Solving Simultaneous Equations Using Matrices

Rewrite the following set of equations in matrix form (also known as \(Ax = b\) form):\begin{equation}\begin{split}-4x + 4y + z & = 3 \\11y - 7x + z & = 4 \\-5x + 3y + 2z & = 5 \\\end{split}\qquad\begin{matrix}(1) \\ (2) \\ (3)\end{matrix}\end{equation}

Solution

STEP 1: Make sure that the variables are in the same order for every equation (i.e. if the order you choose for the first equation is \(x, y, z\) then the variables in every equation that follows must also be in the order \(x, y, z\)).

The order we will choose for this example is \(x, y, z\).

The variables in the second equation are not in the correct order. We correct this by simply rearranging the equation so that the order of the variables is the same as it is for the other two equations.

The second equation will thus look like this:

\[-7x + 11y + z = 4\]

STEP 2: Rewrite the equations in matrix form.

The first matrix we need is the coefficient matrix. It is a square matrix that houses the coefficients of each variable, and its size depends on the number of variables in the given equations.

Each column within the coefficient matrix holds the coefficients for a specific variable. For example, Column 1 holds the coefficients of \(x\). Columns 2 and 3 hold the coefficients for \(y\) and \(z\) respectively. This is why the order of the variables in the given equations is so important; if they are in the wrong order, then the coefficients will be in the wrong columns.

Each row in the coefficient matrix corresponds with one of the given equations. The coefficients in Row 1 are all from Equation 1, Row 2 holds the coefficients from Equation 2, etc.

When we put all of the above together, we will thus get a matrix that looks like this:

\[\begin{bmatrix}-4 & 4 & 1 \\ 11 & -7 & 1 \\ -5 & 3 & 2\end{bmatrix}\]

As you can see, columns 1, 2 and 3 hold the coefficients for \(x, y\) and \(z\), respectively, and rows 1, 2 and 3 hold the coefficients for equations 1, 2 and 3, respectively.

The second matrix we need is a 3 x 1 matrix containing the variables, and it is known as the variable matrix. It must always be placed to the right of the coefficient matrix.

The variables in the matrix are listed from the top down in the order you chose for them, and it should look like the following:

\[\begin{bmatrix}x \\ y \\ z\end{bmatrix}\]

The final matrix needed is a 3 x 1 matrix that lies to the right of the equal sign. It is called the constant matrix, and, similarly to the coefficient matrix, each row contains the constant value belonging to its corresponding equation.

It will look like this:

\[\begin{bmatrix}3 \\ 4 \\ 5\end{bmatrix}\]

Our final answer consists of all three matrices, and it will look as follows:

\[\begin{bmatrix}-4 & 4 & 1 \\11 & -7 & 1 \\-5 & 3 & 2 \\\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}3 \\ 4 \\ 5 \\\end{bmatrix}\]

Using the same equations as from the previous example, rewrite them in an augmented matrix.

\[\begin{equation}\begin{split}-4x + 4y + z & = 3 \\-7x +11y + z & = 4 \\-5x + 3y + 2z & = 5 \\\end{split}\end{equation}\]

Solution

STEP 1: First ensure your equations all have the same order. In this case, there is no equation that needs rearranging.

STEP 2: Write down the coefficients of the variables to begin the matrix.

\[\left[\begin{array}{rrr}-4 & 4 & 1 \\ 11 & -7 & 1 \\ -5 & 3 & 2\end{array}\right.\]

STEP 3: Draw a vertical line to the right of the coefficients.

\[\left[\begin{array}{rrr|}-4 & 4 & 1 \\ 11 & -7 & 1 \\ -5 & 3 & 2\end{array}\right.\]

STEP 4: Write down the constants on the right-hand side of the line and close the brackets

Your answer should look as follows:

\[\left[\begin{array}{rrr|r}-4 & 4 & 1 & 3 \\11 & -7 & 1 & 4 \\-5 & 3 & 2 & 5 \end{array}\right]\]

Calculate the solution to the following system of equations:

\[\begin{align}x + 5y - z & = 1 \\2x + 7y - 4z & = 0 \\4x + 11y -10z & = -2\end{align}\]

Solution

STEP 1: Write down the augmented matrix.

\[\left[\begin{array}{rrr|r}1 & 5 & -1 & 1 \\ 2 & 7 & -4 & 0 \\ 4 & 11 & -10 & -2\end{array}\right]\]

STEP 2: Perform row calculations.

