Solve the system of equations using a matrix: x-y-z=1-x+2y-3z=-43x-2y-7z=0

Q. 4.88 - Intermediate Algebra

Solve the system of equations using a matrix: $\{\begin{cases} x - y - z = 1 \\ - x + 2 y - 3 z = - 4 \\ 3 x - 2 y - 7 z = 0 \end{cases}$

Verified

Answer

The solution for the system of linear equation is $(5 z - 2, 4 z - 3, z)$ where, $z$ is any real number.

1Step 1. Given information.

Consider the given system of equations,

$\{\begin{cases} x - y - z = 1 \\ - x + 2 y - 3 z = - 4 \\ 3 x - 2 y - 7 z = 0 \end{cases}$

2Step 2. Write in augmented form.

The augmented matrix for the given system of equations is

$[\begin{array}{cccr} 1 & - 1 & - 1 & 1 \\ - 1 & 2 & - 3 & - 4 \\ 3 & - 2 & - 7 & 0 \end{array}]$

3Step 3. Apply row operations.

Apply $R_{2} + R_{1} \to R_{2}$ and $R_{3} - 3 \times R_{1} \to R_{3}$ ,

$[\begin{array}{cccr} 1 & - 1 & - 1 & 1 \\ 0 & 1 & - 4 & - 3 \\ 0 & 1 & - 4 & - 3 \end{array}]$

Apply $R_{3} - R_{2} \to R_{3}$ ,

$[\begin{array}{cccr} 1 & - 1 & - 1 & 1 \\ 0 & 1 & - 4 & - 3 \\ 0 & 0 & 0 & 0 \end{array}]$

4Step 4. Write in system of equations.

Writing the corresponding system of equations,

$x - y - z = 1 . . . . . . (i) y - 4 z = - 3 . . . . . . (i i)$

Solve equation (ii) for $y$ ,

$y = 4 z - 3$

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

$x - (4 z - 3) - z = 1 x - 5 z + 3 = 1 x = 5 z - 2$

5Step 5. Check the answers.

Substitute the values in equation (i),

$(5 z - 2) - (4 z - 3) - z = 1 5 z - 2 - 4 z + 3 - z = 1 1 = 1$

This is true.