Q63. - Algebra 1

Question

Use an augmented matrix to solve each system of equations.

$\begin{array}{l}x-3y=24\\ 5x-2y=29\end{array}$

Step-by-Step Solution

Answer

The solution of the system of equations is $\left(3,-7\right)$.

Step 1. Write the augmented matrix.

To write the equations in augmented matrix place the coefficients of the equations and the constant terms into a matrix separated by a dashed line.

 

Here, the augmented matrices are:

$\left[\begin{array}{ccc}1& -3& 24\\ 5& -2& 29\end{array}\right]$

Step 2. Use row operations to solve the system.

To make the first element in the second row a 0, divide the second row by 3 and then subtract row 1 from the resultant row 2.

 

$\left[\begin{array}{ccc}1& -3& 24\\ 5& -2& 29\end{array}\right]\stackrel{R_2\to R_2-5R_1}{\to }\left[\begin{array}{ccc}1& -3& 24\\ 0& 13& -91\end{array}\right]$ 

 

To make the second element in the second row a 1, divide the second row by 13.

 

$\left[\begin{array}{ccc}1& -3& 24\\ 0& 13& -91\end{array}\right]\stackrel{\frac{1}{13}{R}_2}{\to }\left[\begin{array}{ccc}1& -3& 24\\ 0& 1& -7\end{array}\right]$ 

Step 3. Row reduce the matrix.

Further row-reduce the augmented matrix, by making the second element in the first row a zero.

 

$\left[\begin{array}{ccc}1& -3& 24\\ 0& 1& -7\end{array}\right]\stackrel{R_1\to R_1+3R_2}{\to }\left[\begin{array}{ccc}1& 0& 3\\ 0& 1& -7\end{array}\right]$ 

 

Here, the first row will give the solution of x, because the coefficient of y is 0 and the coefficient of x is 1. Similarly, the second row will give the solution of y, because the coefficient of x is 0 and the coefficient of y is 1. Therefore, the solution is $\left(3,-7\right)$.