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$6 x - 5 y + 6 z = - 15$ , $- 6 x + 5 y - 6 z = 15$ , $12 x - 10 y + 12 z = - 30$

Bước 1

Write the system as a matrix.

$[\begin{array}{c} 6 & - 5 & 6 & - 15 \\ - 6 & 5 & - 6 & 15 \\ 12 & - 10 & 12 & - 30 \end{array}]$

Bước 2

Tìm dạng ma trận hàng bậc thang rút gọn.

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Bước 2.1

Multiply each element of $R_{1}$ by $\frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

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Bước 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{6}{6} & \frac{- 5}{6} & \frac{6}{6} & \frac{- 15}{6} \\ - 6 & 5 & - 6 & 15 \\ 12 & - 10 & 12 & - 30 \end{array}]$

Bước 2.1.2

Rút gọn $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{5}{6} & 1 & - \frac{5}{2} \\ - 6 & 5 & - 6 & 15 \\ 12 & - 10 & 12 & - 30 \end{array}]$

Bước 2.2

Perform the row operation $R_{2} = R_{2} + 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Bước 2.2.1

Perform the row operation $R_{2} = R_{2} + 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{5}{6} & 1 & - \frac{5}{2} \\ - 6 + 6 \cdot 1 & 5 + 6 (- \frac{5}{6}) & - 6 + 6 \cdot 1 & 15 + 6 (- \frac{5}{2}) \\ 12 & - 10 & 12 & - 30 \end{array}]$

Bước 2.2.2

Rút gọn $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{5}{6} & 1 & - \frac{5}{2} \\ 0 & 0 & 0 & 0 \\ 12 & - 10 & 12 & - 30 \end{array}]$

Bước 2.3

Perform the row operation $R_{3} = R_{3} - 12 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Bước 2.3.1

Perform the row operation $R_{3} = R_{3} - 12 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{5}{6} & 1 & - \frac{5}{2} \\ 0 & 0 & 0 & 0 \\ 12 - 12 \cdot 1 & - 10 - 12 (- \frac{5}{6}) & 12 - 12 \cdot 1 & - 30 - 12 (- \frac{5}{2}) \end{array}]$

Bước 2.3.2

Rút gọn $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{5}{6} & 1 & - \frac{5}{2} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Bước 3

Use the result matrix to declare the final solution to the system of equations.

$x - \frac{5}{6} y + z = - \frac{5}{2}$

$0 = 0$

Bước 4

The solution is the set of ordered pairs that make the system true.

$(- \frac{5}{2} + \frac{5 y}{6} - z, y, z)$