We perform the following row operations:

\[\begin{align}R_{3} - 2R_{2} & \to R_{3} \\R_{2} - 2R_{1} & \to R_{2} \\R_{3} - R_{2} & \to R_{3} \end{align}\]

to obtain the augmented matrix:

\[\left[\begin{array}{rrr|r}1 & 5 & -1 & 1 \\ 0 & -3 & -2 & -2 \\ 0 & 0 & 0 & 0\end{array}\right]\]

In \(Ax = B\) form, this will look as follows:

\[\begin{bmatrix}1 & 5 & -1 \\ 0 & -3 & -2 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}=\begin{bmatrix}1 \\ -2 \\ 0\end{bmatrix}\]

From this, we know that \(x + 5y - z = 1\) and \(-3y - 2z = -2\).

We now simplify the equations to obtain the following:

\[\begin{align}y & = \frac{2}{3} -\frac{2}{3}z \\\hspace{1cm} \\x & = -5y + 1 + z \qquad \text{Substitute} \; y \; \text{into} \; x \\\hspace{1cm} \\\therefore x & = -\frac{7}{3} + \frac{7}{3}z\end{align}\]

Let \(\frac{z}{3}\) be the free variable \(t\).

The equations will thus look like this:

\[\begin{align}x & = -\frac{7}{3} + 7t \\y & = \frac{2}{3} - 2t\end{align}\]

and when in matrix form will give us:

\[\begin{bmatrix}x \\ y \\ z \end{bmatrix} =\begin{bmatrix}-\frac{7}{3} \\ \frac{2}{3} \\ 0\end{bmatrix} +\begin{bmatrix}7 \\ -2 \\ 3\end{bmatrix} t\]

Both \(x\) and \(y\) depend on \(z\) which is why we can introduce them into the system. \(z\) is a free variable, meaning it can change its value freely, and since \(x\) and \(y\) are dependent on \(z\), this means that there are an infinite number of solutions to this system of equations.

Show that the following system of equations has no solutions:

\[\begin{align}x + 2y - z = -8 \\ 2x - y + z = 4 \\ 8x + y + z = 2\end{align}\]

Solution

STEP 1: Write down the augmented matrix.

\[\begin{array}{rrr|r}1 & 2 & -1 & -8 \\ 2 & -1 & 1 & 4 \\ 8 & 1 & 1 & 2\end{array}\]

STEP 2: Perform row calculations until the necessary zeroes are obtained.

\(R_{3} - R_{2} \to R_{3}, \; R_{2} + R_{1} \to R_{2}, \; R_{3} - 2R_{2} \to R_{3}\) are performed to get:

\[\left[\begin{array}{rrr|r}1 & 2 & -1 & -8 \\ 3 & 1 & 0 & 4 \\ 0 & 0 & 0 & -6\end{array}\right]\]

When you look at the last row, you will see that it states that \(0 = -6\). This is definitely not true, and the system thus has no solutions.

Solve the following equations simultaneously using matrix algebra:

\[\begin{align}4x + y & = -7 \\3x-2y & = 3 \\\end{align}\]

Solution

STEP 1: Rewrite the two equations in the form of a matrix equation. Your answer should look like this:

\[\begin{bmatrix}4 & 1 \\3 & -2 \end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}=\begin{bmatrix}-7 \\ 3\end{bmatrix}\]

STEP 2: Calculate the inverse of the coefficient matrix.

We want to solve for \(x\) and \(y\), so we must isolate those two variables on one side of the equation. To do this we must get the identity matrix involved. We do this by multiplying both sides of the equation by the inverse of the coefficient matrix.

