Chapter 11.4, Problem 41AYU

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

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

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

$\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]$

To determine

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

Answer to Problem 41AYU

Solution:

The solution of the system of equations $\{\begin{array}{l}2x+y=-1\\ x+y=3\end{array}$ is $x=-4,y=7$.

Explanation of Solution

Given information:

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

Explanation:

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

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

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

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

Here, $A=\left|\begin{array}{cc}2& 1\\ 1& 1\end{array}\right|=2-1=1\ne 0$. 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 $\left[A|I_2\right]=\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Use $R_1\to R_2$,

$\left[A|I_2\right]=\left[\begin{array}{cc}1& 1\\ 2& 1\end{array}|\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

Use row operation $R_2\to R_2-2R_1$,

$\left[A|I_2\right]=\left[\begin{array}{cc}1& 1\\ 0& -1\end{array}|\begin{array}{cc}0& 1\\ 1& -2\end{array}\right]$

Use $R_2\to -R_2$

$\left[A|I_2\right]=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}|\begin{array}{cc}0& 1\\ -1& 2\end{array}\right]$

Use $R_1\to R_1-R_2$

$\left[A|I_2\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}|\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]$

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

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

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

$=\left[\begin{array}{c}-4\\ 7\end{array}\right]$

Therefore, $\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-4\\ 7\end{array}\right]$.

Thus, the solution of the system of equations $\{\begin{array}{l}2x+y=8\\ x+y=5\end{array}$ is $x=-4,y=7$.