Chapter 11.4, Problem 53AYU

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

$\{\begin{array}{c}x-y+z=4\\ -2y+z=1\\ -2x-3y=-4\end{array}$

Problems $35-44$, each matrix is nonsingular. Find the inverse of each matrix.

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

To determine

The solution of the system of equations $\{\begin{array}{l}x-y+z=4\\ -2y+z=1\\ -2x-3y=-4\end{array}$ by using inverse of the corresponding matrix.

Answer to Problem 53AYU

Solution:

The solution of the system of equations $\{\begin{array}{l}x-y+z=4\\ -2y+z=1\\ -2x-3y=-4\end{array}$ is $x=5,y=-2,z=-3$.

Explanation of Solution

Given information:

The system of equations $\{\begin{array}{l}x-y+z=4\\ -2y+z=1\\ -2x-3y=-4\end{array}$

Explanation:

Using system of linear equations $\{\begin{array}{l}x-y+z=4\\ -2y+z=1\\ -2x-3y=-4\end{array}$,

$A=\left[\begin{array}{ccc}1& -1& 1\\ 0& -2& 1\\ -2& -3& 0\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ and $B=\left[\begin{array}{c}4\\ 1\\ -4\end{array}\right]$.

Then the original system of equation can be written compactly as $AX=B$.

Hence, $X=A^{-1}B$

Here, $A=\left|\begin{array}{ccc}1& -1& 1\\ 0& -2& 1\\ -2& -3& 0\end{array}\right|=1\left(3\right)-1\left(2\right)+1\left(-4\right)=-3\ne 0$. Thus $A$ is nonsingular.

To find inverse of matrix $A$:

Step 1: Form the matrix $\left[A|I_n\right]$.

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

Step 3: The reduced row echelon form of $\left[A|I_n\right]$ contain the identity matrix $I_n$ on the left of the vertical bar, the $n$ by $n$ matrix on the right vertical bar is an inverse of $A$.

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

Use row operation $R_3\to R_3+2R_1$,

$\left[A|I_3\right]=\left[\begin{array}{ccc}1& -1& 1\\ 0& -2& 1\\ 0& -5& 2\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 2& 0& 1\end{array}\right]$

Use $R_2\to -\frac{1}{2}{R}_2$

$\left[A|I_3\right]=\left[\begin{array}{ccc}1& -1& 1\\ 0& 1& -\frac{1}{2}\\ 0& -5& 2\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& -\frac{1}{2}& 0\\ 2& 0& 1\end{array}\right]$

Use $R_3\to R_3+5R_2$

$\left[A|I_3\right]=\left[\begin{array}{ccc}1& -1& 1\\ 0& 1& -\frac{1}{2}\\ 0& 0& -\frac{1}{2}\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& -\frac{1}{2}& 0\\ 2& -\frac{5}{2}& 1\end{array}\right]$

Use $R_2\to R_2-R_3$

$\left[A|I_3\right]=\left[\begin{array}{ccc}1& -1& 1\\ 0& 1& 0\\ 0& 0& -\frac{1}{2}\end{array}|\begin{array}{ccc}1& 0& 0\\ -2& 2& -1\\ 2& -\frac{5}{2}& 1\end{array}\right]$

Use $R_3\to -2R_3$

$\left[A|I_3\right]=\left[\begin{array}{ccc}1& -1& 1\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{ccc}1& 0& 0\\ -2& 2& -1\\ -4& 5& -2\end{array}\right]$

Use $R_1\to R_1-R_3$

$\left[A|I_3\right]=\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{ccc}5& -5& 2\\ -2& 2& -1\\ -4& 5& -2\end{array}\right]$

Use $R_1\to R_1+R_2$

$\left[A|I_3\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{ccc}3& -3& 1\\ -2& 2& -1\\ -4& 5& -2\end{array}\right]$

Thus, the inverse of the matrix $A$ is $\left[\begin{array}{ccc}3& -3& 1\\ -2& 2& -1\\ -4& 5& -2\end{array}\right]$.

Now, $X=A^{-1}B$.

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

$=\left[\begin{array}{c}5\\ -2\\ -3\end{array}\right]$

Therefore, $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}5\\ -2\\ -3\end{array}\right]$.

Thus, the solution of the system of equations $\{\begin{array}{l}x-y+z=4\\ -2y+z=1\\ -2x-3y=-4\end{array}$ is $x=5,y=-2,z=-3$.