This is an example of a system of linear equations. This problem
can be solved using basic algebra steps by repeatedly isolating
and substituting variables among the three equations.
However, that process can be quite tedious, and it can sometimes be
challenging to figure out exactly what steps will advance you
toward a solution.
In this chapter, we’ll introduce a technique for solving these problems in a
systematic way that’s usually quite faster.
This introduction is for students who have had one year of algebra education.
Sections 13 and beyond are for students who are also familiar with
matrix multiplication.
Index
Solving by algebra steps
Let’s solve this system of two linear equations:
{2x}
{+}
{4y}
{=}
{14}
{5x}
{+}
{3y}
{=}
{21}
The goal is to find values for {x} and {y} that will make both equations true.
Elementary algebra provides us with the techniques for solving this
problem. The algebra steps are easy to apply, although the entire
process is a bit lengthy:
First, pick an equation and isolate one of the variables on one side of the equation:
{2x}
{+}
{4y}
{=}
{14}
{[ 1]}
given
{ x}
{+}
{2y}
{=}
{7}
{[ 2]}
divide by {2} to help isolate {x}
{ x}
{=}
{7}
{-}
{2y}
{[ 3]}
subtract {2y} to isolate {x}
Then, substitute the isolated variable into the other equation, which will solve it:
{ 5x }
{+}
{ 3y}
{=}
{ 21}
{}
{}
{[ 4]}
given
{ 5(7 - 2y) }
{+}
{ 3y}
{=}
{ 21}
{}
{}
{[ 5]}
substitute {x} from {[ 3]}
{ 35 - 10y }
{+}
{ 3y}
{=}
{ 21}
{}
{}
{[ 6]}
distribute
{ 35 }
{-}
{ 7y}
{=}
{ 21}
{}
{}
{[ 7]}
combine the factors of {y}
{ }
{-}
{ 7y}
{=}
{ 21}
{-}
{35}
{[ 8]}
subtract {25}
{ }
{-}
{ 7y}
{=}
{-14}
{}
{}
{[ 9]}
simplify
{ }
{ }
{ y}
{=}
{ 2}
{}
{}
{[10]}
divide by {-7}
Finally, take the known variable ( {y} in this case) and use it to solve for the other variable
using either equation:
{ 2x }
{+}
{ 4y }
{=}
{14}
{}
{}
{[11]}
given, from {[ 1]}
{ 2x }
{+}
{ 4 \mathord{\times} 2}
{=}
{14}
{}
{}
{[12]}
substitute {y} from {[ 10]}
{ 2x }
{+}
{ 8 }
{=}
{14}
{}
{}
{[13]}
simplify
{ 2x }
{ }
{ }
{=}
{14}
{-}
{8}
{[14]}
subtract {8}
{ 2x }
{ }
{ }
{=}
{ 6}
{}
{}
{[15]}
simplify
{ \phantom{2}x }
{ }
{ }
{=}
{ 3}
{}
{}
{[16]}
divide by {2}
And we obtain the solution:
{x}
{=}
{3}
from {[16]}
{y}
{=}
{2}
from {[10]}
Row reduction: the augmented matrix
In this section, we’ll introduce another way of solving
a system of linear equations using a technique called “row reduction”.
The row reduction technique has these advantages:
It minimizes the amount of writing you have to do.
It defines a clear strategy for advancing toward a solution.
It scales up well for solving systems that have more variables.
Over the next four sections, we’ll be applying the row reduction technique to
solve this system of linear equations:
{2x}
{+}
{4y}
{=}
{14}
{5x}
{+}
{3y}
{=}
{21}
The first step in this technique is to define an augmented matrix
that represents the two equations. An augmented matrix is an organized,
rectangular grid of all the numbers used in the equations.
Here’s the augmented matrix for the above equations:
The augmented matrix is designed to minimize the
amount of writing you need to do — you don’t have to
write down any variable names or operators.
For each variable, its coefficients are all lined up in a single column,
and every operator is assumed to be addition.
The row reduction technique is also frequently called the
Gaussian elimination technique.
As an exercise, convert the following problem to an augmented matrix
that represents a system of linear equations:
A farmer has a total of {11} animals, a mix of chickens and cows.
