Question

Solve the following system of equations: \left\{\begin{array}{c l}

-x-6y-4z = 30\\
-4x-3y+z = 20\\
-x+y+5z = 10
\end{array}\right.

Answer 1

To solve the system of equations, we can use the method of Gaussian elimination. Here are the steps:

Step 1: Write the system of equations in matrix form:

`\begin{bmatrix}-1 & -6 & -4 \\ -4 & -3 & 1 \\ -1 & 1 & 5 \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}=\begin{bmatrix} 30 \\ 20 \\ 10 \\ \end{bmatrix}`

Step 2: Apply row operations to transform the matrix into row-echelon form.

`\begin{bmatrix}-1 & -6 & -4 & | & 30 \\ -4 & -3 & 1 & | & 20 \\ -1 & 1 & 5 & | & 10 \\ \end{bmatrix} \\`

First, we will perform row 2 = row 2 + 4 * row 1 and row 3 = row 3 + row 1.

`\begin{bmatrix}-1 & -6 & -4 & | & 30 \\ 0 & 21 & -15 & | & 110 \\ 0 & -5 & 1 & | & 40 \\ \end{bmatrix} \\`

Next, we will perform row 2 = (1/21) * row 2 and row 3 = row 3 + (5/21) * row 2.

`\begin{bmatrix}-1 & -6 & -4 & | & 30 \\ 0 & 1 & -5/7 & | & 20/7 \\ 0 & 0 & -10/7 & | & 70/7 \\ \end{bmatrix} \\`

Finally, we will perform row 3 = (7/10) * row 3.

`\begin{bmatrix}-1 & -6 & -4 & | & 30 \\ 0 & 1 & -5/7 & | & 20/7 \\ 0 & 0 & 1 & | & -1 \\ \end{bmatrix} \\`

Step 3: Back-substitution

From the row-echelon form, we can read the values of x, y, and z:

`\begin{align}\sf\:z &= -1 \\ y - (5)/(7)z &= (20)/(7) \\ y - (5)/(7)(-1) &= (20)/(7) \\ y + (5)/(7) &= (20)/(7) \\ y &= (20)/(7) - (5)/(7) \\ y &= (15)/(7) \\ x - 6y - 4z &= 30 \\ x - 6\left((15)/(7)\right) - 4(-1) &= 30 \\ x - (90)/(7) + (4)/(7) &= 30 \\ x &= 30 + (90)/(7) - (4)/(7) \\ x &= (210)/(7) + (90)/(7) - (4)/(7) \\ x &= (296)/(7) \\ \end{align} \\`

Therefore, the solution to the system of equations is:

`\begin{align}\sf\:x &= (296)/(7) \\ y &= (15)/(7) \\ z &= -1 \\ \end{align} \\`

Please note that the solution is in fractional form. It can be simplified or expressed as a decimal if needed.