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Q57.

Expert-verifiedFound in: Page 392

Book edition
Student Edition

Author(s)
Carter, Cuevas, Day, Holiday, Luchin

Pages
801 pages

ISBN
9780078884801

**Use an augmented matrix to solve each system of equations.**

$\begin{array}{l}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{5}\\ \mathbf{3}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{3}\end{array}$

The solution of the system of equations is$\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{)}$.

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:

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

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

$\begin{array}{r}\left[\begin{array}{ccc}1& -1& 5\\ 3& 3& 3\end{array}\right]\stackrel{\frac{1}{3}{R}_{2}}{\to}\left[\begin{array}{ccc}1& -1& 5\\ 1& 1& 1\end{array}\right]\\ \stackrel{{R}_{2}-{R}_{1}}{\to}\left[\begin{array}{ccc}1& -1& 5\\ 0& 2& -4\end{array}\right]\end{array}$

To make the second element in the second row a 1, divide the second row by 2.

$\left[\begin{array}{ccc}1& -1& 5\\ 0& 2& -4\end{array}\right]\stackrel{\frac{1}{2}{R}_{2}}{\to}\left[\begin{array}{ccc}1& -1& 5\\ 0& 1& -2\end{array}\right]$

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

$\left[\begin{array}{ccc}1& -1& 5\\ 0& 1& -2\end{array}\right]\stackrel{{R}_{1}+{R}_{2}}{\to}\left[\begin{array}{ccc}1& 0& 3\\ 0& 1& -2\end{array}\right]$$\left[\begin{array}{ccc}1& -1& 5\\ 0& 1& -2\end{array}\right]\stackrel{{R}_{1}+{R}_{2}}{\to}\left[\begin{array}{ccc}1& 0& 3\\ 0& 1& -2\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 $(3,-2)$.

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