One day he counts the legs of all of his animals and realizes there’s
a total of {34} legs. How many chickens and how many cows does the farmer have?
Use {x} to represent the number of chickens, and {y} to represent the
number of cows.
Row reduction: the reduced matrix
Here’s the system of linear equations that we’ll be solving using the
row reduction technique:
{2x}
{+}
{4y}
{=}
{14}
{5x}
{+}
{3y}
{=}
{21}
In the previous section, we introduced the first step of the technique,
which is to define an augmented matrix that represents those equations:
The goal of the row reduction technique is to convert this
matrix to another matrix, called the reduced matrix, which
directly contains the solution to the equations.
Here’s the conversion that we’ll be performing on the matrix:
Notice that the reduced matrix represents the following system of equations:
{1x}
{+}
{0y}
{=}
{3}
{0x}
{+}
{1y}
{=}
{2}
which can be simplified to:
{x}
{=}
{3}
{y}
{=}
{2}
which, together, provide the solution to the original equations.
The reduced matrix uses the coefficients {1}
and {0} to isolate exactly one variable per equation.
To accomplish this, the entries on the left side
of the vertical bar must satisfy these conditions:
there must be exactly one {1} in each column,
there must be exactly one {1} in each row,
there must be all {0}s everywhere else.
Sometimes it isn’t possible to satisfy all of these conditions,
but if we can, then the rightmost column
of the reduced matrix will contain the values of the variables
that, together, solve the equations.
The reduced matrix is also frequently called the
reduced row echelon form matrix.
We can now focus on the second column.
We need to create a {1} in the second column, but it cannot be
in the same row as the {1} in the first column. So our only choice
is to change the {-7} to a {1.}
We can change the {-7} to a {1} by dividing the second row by {-7}:
which, together, provide the solution to the original equations.
Row reduction: exercise
Solve this problem using the row reduction technique:
A farmer has a total of {11} animals, a mix of chickens and cows.
One day he counts the legs of all of his animals and realizes there’s
a total of {34} legs. How many chickens and how many cows does the farmer have?
We can describe this problem using a system of linear equations like this:
{ x}
{+}
{ y}
{=}
{11}
{2x}
{+}
{4y}
{=}
{34}
where:
{
\; \left\lbrace \vphantom{ \eqalign { x \cr x \cr } } \right. \space
}
{x}
{\;=} number of chickens
{y}
{\;=} number of cows
Recall the techniques used for row reduction:
Row operations:
Interchange any two rows.
Multiply all the entries in a row by a nonzero constant.
Add or subtract a multiple of one row to another row.
Row reduction strategy:
Work column by column, doing the columns in any order that’s convenient.
For each column, first change one entry to {1,} then change all the rest to {0}s.
Never put two {1}s in the same row on the left side of the vertical bar.
For extra speed, start with columns that already have a {1} or a {0.}
For extra speed, don’t bother interchanging any rows.
Row reduction: exercise solution
In this section, we’ll solve the row reduction exercise from the
previous section, but in a more generalized form, by introducing the
variables {A} and {L} to represent the numbers. Here’s the new version
of the problem:
A farmer has a total of {A} animals, a mix of chickens and cows.
One day he counts the legs of all of his animals and realizes there’s
a total of {L} legs. How many chickens and how many cows does the farmer have?
Here’s the system of linear equations that describes this problem:
{ x}
{+}
{ y}
{=}
{A}
{2x}
{+}
{4y}
{=}
{L}
where:
{
\; \left\lbrace \vphantom{ \eqalign { x \cr x \cr x \cr x \cr } } \right. \space
}
{x}
{\;=} number of chickens
{y}
{\;=} number of cows
{A}
{\;=} total number of animals
{L}
{\;=} total number of legs
We’ll also take this opportunity to write the augmented matrix in an
abbreviated way that minimizes the amount of writing you’ll need to do.