Let the coefficient matrix, \(\begin{bmatrix} 4 & 1 \\ 3 & -2 \end{bmatrix} = A\). Therefore,

\[A^{-1} =-\frac{1}{11}\begin{bmatrix}-2 & -1 \\ -3 & 4\end{bmatrix}\]

STEP 3: Multiply both sides of the equation with \(A^{-1}\).

Your result will look as follows:

\[\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} =-\frac{1}{11}\begin{bmatrix}-2 & -1 \\ -3 & 4\end{bmatrix}\begin{bmatrix}-7 \\ 3 \end{bmatrix} \]

STEP 4: Simplify to get the solutions for \(x\) and \(y\).

\[\begin{align}\begin{bmatrix}x \\ y\end{bmatrix}& =-\frac{1}{11} \times\begin{bmatrix}11 \\ 33\end{bmatrix}\\\hspace{1cm}\\\begin{bmatrix}x \\ y \end{bmatrix}& = \begin{bmatrix} -1 \\ -3 \end{bmatrix}\end{align}\]\(\therefore \quad x = -1\) and \(y = -3\)

Use inverse matrices to solve the following set of simultaneous equations:

\[\begin{align}x - y - z & = 4 \\2x + 3y - z & = 2 \\- x - 2y + 3z & = -3\end{align}\]

Solution

STEP 1: Rewrite the equations in matrix form.

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

STEP 2: Let the coefficient matrix equal A and calculate it's inverse, \(A^{-1}\).

\[A^{-1} =\frac{1}{13} \begin{bmatrix}\tag{*}7 & 5 & 4 \\-5 & 2 & -1 \\-1 & 3 & 5\end{bmatrix}\]

*The full workings are not shown here. Please refer to Inverting Matrices to see how to calculate the inverse of a 3 x 3 matrix.

STEP 3: Multiply by \(A^{-1}\) on the left of both sides of the equation.

\[\frac{1}{13} \begin{bmatrix}7 & 5 & 4 \\-5 & 2 & -1 \\-1 & 3 & 5\end{bmatrix}\begin{bmatrix}1 & -1 & -1 \\ 2 & 3 & -1 \\-1 & -2 & 3 \\\end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix} =\frac{1}{13} \begin{bmatrix}7 & 5 & 4 \\-5 & 2 & -1 \\-1 & 3 & 5\end{bmatrix} \begin{bmatrix}4 \\ 2 \\ -3\end{bmatrix}\]

STEP 4: Simplify.

\[\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= \frac{1}{13} \begin{bmatrix}26 \\ -13 \\ -12 \end{bmatrix}\]

\[\begin{bmatrix}x \\ y\\z \end{bmatrix}= \begin{bmatrix} 2 \\ -1 \\ -1 \end{bmatrix}\]

\(\therefore \quad x = 2,\quad y = -1\quad\) and \(\quad z = -1\).

Use row reduction to solve the following system of equations:

\[\begin{align}2x + y - 3z = -11 \\x - 2y + z = -3 \\-3x + 4y - 2z = 9 \end{align}\qquad \quad \quad \begin{matrix}(1) \\ (2) \\ (3) \end{matrix}\]

Solution

STEP 1: Write down the augmented matrix for these equations. It should look as follows:

\[\left[\begin{array}{rrr|r}2 & 1 & -3 & -11 \\ 1 & -2 & 1 & -3 \\ -3 & 4 & -2 & 9\end{array}\right]\qquad \quad \quad \begin{array}{r}(1) \\ (2) \\ (3)\end{array}\]

STEP 2: Perform row calculations.

First, write down your augmented matrix.

\[\left[\begin{array}{rrr|r}2 & 1 & -3 & -11 \\ 1 & -2 & 1 & -3 \\-3 & 4 & -2 & 9\end{array}\right]\]

What calculation can we do to obtain our first zero?

\[\left[\begin{array}{rrr|r}5 & 0 & -5 & -25 \\ 1 & -2 & 1 & -3 \\-3 & 4 & -2 & 9\end{array}\right]\quad \begin{array}{r}2 \times R_{1} + R_{2} \to R_{1} \\ \hspace{1cm} \\ \hspace{1cm}\end{array}\]

We always write down the row calculations we're doing next to the matrix.

