Chapter D, Problem 1P

To determine

To find:

The solutions of the system of equations.

$x+4y=-2$

$3x-y=7$

Answer to Problem 1P

Solution:

$x=2;y=-1$

Explanation of Solution

1) Concept:

To solve a system of two equations in $x$ and $y$ using matrices:

Step 1. Write the corresponding matrix associated with the system of equations.

Step 2. Use elementary row operations to get equivalent matrix of the form:

$\left[\begin{array}{cc}1& a\\ 0& 1\end{array}⋮\begin{array}{c}b\\ c\end{array}\right];$ where $a,b,c$ are constants.

Step 3. Solve for $x$ and $y.$

2) Calculation:

$x+4y=-2$

$3x-y=7$

The corresponding matrix associated with the above system of equations is

$\left[\begin{array}{cc}1& 4\\ 3& -1\end{array}⋮\begin{array}{c}-2\\ 7\end{array}\right]$

The element in (row1, column1) position is already 1.

To get the (row2, column1) position as zero, multiply row1 by $-3$ and add to row2:

$\left[\begin{array}{cc}1& 4\\ -3\left(1\right)+3& -3\left(4\right)-1\end{array}⋮\begin{array}{c}-2\\ -3\left(-2\right)+7\end{array}\right]$

Simplify further:

$\left[\begin{array}{cc}1& 4\\ 0& -13\end{array}⋮\begin{array}{c}-2\\ 13\end{array}\right]$

Use elementary row operation to get the (row2, column2) position as 1.

So, divide row 2 by $-13$

$\left[\begin{array}{cc}1& 4\\ \frac{0}{-13}& \frac{-13}{-13}\end{array}⋮\begin{array}{c}-2\\ \frac{13}{-13}\end{array}\right]$

Simplify further,

$\left[\begin{array}{cc}1& 4\\ 0& 1\end{array}⋮\begin{array}{c}-2\\ -1\end{array}\right]$

The above matrix corresponds to the system of equations:

$x+4y=-2$

$y=-1$

Substitute $y=-1$ in the first equation, $x+4y=-2.$

$x+4\left(-1\right)=-2$

$x-4=-2$

$x=-2+4$

$x=2$

Thus, the solutions of the given system of equations are: $x=2;y=-1.$

Conclusion:

Therefore, the required solutions are: $x=2;y=-1.$