You can solve it a little faster if you write it this way:
{x \quad}
{y \quad}
{1 \quad}
{1 \quad}
{ A \quad }
{[ 1]}
write the first equation’s row
{2 \quad}
{4 \quad}
{ L \quad }
{[ 2]}
write the second equation’s row
{2 \quad}
{2 \quad}
{ 2A \quad }
{[ 3]}
multiply {[ 1]} by {2} so subtraction can zero out the {2}
{0 \quad}
{2 \quad}
{ L - 2A \quad }
{[ 4]}
subtract {[ 3]} from {[ 2]} to zero out the {2} for {x}
{0 \quad}
{1 \quad}
{ {\Large{L \over 2}} - A \quad }
{[ 5]}
divide {[ 4]} by {2} to get a {1} for {y}
Every time you rewrite a row, be sure to cross out the old row that
it replaces. Also, cross out any temporary rows {(}like line {[ 3])}
once you’re done with them. This ensures that you’ll always know
which two rows are the active ones.
After finishing step {[ 5],} this is our current matrix
(which you don’t need to write down):
We haven’t completely finished isolating both variables, but notice that
we’ve found a solution for {y} already:
{y = {\Large{L \over 2}} - A }
At this point, we can simply back substitute {y} into to
the first equation to obtain a solution for {x}:
{x + y = A} 🡇
{x = A - y}
We’ve now got a solution for both {x} and {y,} so we’re done.
Notice that we were able to solve the problem
even though we never did isolate {x} in the matrix, and we ever
did figure out exactly how to calculate {x} directly from {A} and {L.}
It turns out that { x = 2A - {\large{L \over 2}},} but that formula
isn’t needed because we found a simpler way to solve for {x.}
Row reduction: triangular matrix and back substitution
In the previous section, we stopped the row reduction process early
and used back substitution to find the value of {x.}
That shortcut saved us some time.
If you take this shortcut to the extreme, you can significantly reduce
the number of row operations you need to perform.
For example, here is the result of performing several row reduction
steps on a system of four linear equations:
{ \hspace8pt w \hspace18pt x \hspace19pt y \hspace20pt z }
Looking at the resulting matrix, we clearly didn’t
achieve the goal of isolating just one variable per equation.
So the values of most of the variables are not yet obvious.
Nevertheless, we can immediately see that {z = 4} by looking
at the bottom row.
And now that we know {z = 4}, it’s relatively easy to solve for {y} by
back substituting {z} with {4} in the third row, and solving
{5y - 2z = -3} to find that {y = 1.}
The reason we found {z} and {y} so quickly is because there are numerous
{0}s in the lower-left portion of the matrix.
Notice that those {0}s form a triangular pattern
that fills up one whole side of the diagonal. A matrix that contains
this triangular pattern of {0}s is called a triangular matrix.
That triangular pattern of {0}s allows us to perform a
chain of back substitutions to solve for all the variables in the system.
For example, if we look at the general triangular matrix for four variables:
{ \hspace9pt w \hspace15pt x \hspace14pt y \hspace15pt z }
However, if you don’t want to do any back substitutions, then you can
continue to perform more row operations on the matrix until you
finally obtain a reduced matrix, like this:
{ \hspace9pt w \hspace15pt x \hspace14pt y \hspace15pt z }
To solve this system, we’ll need to change the {2} to a {0.}
We can do this by multiplying the first row by {2} and subtracting
it from the second row like this:
The second row now contains all {0}s, so it represents this
linear equation:
{0x + 0y = 0}
which, clearly, can be satisfied by every pair of numbers {x} and {y.}
Since the second row provides us with no additional information
to identify a single solution, we must conclude that this system
has an infinite number of solutions, all of them lying along the
line {x + 5y = 3.}
2. A row with all zeros, except for a non-zero in the rightmost column.
For example, let’s perform row reduction on this augmented matrix:
To solve this system, we’ll need to change the {2} to a {0.}
We can do this by multiplying the first row by {2} and subtracting
it from the second row like this:
which, clearly, cannot be satisfied by any pair of numbers {x} and {y.}
If row reduction produces an always-false equation like this,
it shows that the system of equations has no solution.