The second zero must be in the same column as the first.

\[\left[\begin{array}{rrr|r}5 & 0 & -5 & -25 \\ -1 & 0 & 0 & 3 \\-3 & 4 & -2 & 9\end{array}\right]\quad \begin{array}{r}\hspace{1cm} \\ 2 \times R_{2} + R_{3} \to R_{2} \\ \hspace{1cm}\end{array}\]

It just so happened that both the second and third zero were obtained through this second-row calculation. This is not always the case, and you will often have to perform a third-row calculation to get the third zero.

STEP 3: Calculate the values of the unknowns.

From Row 2:

\[\begin{align}-x & = 3 \\\therefore \qquad x & = -3 \end{align}\]

From Row 1 (substitute \(x\) in):

\[\begin{align}5x - 5z & = -25 \\5(-3) - 5z & = -25 \\-5z & = -10 \\\therefore \qquad z & = 2\end{align}\]

From Row 3 (substitute \(x\) and \(z\) in):

\[\begin{align}-3x + 4y - 2z & = 9 \\-3(-3) + 4y - 2(2) & = 9 \\4y & = 4 \\\therefore \qquad y & = 1 \end{align}\]

\(\therefore \qquad x = -3; \; y = 1\) and \(z = 2\)

Solve for \(x\) and \(y\) using inverse matrices.

\[\begin{align}6x - 5y & = 7 \\12x + 20y & = -4 \end{align}\]

Solution

First, we must rewrite the equations in matrix form.

\[\begin{bmatrix}6 & -5 \\ 12 & 20 \end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}=\begin{bmatrix}7 \\ -4\end{bmatrix}\]

Next, we have to calculate the inverse of the coefficient matrix

\[\begin{align}\begin{vmatrix}6 & -5 \\ 12 & 20\end{vmatrix}= (6)(20) - (-5)(12)= 180 \\\hspace{1cm} \\\therefore \qquad \begin{bmatrix}6 & -5 \\ 12 & 20 \end{bmatrix}^{-1}=\frac{1}{180}\begin{bmatrix}20 & 5 \\ -12 & 6\end{bmatrix}\end{align}\]

Multiply by the inverse matrix on the left of both sides of the equation to get:

\[\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \frac{1}{180} \begin{bmatrix}20 & 5 \\ -12 & 6\end{bmatrix}\begin{bmatrix}7 \\ -4\end{bmatrix}\]

Simplify further to get the final result:

\[\begin{bmatrix}x \\ y\end{bmatrix}=\frac{1}{180}\begin{bmatrix}120 \\ -108\end{bmatrix}\]

\(\therefore \qquad x = \frac{2}{3} \; \text{and} \; y = -\frac{3}{5}\)

Solve for \(x\) and \(y\) using row reduction.

\[\begin{align}8x - 3y & = 6 \\-16x + y & = -\frac{31}{3}\end{align}\]

Solution

Write down the augmented matrix for the system.

\[\left[\begin{array}{rr|r}8 & -3 & 6 \\ -16 & 1 & -\frac{31}{3}\end{array}\right]\]

Next, perform row operations until a zero is obtained in one of the rows. Because there are only two variables to solve for, we only need one zero as that will allow us to calculate the value of one of the variables. We can then use the calculated variable to find the value of the other unknown by substituting it back into the equation.

\[\left[\begin{array}{rr|r}8 & -3 & 6 \\ 0 & -5 & \frac{5}{3}\end{array}\right]\qquad\begin{matrix}\hspace{1cm} \\ R_{2} + 2R_{1} \to R_{2}\end{matrix}\]

Solve for \(x\) and \(y\).

\[\begin{align}-5y & = \frac{5}{3} \\y & = -\frac{1}{3} \\\hspace{1cm} \\\therefore \qquad8x - 3y & = 6 \\8x - 3(-\frac{1}{3}) & = 6 \\x & = \frac{5}{8}\end{align}\]

