Question

Solve the following sets of simultaneous equations by reducing the matrix to row echelon form. $$ \left\\{\begin{array}{r} 2 x+y-5 z=7 \\ x-2 y=1 \\ 3 x-5 y-z=4 \end{array}\right. $$

Short answer

x = 4, y = 2, z = 1

Step-by-step solution

Write the augmented matrix

First, write the system of equations as an augmented matrix.equations:\[ \begin{align*} 2x + y - 5z &= 7 \ x - 2y + 0z &= 1 \ 3x - 5y - z &= 4 \end{align*} \]matrix:\[ \left[ \begin{array}{ccc|c} 2 & 1 & -5 & 7 \ 1 & -2 & 0 & 1 \ 3 & -5 & -1 & 4 \end{array} \right] \]

Make the leading coefficient of row 1 equal to 1

Divide the first row by 2.\[ \left[ \begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{5}{2} & \frac{7}{2} \ 1 & -2 & 0 & 1 \ 3 & -5 & -1 & 4 \end{array} \right] \]

Eliminate the first element of row 2 and row 3

Subtract row 1 from row 2 and subtract 3 times row 1 from row 3.\[ R_2 = R_2 - R_1 \]\[ R_3 = R_3 - 3R_1 \]results in:\[ \left[ \begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{5}{2} & \frac{7}{2} \ 0 & -\frac{5}{2} & \frac{5}{2} & -\frac{5}{2} \ 0 & -\frac{11}{2} & \frac{13}{2} & -\frac{9}{2} \end{array} \right] \]

Make the leading coefficient of row 2 equal to 1

Multiply row 2 by -2/5.\[ R_2 = -\frac{2}{5}R_2 \]results in:\[ \left[ \begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{5}{2} & \frac{7}{2} \ 0 & 1 & -1 & 1 \ 0 & -\frac{11}{2} & \frac{13}{2} & -\frac{9}{2} \end{array} \right] \]

Eliminate the second element of row 3

Add 11/2 times row 2 to row 3.\[ R_3 = R_3 + \frac{11}{2} R_2 \]results in:\[ \left[ \begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{5}{2} & \frac{7}{2} \ 0 & 1 & -1 & 1 \ 0 & 0 & 1 & 1 \end{array} \right] \]

Eliminate above rows to achieve row echelon form

Subtract appropriate multiples of row 3 from row 1 and row 2 to eliminate the third column elements.For row 2 (add row 3 to row 2):\[ R_2 = R_2 + R_3 \]For row 1 (add 5/2 times row 3 to row 1):\[ R_1 = R_1 + (\frac{5}{2} R_3) \]results in:\[ \left[ \begin{array}{ccc|c} 1 & \frac{1}{2} & 0 & 5 \ 0 & 1 & 0 & 2 \ 0 & 0 & 1 & 1 \end{array} \right] \]

Interpret the row echelon form

Each row corresponds to an equation:\[ \begin{align*} x + \frac{1}{2}y &= 5 \ y &= 2 \ z &= 1 \end{align*} \]Substitute back to find final values of x and y.

Solve for x

From \(y = 2\), substitute into the first equation to solve for x:\[ x + \frac{1}{2} \times 2 = 5 \ x + 1 = 5 \ x = 4 \]

Final Solution

The solution is: \( x = 4 \), \( y = 2 \), \( z = 1 \)

Key concepts

Row Echelon Form

To solve systems of linear equations, we often transform the system into a simpler form known as 'Row Echelon Form' (REF). This is a form where:

• All nonzero rows (rows with at least one non-zero element) are above any rows of all zeros.
• The leading coefficient (the first non-zero number from the left) of a nonzero row is always to the right of the leading coefficient of the row above it.
• The leading coefficient of each nonzero row is 1.

Here is why we do this:
Transforming a matrix into REF makes it easier to solve for the variables as it isolates each variable step by step. Let's take a closer look at our example:
Starting off with the system of equations provided: \[ \begin{align*} 2x + y - 5z &= 7 \ x - 2y &= 1 \ 3x - 5y - z &= 4 \end{align*} \] we first converted this into an augmented matrix (more on that in the next section). As we performed various row operations, we aimed for the matrix to look like this: \[ \begin{array}{ccc|c} 1 & a & b & c \ 0 & 1 & d & e \ 0 & 0 & 1 & f \ \end{array} \] This form clearly identifies the values of variable z, then y, and subsequently x.

Augmented Matrix

In solving simultaneous equations, an efficient way to handle multiple equations is by expressing them in an 'augmented matrix'. This matrix includes all the coefficients of the variables and the constants from the equations as an additional column.
Let's see how we do it. From our original equations: \[ \begin{align*} 2x + y - 5z &= 7 \ x - 2y &= 1 \ 3x - 5y - z &= 4 \ \end{align*} \ \] We write the coefficients of x, y, and z in matrix form along with the constants: \[ \begin{array}{ccc|c} 2 & 1 & -5 & 7 \ 1 & -2 & 0 & 1 \ 3 & -5 & -1 & 4 \ \end{array} \ \] Each row represents an equation, and each column represents the coefficients of x, y, z, and the constants. This compact form allows for better visualization and streamlined row operations.

Gaussian Elimination

Gaussian Elimination is a method used to solve systems of linear equations. It involves transforming the augmented matrix into a row echelon form using row operations.
There are three types of row operations:

• Swapping two rows
• Multiplying a row by a nonzero scalar
• Adding or subtracting a multiple of one row to another row

Our step-by-step solution used Gaussian Elimination as follows:

• First, we wrote the augmented matrix and then made the leading coefficient of row 1 a 1 by dividing the entire row by 2.
• Next, we eliminated the first element of rows 2 and 3 by subtracting appropriate multiples of row 1.
• We then made the leading coefficient of row 2 a 1 and repeated the elimination process for the subsequent rows.

The aim here is to achieve a matrix that can easily be interpreted back into equations, ideally resembling the identity matrix on the left side and the solutions in the rightmost column as seen in our final form: \[ \begin{array}{ccc|c} 1 & \frac{1}{2} & 0 & 5 \ 0 & 1 & 0 & 2 \ 0 & 0 & 1 & 1 \ \end{array} \] This systematic process ensures we deduce the values of x, y, and z step-by-step, leading us to the final solution: x = 4, y = 2, z = 1.