Cramer’s rule
Let’s look at the general system of linear equations in two variables:
{Ax}
{+}
{By}
{=}
{C}
{Dx}
{+}
{Ey}
{=}
{F}
It’s interesting to solve this system for {x} and {y} using the row reduction
technique. When we do, we find that:
[[ x = {{CE - BF} \over {AE - BD}} ]] and [[ y = {{AF - CD} \over {AE - BD}} ]]
This result is called Cramer’s rule.
Notice that all the numerators and denominators have a similar
“cross multiplication” pattern over two different columns:
the denominator
{AE-BD}
has this pattern:
the {x} numerator
{CE-BF}
has this pattern:
the {y} numerator
{AF-CD}
has this pattern:
Determinant
In the previous section on Cramer’s rule, we found that the solution to
this pair of equations:
{Ax}
{+}
{By}
{=}
{C}
{Dx}
{+}
{Ey}
{=}
{F}
is:
[[ x = {{CE - BF} \over {AE - BD}} ]]
and
[[ y = {{AF - CD} \over {AE - BD}} ]]
It’s interesting that these formulas have a strong
recurring pattern that involves calculating the
difference of two “cross multiplications”.
This pattern occurs so frequently in linear algebra
that we have a special name for it:
the determinant.
The determinant of the matrix
{
\hspace+2pt
\left[\hspace+2pt
\begin{array}{rr|r}
\hfil a &\hfil\hspace-5pt b \cr
\hfil c &\hfil\hspace-5pt d \cr
\end{array}
\hspace+2pt\right]
\space
}
is
{
\space ad - bc.
}
The determinant is an operator, and it’s commonly written either like this:
{
\det
\left[\hspace+2pt
\begin{array}{rr|r}
\hfil a &\hfil\hspace-5pt b \cr
\hfil c &\hfil\hspace-5pt d \cr
\end{array}
\hspace+2pt\right]
=
ad - bc
}
or like this:
{
\left|
\hspace+2pt
\begin{array}{rr|r}
\hfil a &\hfil\hspace-5pt b \cr
\hfil c &\hfil\hspace-5pt d \cr
\end{array}
\hspace+2pt
\right|
=
ad - bc
}
The determinant allows us to define Cramer’s rule in a
way that better preserves the visual structure of the coefficients
in the equations:
{
\left.
\eqalign
{
\hfil Ax &\hfil + &\hfil By &\hfil = &\hfil C \cr
\hfil Dx &\hfil + &\hfil Ey &\hfil = &\hfil F \cr
}
\space
\right\rbrace
}
is solved by:
[[ x =
{
{
\vphantom{ \large{ \left| \begin{array} 0 & 0 \cr 0 & 0 \end{array} \right| } }
{
-{
\left|
\hspace+2pt
\begin{array}{rr|r}
\hfil B &\hfil\hspace-5pt C \cr
\hfil E &\hfil\hspace-5pt F \cr
\end{array}
\hspace+2pt
\right|
}
}
}
\over
{
\vphantom{ \large{ \left| \begin{array} 0 & 0 \cr 0 & 0 \end{array} \right| } }
{
\phantom{-}{
\left|
\hspace+2pt
\begin{array}{rr|r}
\hfil A &\hfil\hspace-5pt B \cr
\hfil D &\hfil\hspace-5pt E \cr
\end{array}
\hspace+2pt
\right|
}
}
}
}
]]
and
[[ y =
{
{
\vphantom{ \large{ \left| \begin{array} 0 & 0 \cr 0 & 0 \end{array} \right| } }
\left|
\hspace+2pt
\begin{array}{rr|r}
\hfil A &\hfil\hspace-5pt C \cr
\hfil D &\hfil\hspace-5pt F \cr
\end{array}
\hspace+2pt
\right|
}
\over
{
\vphantom{ \large{ \left| \begin{array} 0 & 0 \cr 0 & 0 \end{array} \right| } }
\left|
\hspace+2pt
\begin{array}{rr|r}
\hfil A &\hfil\hspace-5pt B \cr
\hfil D &\hfil\hspace-5pt E \cr
\end{array}
\hspace+2pt
\right|
}
}
]]
As an exercise, solve this system of linear equations
using the determinant version of Cramer’s rule:
{2x}
{+}
{4y}
{=}
{14}
{5x}
{+}
{3y}
{=}
{21}
Using matrix multiplication: introduction
It’s possible to use matrix multiplication to solve a system of
linear equations. The next six sections will show how to do this.