Solve the following system of equations using row reduction:

\[\begin{align}-4x - 5y +3z & = 7\\-2x + 3y - z & = -25 \\3x - 2y - 4z & = -6\end{align}\]

Solution

STEP 1: Write down the augmented matrix.

\[\left[\begin{array}{rrr|r}-4 & - 5 & 3 & 7 \\ -2 & 3 & -1 & -25 \\ 3 & -2 & -4 & -6 \end{array}\right]\]

STEP 2: Perform row calculations.

\[\left[\begin{array}{rrr|r}-10 & 4 & 0 & -68 \\ -2 & 3 & -1 & -25 \\ 3 & -2 & -4 & -6\end{array}\right]\quad\begin{array}{r}R_{1} + 3R_{2} \to R_{1} \\\hspace{1cm} \\\hspace{1cm}\end{array}\]\[\left[\begin{array}{rrr|r}-10 & 4 & 0 & -68 \\ -11 & 14 & 0 & -94 \\ 3 & -2 & -4 & -6\end{array}\right]\quad\begin{array}{r}4R_{2} - R_{3} \to R_{2}\end{array}\]\[\left[\begin{array}{rrr|r}-48 & 0 & 0 & -288 \\ -11 & 14 & 0 & -94 \\ 3 & -2 & -4 & -6\end{array}\right]\quad\begin{array}{r}7R_{1} - 2R_{2} \to R_{1} \\\hspace{1cm} \\\hspace{1cm}\end{array}\]

STEP 3: Calculate the values of the variables.

\[\begin{align}-48x & = - 288 \\x & = 6\end{align}\]

\[\begin{align}-11x + 14y & = -94 \\\qquad -11(6) + 14y & = -94 \\14y & = -28 \\y & = -2\end{align}\]

\[\begin{align}3x - 2y - 4z & = -6 \\3(6) - 2(-2) - 4z & = -6 \\-4z & = -28 \\z &= 7\end{align}\]

\(\therefore \qquad x = 6, \; y = -2\) and \(z = 7\)

Solve the following simultaneous equations using row reduction:

\[\begin{align}2b + c & = - 8 \\a - 2b - 3c & = 0 \\-a + b + 2c & = 3\end{align}\]

Solution

STEP 1: Write down the augmented matrix.

\(2b + c= - 8\) is the same thing as saying \(0a + 2b + c\) so we therefore say \(a\)'s coefficient is 0 for equation 1.

\[\left[\begin{array}{rrr|r}0 & 2 & 1 & -8 \\ 1 & -2 & -3 & 0 \\ -1 & 1 & 2 & 3\end{array}\right]\]

STEP 2: Perform row calculations.

The zero actually makes it easier to solve the system of equations, as there are fewer steps needed to obtain the three necessary zeroes.

\[\left[\begin{array}{rrr|r}0 & 2 & 1 & -8 \\ 0 & -1 & -1 & 3 \\ -1 & 1 & 2 & 3 \end{array}\right]\quad\begin{array}{r}\hspace{1cm} \\R_{2} + R_{3} \to R_{2} \\\hspace{1cm}\end{array}\]\[\left[\begin{array}{rrr|r}0 & 1 & 0 & -5 \\ 0 & -1 & -1 & 3 \\ -1 & 1 & 2 & 3\end{array}\right]\quad\begin{array}{r}R_{1} + R_{2} \to R_{1} \\\hspace{1cm} \\\hspace{1cm}\end{array}\]

STEP 3: Calculate the values of the variables.

\[\begin{align}b & = -5 \end{align}\]

\[\begin{align}-b - c & = 3 \\-(-5) - c & = 3 \\c & = 2 \end{align}\]

\[\begin{align}-a + b + 2c & = 3 \\-a + (-5) + 2(2) & = 3 \\a & = -4 \end{align}\]

\( \therefore \qquad a = -4, \; b = -5\) and \(c = 2\)