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We concluded Section MatArithmetic by showing how we can rewrite a system of linear equations as the matrix equation \(AX=B\) where \(A\) and \(B\) are
known matrices and the solution matrix \(X\) of the equation corresponds to the solution of the system. In this section, we develop
the method for solving such an equation. To that end, consider the system
To write this as a matrix equation, we follow the
procedure outlined on page ??. We find the coefficient matrix \(A\), the unknowns matrix \(X\) and constant matrix \(B\) to be
\[ \begin{array}{ccc}
A = \left [ \begin{array}{rr} 2 & -3 \\ 3 & 4 \\ \end{array} \right ]
&
X = \left [ \begin{array}{r} x \\ y \\ \end{array} \right ]
&
B = \left [ \begin{array}{r} 16 \\ 7 \\ \end{array} \right ] \end{array}\]
In order to motivate how we solve a matrix equation like \(AX = B\), we revisit solving a similar equation involving real numbers. For
instance, to solve \(3x = 5\), we divide both sides by \(3\) and obtain \(x = \frac {5}{3}\). How can we go about defining an analogous process
for matrices? To answer this question, we solve \(3x=5\) again, but this time, we pay attention to the properties of
real numbers being used at each step. Recall that dividing by \(3\) is the same as multiplying by \(\frac {1}{3} = 3^{-1}\), the so-called
multiplicative inverse(Every nonzero real number \(a\) has a multiplicative inverse, denoted \(a^{-1}\), such that \(a^{-1} \cdot a = a \cdot a^{-1} = 1\).) of
\(3\).
\[ \begin{array}{rclr}
3x & = & 5 \\ 3^{-1}(3x) & = & 3^{-1}(5) & \text {Multiply by the (multiplicative) inverse of $3$} \\ \left (3^{-1}\cdot 3\right ) x & = & 3^{-1}(5) & \text {Associative property of multiplication} \\ 1 \cdot x & = & 3^{-1}(5) & \text {Inverse property} \\ x & = & 3^{-1}(5) & \text {Multiplicative Identity} \\
\end{array} \]
If we wish to check our answer, we substitute \(x = 3^{-1}(5)\) into the original equation
Thinking back to Theorem matrixmultprops, we know that matrix multiplication enjoys both an associative property and a multiplicative identity.
What’s missing from the mix is a multiplicative inverse for the coefficient matrix \(A\). Assuming we can find such a beast, we can
mimic our solution (and check) to \(3x=5\) as follows
\[ \begin{array}{cc}
\text {Solving $AX = B$} & \text {Checking our answer} \\
\begin{array}{rcl}
AX & = & B \\
A^{-1}(AX) & = & A^{-1}B \\
\left (A^{-1}A\right ) X & = & A^{-1}B \\
I_2X & = & A^{-1}B \\
X & = & A^{-1}B \\
\end{array}
&
\begin{array}{rcl}
AX & \stackrel {?}{=} & B \\
A \left (A^{-1}B\right ) & \stackrel {?}{=} & B \\
\left (AA^{-1}\right ) B & \stackrel {?}{=}& B \\
I_2B & \stackrel {?}{=} & B \\
B & \stackrel {\checkmark }{=}& B \\
\end{array} \\
\end{array}\]
The matrix \(A^{-1}\) is read ‘\(A\)-inverse’ and we will define it formally later in the section. At this stage, we have no idea if such a matrix \(A^{-1}\)
exists, but that won’t deter us from trying to find it. (Much like Carl’s quest to find Sasquatch.)
We want \(A^{-1}\) to satisfy two equations, \(A^{-1}A = I_2\) and \(AA^{-1} = I_2\), making \(A^{-1}\) necessarily a \(2 \times 2\) matrix. (Since matrix multiplication isn’t necessarily
commutative, at this stage, these are two different equations.) Hence, \(A^{-1}\) has the form
for real numbers \(x_1\), \(x_2\), \(x_3\) and \(x_4\). For reasons which will become clear later, we focus our attention on the equation \(AA^{-1} = I_2\). We
have
At this point, it may seem absurd to continue with this venture. After all, the intent was to solve one system of equations, and
in doing so, we have produced two more to solve. Remember, the objective of this discussion is to develop a
general method which, when used in the correct scenarios, allows us to do far more than just solve a system of
equations.
If we set about solving these systems using augmented matrices using the techniques in Section AugMatrices, we see that not only do
both systems have the same coefficient matrix, this coefficient matrix is none other than the matrix \(A\) itself! (We will come back
to this observation in a moment.)
To solve these two systems, we use Gauss-Jordan Elimination to put the augmented matrices into reduced row echelon form.
(We leave the details to the reader.) For the first system, we get
which gives \(x_1 = \frac {4}{17}\) and \(x_3 = -\frac {3}{17}\). To solve the second system, we use the exact same row operations, in the same order, and we
obtain
so that our final solution to the system is \((x,y) = (5,-2)\).