First, in this section, we’ll see how a system of linear equations can be
written using matrix multiplication notation.
Let’s start with a single linear equation:
{x - 5y + 2z \ = \ 1}
This equation can be written equivalently in vector dot product notation like this:
{(1, -5, 2) \cdot (x, y, z) \ = \ 1}
Matrix multiplication is also capable of performing this dot product,
but it uses a different notation:
{
\mathbf{[}\hspace+2pt
\begin{matrix}
\hfil 1 &\hfil\hspace-5pt -5 &\hfil\hspace-5pt 2 \cr
\end{matrix}
\hspace+2pt\mathbf{]}
\hspace-3pt
\left[\hspace+2pt
\begin{matrix}
x \cr
y \cr
z \cr
\end{matrix}
\hspace+2pt\right]
= \ \mathbf{[}\hspace+2pt 1 \hspace+2pt\mathbf{]}
}
The advantage of the matrix notation is that it can be extended
to multiple equations in a concise way. For example, these three linear equations:
As you know, if a number {x} is multiplied by {1,} the result is unchanged:
{1 \times x = x}
This can also occur in matrix multiplication. An identity matrix is
a matrix that, when multiplied by another matrix, produces a result that
leaves the other matrix unchanged.
There’s a unique identity matrix for each dimension {D}:
{
D = 1\mathord{:}
\hspace+13pt
\mathbf{[}\hspace+2pt
\begin{matrix}
\hfil 1 \cr
\end{matrix}
\hspace+2pt\mathbf{]}
\mathbf{[}\hspace+2pt
\begin{matrix}
x \cr
\end{matrix}
\hspace+2pt\mathbf{]}
=
\mathbf{[}\hspace+2pt
\begin{matrix}
x \cr
\end{matrix}
\hspace+2pt\mathbf{]}
}
{
D = 2\mathord{:}
\hspace+10pt
\left[\hspace+2pt
\begin{matrix}
\hfil 1 &\hfil\hspace-5pt 0 \cr
\hfil 0 &\hfil\hspace-5pt 1 \cr
\end{matrix}
\hspace+2pt\right]
\hspace-3pt
\left[\hspace+2pt
\begin{matrix}
x \cr
y \cr
\end{matrix}
\hspace+2pt\right]
=
\left[\hspace+2pt
\begin{matrix}
x \cr
y \cr
\end{matrix}
\hspace+2pt\right]
}
{
D = 3\mathord{:}
\hspace+10pt
\left[\hspace+2pt
\begin{matrix}
\hfil 1 &\hfil\hspace-5pt 0 &\hfil\hspace-5pt 0 \cr
\hfil 0 &\hfil\hspace-5pt 1 &\hfil\hspace-5pt 0 \cr
\hfil 0 &\hfil\hspace-5pt 0 &\hfil\hspace-5pt 1 \cr
\end{matrix}
\hspace+2pt\right]
\hspace-3pt
\left[\hspace+2pt
\begin{matrix}
x \cr
y \cr
z \cr
\end{matrix}
\hspace+2pt\right]
=
\left[\hspace+2pt
\begin{matrix}
x \cr
y \cr
z \cr
\end{matrix}
\hspace+2pt\right]
}
{
D = 4\mathord{:}
\hspace+10pt
\left[\hspace+2pt
\begin{matrix}
\hfil 1 &\hfil\hspace-5pt 0 &\hfil\hspace-5pt 0 &\hfil\hspace-5pt 0 \cr
\hfil 0 &\hfil\hspace-5pt 1 &\hfil\hspace-5pt 0 &\hfil\hspace-5pt 0 \cr
\hfil 0 &\hfil\hspace-5pt 0 &\hfil\hspace-5pt 1 &\hfil\hspace-5pt 0 \cr
\hfil 0 &\hfil\hspace-5pt 0 &\hfil\hspace-5pt 0 &\hfil\hspace-5pt 1 \cr
\end{matrix}
\hspace+2pt\right]
\hspace-3pt
\left[\hspace+2pt
\begin{matrix}
w \cr
x \cr
y \cr
z \cr
\end{matrix}
\hspace+2pt\right]
=
\left[\hspace+2pt
\begin{matrix}
w \cr
x \cr
y \cr
z \cr
\end{matrix}
\hspace+2pt\right]
}
etc.
