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Solving Simultaneous Equations Using Matrices

- Calculus
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Matrices are used for many things in life, from encrypting data to 3D game design, but they are also extremely useful in other mathematical contexts such as solving simultaneous equations. A set of simultaneous equations (also known as a system of equations) can be solved through various applications of matrix algebra by making use of inverse matrices and methods such as row reduction.

This article will address solving simultaneous equations using matrices by first writing them in matrix form and then exploring different methods of solving them, like using inverse matrices and row reduction.

To solve a set of equations using matrices, you have to be able to rewrite said equations in the form of matrices.

The following example explains the process in detail.

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:

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

What would happen if you were to multiply the matrices?

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

The result gives us the original equations. This is the reason why these specific matrices are set out in such a particular way.

The most common method of solving simultaneous equations with matrices involves using row reduction. To do row reduction, one must be able to write the given equations in an augmented matrix.

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:

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

It is possible that a system of equations might not have a unique solution. One can use augmented matrices to determine the number of solutions, if any, to a system of equations.

If given a system of equations with three equations in three unknowns, we can model the three equations as planes. One unique solution would be where the planes all intersect. If the system has an infinite number of solutions, then it means all three planes meet on a line of intersection. A system with no solutions will have no point of intersection between all three planes.

Some systems of equations may result in an infinite number of solutions. This comes as a result of a **free variable** being a part of the equations.

A **free variable **is a variable that can change in value freely.

The following example shows how to determine the infinitely many solutions a system of equations may have.

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.

Remember that we designated \(\frac{z}{3} = t\), so \(z = 3t\).

An example of what a system of equations with infinitely many solutions could look like is shown below:

The following example shows how to determine if a system of equations has no solutions.

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.

When modelled using planes, a system with no solutions could look like this:

Inverse matrices can be used to solve simultaneous equations by multiplying both sides of the equation with the inverse of the coefficient matrix and simplifying to calculate the values of the variables.

There are two important things to remember when solving simultaneous equations using inverse matrices:

**\(A^{-1} \times A = I\)**where \(I\) is the identity matrix. Multiplying by \(I\) will have the same effect as multiplying the matrix by 1;The inverse matrix, \(A^{-1}\), must be on the

**left side**of the other matrices; otherwise, you won't be able to multiply them.

The following example shows how to solve simultaneous equations using inverse matrices.

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.

Remember that \(A^{-1} \times A = I\) and \(I\) multiplied by any compatible matrix will have the same effect as multiplying any number by 1.

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 \(A^{-1}\).

The order in which you multiply matrices is important. In this case, we want the result to be the identity matrix, and so we must multiply both sides of the equation by \(A^{-1}\) on the left-hand side to get that.

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 \(y = -3\)

The following example shows how to use inverse matrices to solve simultaneous equations that contain three variables.

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 \(\quad z = -1\).

Although it is possible to solve for three unknowns using inverse matrices (as shown above), this method is not recommended as it quickly gets tedious and overly complicated. One should rather make use of row reduction, which will be covered in detail in the next section.

**Row reduction** is another method of solving simultaneous equations using matrices. Also known as **Gaussian elimination**, row reduction makes use of row calculations within an augmented matrix to solve for the unknowns in a set of simultaneous equations.

You want to get a row with two zeroes so that you can calculate the value of one of the variables, and then you want a row with one zero so that you can calculate the value of the second variable.

There are a few important things to remember when performing these row calculations:

The second zero you calculate must always be in the same column as the first zero, and must be either directly above or below the first zero;

You must always use the rows already containing zeros to calculate the third zero;

The zeroes you calculate must be in the first three columns of the augmented matrix. A zero in the fourth column does not help solve any of the unknowns;

Row calculations are applied to the whole row, including the fourth column.

The example below shows how to do this.

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 \(z = 2\)

There are many different ways to solve simultaneous equations using row reduction, and the row calculations you choose to do may be different to the ones used here.

The following examples focus on solving simultaneous equations of order 2, meaning you'll have to make use of 2 x 2 matrices.

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:

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

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

It is also possible to solve a set of simultaneous with only 2 unknowns using row reduction. This next example shows how this is done.

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 \(y\).

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

In some cases, it is impossible to avoid doing calculations that result in large numbers. The following example illustrates this.

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 \(z = 7\)

Some equations might only contain two variables. In these cases, you simply use zero as the coefficient for the missing variable.

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 \(c = 2\)

- To write simultaneous equations in matrix form, you must first write the square matrix containing the coefficients, followed by the variable matrix and, to the right of the equal sign, you must write the constant matrix.
- Augmented matrices contain only the coefficients and constants.
- Inverse matrices can be used to solve simultaneous equations by multiplying the inverse coefficient matrix on the left of both sides of the matrix equation and simplifying.
- Row reduction is the preferred method of solving simultaneous equations with three unknowns, and involves the use of row operations to calculate the values of the variables.
- Equations can have infinitely many solutions, no solutions and one unique solution, and these can all be modelled using planes.

Follow these steps:

1) Make sure that the variables are in the same order for every equation;

2) Rewrite the equations in matrix form by

a) first writing the coefficient matrix - a square matrix that houses the coefficients of each variable;

b) second writing the variable matrix close to the coefficient matrix - a matrix containing all the variables in order;

c) third writing the constant matrix to the right of the equal sign.

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