As mentioned, the point of this exercise was not just to solve the system of linear equations, but to develop a general method
for finding \(A^{-1}\). To that end, we analyze the foregoing discussion in a more general context. To find \(A^{-1}\), we used two augmented
matrices, both containing the entries as \(A\):
where we have identified the entries to the left of the vertical bar as the identity \(I_2\) and the entries to the right of the
vertical bar as the solutions to our systems. The long and short of the solution process can be summarized
as
Since the row operations for both processes are the same, all of the arithmetic on the left hand side of the vertical bar is
identical in both problems. The only difference between the two processes is what happens to the constants to the right of the
vertical bar. As long as we keep these separated into columns, we can combine our efforts into one ‘super-sized’ augmented
matrix and describe the above process as
We have the identity matrix \(I_2\) appearing as the right hand side of the first super-sized augmented matrix and the left hand
side of the second super-sized augmented matrix. To our surprise and delight, the elements on the right hand
side of the second super-sized augmented matrix are none other than those which comprise \(A^{-1}\). Hence, we
have
In other words, the process of finding \(A^{-1}\) for a matrix \(A\) can be viewed as performing a series of row operations which transform \(A\)
into the identity matrix of the same dimension. We can view this process as follows. In trying to find \(A^{-1}\), we are trying to ‘undo’
multiplication by the matrix \(A\). The identity matrix in the super-sized augmented matrix \([A | I]\) keeps a running memory of all of the
moves required to ‘undo’ \(A\). This results in exactly what we want, \(A^{-1}\). We are now ready to formalize and generalize the foregoing
discussion. We begin with the formal definition of an invertible matrix. (If this rhetoric sounds familiar, it should. See
section InverseFunctions.)
An \(n \times n\) matrix \(A\) is said to be invertible if there exists a matrix \(A^{-1}\), read ‘\(A\) inverse’, such that \(A^{-1}A = AA^{-1}=I_{n}\).
Note that, as a consequence of our definition, invertible matrices are square, and as such, the conditions in
Definition matrixinverse force the matrix \(A^{-1}\) to be same dimensions as \(A\), that is, \(n \times n\). Since not all matrices are square, not all
matrices are invertible. However, just because a matrix is square doesn’t guarantee it is invertible. (See the
exercises.)
Our first result summarizes some of the important characteristics of invertible matrices and their inverses.
Suppose \(A\) is an \(n \times n\) matrix.
If \(A\) is invertible then \(A^{-1}\) is unique.
\(A\) is invertible if and only if \(AX = B\) has a unique solution for every \(n \times r\) matrix \(B\).
The proofs of the properties in Theorem inversematrixprops rely on a healthy mix of definition and matrix arithmetic.
To establish the first property, we assume that \(A\) is invertible and suppose the matrices \(B\) and \(C\) act as inverses for \(A\). That is, \(BA = AB = I_{n}\) and \(CA = AC = I_{n}\).
We need to show that \(B\) and \(C\) are, in fact, the same matrix. To see this, we note that \(B = I_{n}B = (CA)B = C(AB) = CI_{n} = C\). Hence, any two matrices that act like \(A^{-1}\) are,
in fact, the same matrix. (If this proof sounds familiar, it should. See the discussion following Theorem
inversefunctionprops.)
To prove the second property of Theorem inversematrixprops, we note that if \(A\) is invertible then the discussion on page 1 shows the solution to \(AX=B\) to
be \(X = A^{-1}B\), and since \(A^{-1}\) is unique, so is \(A^{-1}B\).
Conversely, if \(AX = B\) has a unique solution for every \(n \times r\) matrix \(B\), then, in particular, there is a unique solution \(X_{0}\) to the equation \(AX = I_{n}\). The
solution matrix \(X_{0}\) is our candidate for \(A^{-1}\). We have \(AX_{0} = I_{n}\) by definition, but we need to also show \(X_{0}A = I_{n}\). To that end, we note that \(A\left (X_{0}A\right ) = \left (AX_{0}\right )A = I_{n}A = A\). In
other words, the matrix \(X_{0}A\) is a solution to the equation \(AX = A\). Clearly, \(X=I_{n}\) is also a solution to the equation \(AX = A\), and since we
are assuming every such equation as a unique solution, we must have \(X_{0}A = I_{n}\). Hence, we have \(X_{0}A = AX_{0} = I_{n}\), so that \(X_{0} = A^{-1}\) and \(A\) is
invertible.
The foregoing discussion justifies our quest to find \(A^{-1}\) using our super-sized augmented matrix approach
We are, in essence, trying to find the unique solution to the equation \(AX = I_{n}\) using row operations.
