Question

in exercises 55-60, use matrices to solve the system of equations, if possible. use gauss-jordan elimination.

1. $\

$$\begin{cases} x - 3z = -2 \\\\ 3x + y - 2z = 5 \\\\ 2x + 2y + z = 4 \\end{cases}$$

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1. $\

$$\begin{cases} 2x - y + 3z = 24 \\\\ 2y - z = 14 \\\\ 7x - 5y = 6 \\end{cases}$$

$

Explanation:

For Problem 55:

Step1: Write augmented matrix

$$\begin{bmatrix} 1 & 0 & -3 & -2 \\ 3 & 1 & -2 & 5 \\ 2 & 2 & 1 & 4 \end{bmatrix}$$

Step2: Eliminate $x$ from row2/3

Row2 = Row2 - 3×Row1; Row3 = Row3 - 2×Row1

$$\begin{bmatrix} 1 & 0 & -3 & -2 \\ 0 & 1 & 7 & 11 \\ 0 & 2 & 7 & 8 \end{bmatrix}$$

Step3: Eliminate $y$ from row3

Row3 = Row3 - 2×Row2

$$\begin{bmatrix} 1 & 0 & -3 & -2 \\ 0 & 1 & 7 & 11 \\ 0 & 0 & -7 & -14 \end{bmatrix}$$

Step4: Normalize row3

Row3 = $\frac{1}{-7}$×Row3

$$\begin{bmatrix} 1 & 0 & -3 & -2 \\ 0 & 1 & 7 & 11 \\ 0 & 0 & 1 & 2 \end{bmatrix}$$

Step5: Back-substitute $z$ to row1/2

Row1 = Row1 + 3×Row3; Row2 = Row2 - 7×Row3

$$\begin{bmatrix} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 2 \end{bmatrix}$$

Step1: Write augmented matrix

$$\begin{bmatrix} 2 & -1 & 3 & 24 \\ 0 & 2 & -1 & 14 \\ 7 & -5 & 0 & 6 \end{bmatrix}$$

Step2: Normalize row1

Row1 = $\frac{1}{2}$×Row1

$$\begin{bmatrix} 1 & -\frac{1}{2} & \frac{3}{2} & 12 \\ 0 & 2 & -1 & 14 \\ 7 & -5 & 0 & 6 \end{bmatrix}$$

Step3: Eliminate $x$ from row3

Row3 = Row3 - 7×Row1

$$\begin{bmatrix} 1 & -\frac{1}{2} & \frac{3}{2} & 12 \\ 0 & 2 & -1 & 14 \\ 0 & -\frac{3}{2} & -\frac{21}{2} & -78 \end{bmatrix}$$

Step4: Normalize row2

Row2 = $\frac{1}{2}$×Row2

$$\begin{bmatrix} 1 & -\frac{1}{2} & \frac{3}{2} & 12 \\ 0 & 1 & -\frac{1}{2} & 7 \\ 0 & -\frac{3}{2} & -\frac{21}{2} & -78 \end{bmatrix}$$

Step5: Eliminate $y$ from row1/3

Row1 = Row1 + $\frac{1}{2}$×Row2; Row3 = Row3 + $\frac{3}{2}$×Row2

$$\begin{bmatrix} 1 & 0 & \frac{5}{4} & \frac{31}{2} \\ 0 & 1 & -\frac{1}{2} & 7 \\ 0 & 0 & -\frac{45}{4} & -\frac{135}{2} \end{bmatrix}$$

Step6: Normalize row3

Row3 = $-\frac{4}{45}$×Row3

$$\begin{bmatrix} 1 & 0 & \frac{5}{4} & \frac{31}{2} \\ 0 & 1 & -\frac{1}{2} & 7 \\ 0 & 0 & 1 & 6 \end{bmatrix}$$

Step7: Back-substitute $z$ to row1/2

Row1 = Row1 - $\frac{5}{4}$×Row3; Row2 = Row2 + $\frac{1}{2}$×Row3

$$\begin{bmatrix} 1 & 0 & 0 & 8 \\ 0 & 1 & 0 & 10 \\ 0 & 0 & 1 & 6 \end{bmatrix}$$

Answer:

$x=4$, $y=-3$, $z=2$

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For Problem 56: