Question

Subscript notation is frequently used for working with larger systems of equations. Use a matrix approach to solve the system. Express the solutions as 4-tuples of the form (x1, x2, x3, x4).X1 – 3x2 – 2x3 + x4 = -3 -2x1 + 7x2 + x3 – 2X4 = -1 3x1 – 7x2 – 3x3 + 3x4 = -5 5x1 + x2 + 4x3 – 2x4 = 18(X1, X2, X3, X4) =_________

Answer 1

The solution for the system of equations is
`(x_1, x_2, x_3, x_4)=(1, -1, 2, -3)`

Explanation:

Matrix Method for solving systems of equations is also known as Row Echelon Method.

We have the following system

`\left\begin{array}{cccccc}x_1&-3x_2&-2x_3&+x_4&=&-3\\-2x_1&+7x_2&+1x_3&-2x_4&=&-1\\3x_1&-7x_2&-3x_3&+3x_4&=&-5\\5x_1&+x_2&+4x_3&-2x_4&=&18\end{array}\right`

Step 1: Find the augmented matrix.

An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.

`\left[ \begin{array}c 1 & -3 & -2 & 1 & -3 \\\\ -2 & 7 & 1 & -2 & -1 \\\\ 3 & -7 & -3 & 3 & -5 \\\\ 5 & 1 & 4 & -2 & 18 \end{array} \right]`

Step 2: Transform the augmented matrix to the reduced row echelon form. For this use elementary row operations.

• Add row 1 multiplied by 2 to row 2

`\left[ \begin{array}c 1 & -3 & -2 & 1 & -3 \\\\ 0 & 1 & -3 & 0 & -7 \\\\ 3 & -7 & -3 & 3 & -5 \\\\ 5 & 1 & 4 & -2 & 18 \end{array} \right]`

• Subtract row 1 multiplied by 3 from row 3

`\left[ \begin{array}c 1 & -3 & -2 & 1 & -3 \\\\ 0 & 1 & -3 & 0 & -7 \\\\ 0 & 2 & 3 & 0 & 4 \\\\ 5 & 1 & 4 & -2 & 18 \end{array} \right]`

• Subtract row 1 multiplied by 5 from row 4

`\left[ \begin{array}c 1 & -3 & -2 & 1 & -3 \\\\ 0 & 1 & -3 & 0 & -7 \\\\ 0 & 2 & 3 & 0 & 4 \\\\ 0 & 16 & 14 & -7 & 33 \end{array} \right]`

• Add row 2 multiplied by 3 to row 1

`\left[ \begin{array}c 1 & 0 & -11 & 1 & -24 \\\\ 0 & 1 & -3 & 0 & -7 \\\\ 0 & 2 & 3 & 0 & 4 \\\\ 0 & 16 & 14 & -7 & 33 \end{array} \right]`

• Subtract row 2 multiplied by 2 from row 3

`\left[ \begin{array}c 1 & 0 & -11 & 1 & -24 \\\\ 0 & 1 & -3 & 0 & -7 \\\\ 0 & 0 & 9 & 0 & 18 \\\\ 0 & 16 & 14 & -7 & 33 \end{array} \right]`

• Divide row 3 by 9

`\left[ \begin{array}c 1 & 0 & -11 & 1 & -24 \\\\ 0 & 1 & -3 & 0 & -7 \\\\ 0 & 0 & 1 & 0 & 2 \\\\ 0 & 0 & 62 & -7 & 145 \end{array} \right]`

• Add row 3 multiplied by 11 to row 1

`\left[ \begin{array}c 1 & 0 & 0 & 1 & -2 \\\\ 0 & 1 & -3 & 0 & -7 \\\\ 0 & 0 & 1 & 0 & 2 \\\\ 0 & 0 & 62 & -7 & 145 \end{array} \right]`

• Add row 3 multiplied by 3 to row 2

`\left[ \begin{array}cccc 1 & 0 & 0 & 1 & -2 \\\\ 0 & 1 & 0 & 0 & -1 \\\\ 0 & 0 & 1 & 0 & 2 \\\\ 0 & 0 & 62 & -7 & 145 \end{array} \right]`

• Subtract row 3 multiplied by 62 from row 4

`\left[ \begin{array}c 1 & 0 & 0 & 1 & -2 \\\\ 0 & 1 & 0 & 0 & -1 \\\\ 0 & 0 & 1 & 0 & 2 \\\\ 0 & 0 & 0 & -7 & 21 \end{array} \right]`

• Divide row 4 by −7

`\left[ \begin{array}cccc 1 & 0 & 0 & 1 & -2 \\\\ 0 & 1 & 0 & 0 & -1 \\\\ 0 & 0 & 1 & 0 & 2 \\\\ 0 & 0 & 0 & 1 & -3 \end{array} \right]`

• Subtract row 4 from row 1

`\left[ \begin{array}c 1 & 0 & 0 & 0 & 1 \\\\ 0 & 1 & 0 & 0 & -1 \\\\ 0 & 0 & 1 & 0 & 2 \\\\ 0 & 0 & 0 & 1 & -3 \end{array} \right]`

Step 3: Interpret the reduced row echelon form.

The reduced row echelon form of the augmented matrix is

`\left[ \begin{array}c 1 & 0 & 0 & 0 & 1 \\\\ 0 & 1 & 0 & 0 & -1 \\\\ 0 & 0 & 1 & 0 & 2 \\\\ 0 & 0 & 0 & 1 & -3 \end{array} \right]`

which corresponds to the system

`\left\begin{array}{cccccc}x_1&&&&=&1\\&x_2&&&=&-1\\&&x_3&&=&2\\&&&x_4&=&-3\end{array}\right`

Since every column in the coefficient part of the matrix has a leading entry that means our system has a unique solution:

`\left\begin{array}{ccc}x_1&=&1\\x_2&=&-1\\x_3&=&2\\x_4&=&-3\end{array}\right`

or

`(x_1, x_2, x_3, x_4)=(1, -1, 2, -3)`