An identity matrix is a square {D \mathord{\times} D} matrix that contains {1}s
along the main diagonal that extends from the upper left corner to the lower
right corner, and contains {0}s everywhere else.
Each row vector of an identity matrix
is a unit vector that points in the positive direction of one of the
axes in {D}-dimensional space.
For example, on the {2}-dimensional plane, {(1, 0)} is the unit vector
that points in the direction of the positive {x} axis, and {(0, 1)} is the unit vector
that points in the direction of the positive {y} axis. Therefore, the two
row vectors of the {2 \mathord{\times} 2} identity matrix are {\mathbf{[}\hspace+2pt 1 \ \ 0 \hspace+2pt\mathbf{]}}
and {\mathbf{[}\hspace+2pt 0 \ \ 1 \hspace+2pt\mathbf{]}.}
Using matrix multiplication: inverse matrix
If the product of two square matrices is the identity matrix, then the two matrices are inverses of each other.
Here’s an example of two matrices that are inverses of each other:
Here’s the inverse of a general {2 \mathord{\times} 2} matrix:
Given a {2 \mathord{\times} 2} matrix {M}:
{
M =
\left[\hspace+2pt
\begin{matrix}
\hfil a &\hfil\hspace-2pt b \cr
\hfil c &\hfil\hspace-2pt d \cr
\end{matrix}
\hspace+2pt\right]
}
The inverse of {M} is called {M^{\large -1},} and is defined as:
[[
M^{\large -1} \ =
\ { {1} \over { \det M } }
\left[\hspace+2pt
\begin{matrix}
\hfil d &\hfil\hspace-2pt -b \cr
\hfil -c &\hfil\hspace-2pt a \cr
\end{matrix}
\hspace+2pt\right]
]]
where:
{
\det M \ =
\ \det
\left[\hspace+2pt
\begin{array}{rr|r}
\hfil a &\hfil\hspace-2pt b \cr
\hfil c &\hfil\hspace-2pt d \cr
\end{array}
\hspace+2pt\right]
\ =
\ ad - bc
}
If {I} is the {2 \mathord{\times} 2} identity matrix:
If the determinant of a matrix is {0,} then it does not have an inverse.
In matrix mathematics, there is no division operator that divides one
matrix by another. To emulate division, we can multiply by the inverse of a matrix.
For example, we can emulate {\text{“}}matrix {A} divided by matrix {B \hspace+1pt \text{”}}
by calculating {A B^{\large -1}.}
Exercise: Recall the method for calculating the inverse of a {2 \mathord{\times} 2} matrix:
If
{
M =
\left[\hspace+2pt
\begin{matrix}
\hfil a &\hfil\hspace-2pt b \cr
\hfil c &\hfil\hspace-2pt d \cr
\end{matrix}
\hspace+2pt\right]
}
then
[[
M^{\large -1} \ =
\ { {1} \over { \det M } }
\left[\hspace+2pt
\begin{matrix}
\hfil d &\hfil\hspace-2pt -b \cr
\hfil -c &\hfil\hspace-2pt a \cr
\end{matrix}
\hspace+2pt\right]
]]
where: { \ \det M = ad - bc }
In this section, we’ll describe the technique for performing
that multiplication.
There are three key things to remember when performing matrix multiplication:
Move the second matrix physically upward,
high enough to create space for the body of the table.
Slice each matrix into its individual vectors:
Slice the first matrix into row vectors.
Slice the second matrix into column vectors.
Create internal grid lines in the body of the table — they
will hold the result.
Each entry in the grid is the dot product of its
row vector and column vector.
Using matrix multiplication: solving a system of three linear equations
Let’s apply the technique from the previous section to
a system of three linear equations, fully generalized with variables:
{ax}
{+}
{by}
{+}
{cz}
{\, = \,}
{p}
{dx}
{+}
{ey}
{+}
{fz}
{\, = \,}
{q}
{gx}
{+}
{hy}
{+}
{iz}
{\, = \,}
{r}
Here, {x,} {y,} and {z} are the three unknown variables, and all the remaining
variables are known constant values.
