Q. 4.87 - Intermediate Algebra

Question

Solve the system of equations using a matrix: $\left\{\begin{array}{l}x+y-z=0\\ 2x+4y-2z=6\\ 3x+6y-3z=9\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of linear equation is $\left(z-3,3,z\right)$ where, $z$is any real numbers.

Step 1. Given information.

Consider the given system of equations,

$\left\{\begin{array}{l}x+y-z=0\\ 2x+4y-2z=6\\ 3x+6y-3z=9\end{array}\right.$

Step 2. Write in augmented form.

The augmented matrix for the given system of equations is

$\left[\begin{array}{cccr}1& 1& -1& 0\\ 2& 4& -2& 6\\ 3& 6& -3& 9\end{array}\right]$

Step 3. Apply row operations.

Apply $R_2-2\times R_1\to R_2$ and $R_3-3\times R_1\to R_3$,

$\left[\begin{array}{cccr}1& 1& -1& 0\\ 0& 2& 0& 6\\ 0& 3& 0& 9\end{array}\right]$

Apply $\frac{R_2}{2}\to R_2$ and $R_3-3\times R_2\to R_3$,

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

Now, the matrix is in row-echelon form.

Step 4. Write in system of equations.

Writing the corresponding system of equations,

$$\begin{gathered} x+y-z=0 ...... \left(i\right) \\ y=3 ...... \left(ii\right) \end{gathered}$$

Substitute the value of $y$ in equation (i),

$$\begin{gathered} \mathrm{x}+3-\mathrm{z}=0 \\ \mathrm{x}-\mathrm{z}=-3 ...... \left(iii\right) \end{gathered}$$

Solve equation (iii) for $x$,

$x=z-3$

where, $z$ is any real numbers.

Step 5. Check the answers.

Substitute the values in equation (i),

$$\begin{gathered} \left(z-3\right)+3-z=0 \\ 0=0 \end{gathered}$$

This is true.