What does all of this mean for a system of linear equations? Theorem inversematrixprops tells us that if we write the system in the form \(AX=B\), then if
the coefficient matrix \(A\) is invertible, there is only one solution to the system \(-\) that is, if \(A\) is invertible, the system is
consistent and independent. (It can be shown that a matrix is invertible if and only if when it serves as a
coefficient matrix for a system of equations, the system is always consistent independent. It amounts to the
second property in Theorem inversematrixprops where the matrices \(B\) are restricted to being \(n \times 1\) matrices. We note that, owing to
how matrix multiplication is defined, being able to find unique solutions to \(AX = B\) for \(n \times 1\) matrices \(B\) gives you the same
statement about solving such equations for \(n \times r\) matrices \(-\) since we can find a unique solution to them one column at a
time.)
We also know that the process by which we find \(A^{-1}\) is determined completely by \(A\), and not by the constants in \(B\). This
answers the question as to why we would bother doing row operations on a super-sized augmented matrix
to find \(A^{-1}\) instead of an ordinary augmented matrix to solve a system; by finding \(A^{-1}\) we have done all of the row
operations we ever need to do, once and for all, since we can quickly solve any equation \(AX = B\) using one multiplication,
\(A^{-1}B\).
Each of the systems in this part has \(A\) as its coefficient matrix. The only difference between the systems is the
constants which is the matrix \(B\) in the associated matrix equation \(AX=B\). We solve each of them using the formula
\(X = A^{-1}B\).
\(X = A^{-1}B = \left [ \begin{array}{rrr} \frac {9}{13} & \frac {2}{13} & -\frac {7}{13} \\ -\frac {10}{13} & -\frac {8}{13} & \frac {15}{13} \\ -\frac {2}{13} & \frac {1}{13} & \frac {3}{13} \end{array} \right ] \left [ \begin{array}{r} 1 \\ 0 \\ 0 \end{array}\right ] = \left [ \begin{array}{r} \frac {9}{13} \\ -\frac {10}{13} \\ -\frac {2}{13} \end{array}\right ]\). We find \(\left ( \frac {9}{13}, -\frac {10}{13}, -\frac {2}{13} \right )\). (Note that the solution is the first column of the \(A^{-1}\). The reader is encouraged to meditate
on this ‘coincidence’.)
□
In Example matrixinverseex, we see that finding one inverse matrix can enable us to solve an entire family of systems of linear equations.
There are many examples of where this comes in handy ‘in the wild’, and we chose our example for this section from the field
of electronics. We also take this opportunity to introduce the student to how we can compute inverse matrices using the
calculator.
Consider the circuit diagram below. (The authors wish to thank Don Anthan of Lakeland Community College for the
design of this example.) We have two batteries with source voltages \(V\!B_1\) and \(V\!B_2\), measured in volts \(V\), along with six resistors with
resistances \(R_1\) through \(R_6\), measured in kiloohms, \(k\Omega \). Using Ohm’s Law and Kirchhoff’s Voltage Law, we can relate the voltage
supplied to the circuit by the two batteries to the voltage drops across the six resistors in order to find the four ‘mesh’
currents: \(i_1\), \(i_2\), \(i_3\) and \(i_4\), measured in milliamps, \(mA\). If we think of electrons flowing through the circuit, we can think
of the voltage sources as providing the ‘push’ which makes the electrons move, the resistors as obstacles
for the electrons to overcome, and the mesh current as a net rate of flow of electrons around the indicated
loops.
[We will replace this circuit diagram this summer]
The system of linear equations associated with this circuit is
For \(V\!B_1 = 10 V\) and \(V\!B_2 = 5 V\), the calculator gives \(i_1 = 10.625 \, \, mA\), \(i_2 = 6.25 \, \, mA\), \(i_3 = 3.125 \, \, mA\), and \(i_4 = 5 \, \, mA\).
We include screenshot below for this part (and this part only!) for reference, as the other solutions are given
similarly.
By keeping \(V\!B_1 = 10 V\) and setting \(V\!B_2 = 0 V\), we are removing the effect of the second battery. We get \(i_1 = 16.25 \, \, mA\), \(i_2 = 12.5 \, \, mA\), \(i_3 = 11.25 \, \, mA\), and \(i_4 = 10 \, \, mA\).
Here, we are zeroing out \(V\!B_1\) and making \(V\!B_2 = 10\), essentially removing the effect of the first battery. We find \(i_1 = -11.25 \, \, mA\), \(i_2 = -12.5 \, \, mA\), \(i_3 = -16.25 \, \, mA\), and \(i_4 = -10 \, \, mA\),
where the negatives indicate that the current is flowing in the opposite direction as is indicated on the diagram.
The reader is encouraged to study the symmetry here, and if need be, hold up a mirror to the diagram to literally
‘see’ what is happening.
For \(V\!B_1 = 10 V\) and \(V\!B_2 = 10 V\), we get \(i_1 = 5 \, \, mA\), \(i_2 = 0 \, \, mA\), \(i_3 = -5 \, \, mA\), and \(i_4 = 0 \, \, mA\). The mesh currents \(i_2\) and \(i_4\) being zero is a consequence of both batteries ‘pushing’ in
equal but opposite directions, causing the net flow of electrons in these two regions to cancel out. □
PreCalculus