Let’s solve these equations for {x,} {y,} and {z} using matrix multiplication:
Express the equations in matrix multiplication notation:
{
\left[\hspace+2pt
\begin{matrix}
\hfil a &\hfil\hspace-5pt b &\hfil\hspace-5pt c \cr
\hfil d &\hfil\hspace-5pt e &\hfil\hspace-5pt f \cr
\hfil g &\hfil\hspace-5pt h &\hfil\hspace-5pt i \cr
\end{matrix}
\hspace+2pt\right]
\hspace-3pt
\left[\hspace+2pt
\begin{matrix}
x \cr
y \cr
z \cr
\end{matrix}
\hspace+2pt\right]
=
\left[\hspace+2pt
\begin{matrix}
\hfil p \cr
\hfil q \cr
\hfil r \cr
\end{matrix}
\hspace+2pt\right]
}
For simplicity, let’s use {M} to represent the entire {3 \mathord{\times} 3} coefficient matrix:
{
M
\hspace-2pt
\left[\hspace+2pt
\begin{matrix}
x \cr
y \cr
z \cr
\end{matrix}
\hspace+2pt\right]
=
\left[\hspace+2pt
\begin{matrix}
\hfil p \cr
\hfil q \cr
\hfil r \cr
\end{matrix}
\hspace+2pt\right]
}
{
\hspace+20pt
\color{#008800}
\text{(where }
M =
\left[\hspace+2pt
\begin{matrix}
\hfil a &\hfil\hspace-5pt b &\hfil\hspace-5pt c \cr
\hfil d &\hfil\hspace-5pt e &\hfil\hspace-5pt f \cr
\hfil g &\hfil\hspace-5pt h &\hfil\hspace-5pt i \cr
\end{matrix}
\hspace+2pt\right]
\text{)}
}
Multiply both sides by {M^{\large {-1}}}:
{
{M^{\large {-1}}}
M
\hspace-2pt
\left[\hspace+2pt
\begin{matrix}
x \cr
y \cr
z \cr
\end{matrix}
\hspace+2pt\right]
=
{M^{\large {-1}}}
\hspace-2pt
\left[\hspace+2pt
\begin{matrix}
\hfil p \cr
\hfil q \cr
\hfil r \cr
\end{matrix}
\hspace+2pt\right]
}
Simplify the left side:
{
\left[\hspace+2pt
\begin{matrix}
x \cr
y \cr
z \cr
\end{matrix}
\hspace+2pt\right]
=
{M^{\large {-1}}}
\hspace-2pt
\left[\hspace+2pt
\begin{matrix}
\hfil p \cr
\hfil q \cr
\hfil r \cr
\end{matrix}
\hspace+2pt\right]
}
Calculate the matrix multiplication on the right side. This will produce
a column vector that contains the solutions for {x,} {y,} and {z.}
We haven’t discussed the technique for finding the inverse of a {3 \mathord{\times} 3}
matrix. The {3 \mathord{\times} 3} inverse matrix is tedious to compute by hand,
so typically, you would use a calculator to compute it.
Using matrix multiplication: exercises
If you have a calculator that supports matrix inverse, then you can do
either of these exercises. If not, then you can do the first exercise.
Recall the method for calculating the inverse of a {2 \mathord{\times} 2} matrix:
If
{
M =
\left[\hspace+2pt
\begin{matrix}
\hfil a &\hfil\hspace-2pt b \cr
\hfil c &\hfil\hspace-2pt d \cr
\end{matrix}
\hspace+2pt\right]
}
then
[[
M^{\large -1} \ =
\ { {1} \over { \det M } }
\left[\hspace+2pt
\begin{matrix}
\hfil d &\hfil\hspace-2pt -b \cr
\hfil -c &\hfil\hspace-2pt a \cr
\end{matrix}
\hspace+2pt\right]
]]
Use matrix multiplication to solve the following system of two linear equations:
