{-1}=\left[\begin{array}{ccc}-\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ \frac{9}{7}& \frac{1}{7}& -\frac{4}{7}\\ \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right]$

Solving $x$, $y$ and $z$.

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

$=\left[\begin{array}{ccc}-\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ \frac{9}{7}& \frac{1}{7}& -\frac{4}{7}\\ \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right]\left[\begin{array}{c}9\\ 8\\ 1\end{array}\right]$

$=\left[\begin{array}{c}-\frac{45}{7}+\frac{8}{7}+\frac{3}{7}\\ \frac{81}{7}+\frac{8}{7}-\frac{4}{7}\\ \frac{27}{7}-\frac{16}{7}+\frac{1}{7}\end{array}\right]$

$=\left[\begin{array}{c}-\frac{34}{7}\\ \frac{85}{7}\\ \frac{12}{7}\end{array}\right]$

The solution for the given system of equations is $x=-\frac{34}{7},y=\frac{85}{7}$ and $z=\frac{12}{7}$.