Question

Solve the system of equations by transforming a matrix representing the system of equation into reduced row echelon form. 2x+y-3z=-19 x+2y+z=-4 x-y+5z=21 what is the solution to the system of equations? Drag a choice into each box to correctly complete the table.

Answer 1

We want to write the following system

`\begin{gathered} 2x+y-3z=-19 \\ x+2y+z=-4 \\ x-y+5z=21 \end{gathered}`

As a matrix. To do that, we just take the coefficients and plug them in the same order in a matrix. Our system can be rewritten as:

`\begin{bmatrix}{2} & 1 & {}-3 \\ {1} & {2} & 1{} \\ {1} & {-1} & {5}\end{bmatrix}=\begin{bmatrix}{-19} \\ {-4} \\ {21}\end{bmatrix}`

Now, to solve this, we can use the gaussian elimination. It consists of adding, multiplying, and changing the order of the rows, until we have an identity matrix on the left side.

Let's start by subtracting the second row from the third row, and subtracting 2 times the second row from the first row.

`\begin{gathered} \begin{bmatrix}{2-2*1} & 1-2*2 & {}-3-2*1 \\ {1} & {2} & 1{} \\ {1-1} & {-1-2} & {5-1}\end{bmatrix}=\begin{bmatrix}{-19-2*(-4)} \\ {-4} \\ {21-(-4)}\end{bmatrix} \\ \begin{bmatrix}{0} & -3 & {}-5 \\ {1} & {2} & 1{} \\ {0} & {-3} & {4}\end{bmatrix}=\begin{bmatrix}{-11} \\ {-4} \\ {25}\end{bmatrix} \end{gathered}`

Multiplying the first row by (- 1), we have:

`\begin{bmatrix}{0} & 3 & {}5 \\ {1} & {2} & 1{} \\ {0} & {-3} & {4}\end{bmatrix}=\begin{bmatrix}{11} \\ {-4} \\ {25}\end{bmatrix}`

Adding the first row to the third row, we have:

`\begin{gathered} \begin{bmatrix}{0} & 3 & {}5 \\ {1} & {2} & 1{} \\ {0} & {-3+3} & {4+5}\end{bmatrix}=\begin{bmatrix}{11} \\ {-4} \\ {25+11}\end{bmatrix} \\ \begin{bmatrix}{0} & 3 & {}5 \\ {1} & {2} & 1{} \\ {0} & {0} & {9}\end{bmatrix}=\begin{bmatrix}{11} \\ {-4} \\ {36}\end{bmatrix} \end{gathered}`

Multiplying the last row by 1/9:

`\begin{bmatrix}{0} & 3 & {}5 \\ {1} & {2} & 1{} \\ {0} & {0} & {1}\end{bmatrix}=\begin{bmatrix}{11} \\ {-4} \\ {4}\end{bmatrix}`

Now, subtracting the third row from the second row, and subtracting 5 times the third row from the first row, we have:

`\begin{gathered} \begin{bmatrix}{0} & 3 & {}5-5*1 \\ {1} & {2} & 1-1{} \\ {0} & {0} & {1}\end{bmatrix}=\begin{bmatrix}{11-5*4} \\ {-4-4} \\ {4}\end{bmatrix} \\ \begin{bmatrix}{0} & 3 & {0} \\ {1} & {2} & 0{} \\ {0} & {0} & {1}\end{bmatrix}=\begin{bmatrix}{-9} \\ {-8} \\ {4}\end{bmatrix} \end{gathered}`

Dividing the first row by 3:

`\begin{bmatrix}{0} & 1 & {0} \\ {1} & {2} & 0{} \\ {0} & {0} & {1}\end{bmatrix}=\begin{bmatrix}{-3} \\ {-8} \\ {4}\end{bmatrix}`

Subtracting the twice the first row from the second row:

`\begin{gathered} \begin{bmatrix}{0} & 1 & {0} \\ {1} & {2-2*1} & 0{} \\ {0} & {0} & {1}\end{bmatrix}=\begin{bmatrix}{-3} \\ {-8-2*(-3)} \\ {4}\end{bmatrix} \\ \begin{bmatrix}{0} & 1 & {0} \\ {1} & {0} & 0{} \\ {0} & {0} & {1}\end{bmatrix}=\begin{bmatrix}{-3} \\ {-2} \\ {4}\end{bmatrix} \end{gathered}`

And finally, permuting the first and the second row

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

And this is our answer.

`\begin{bmatrix}{1} & 0 & {0} \\ {0} & {1} & 0{} \\ {0} & {0} & {1}\end{bmatrix}\begin{bmatrix}{x} \\ {y} \\ {z}\end{bmatrix}=\begin{bmatrix}{-2} \\ {-3} \\ {4}\end{bmatrix}`