Inverse Matrices

Sometimes, we need to find the inverse of a matrix. The concept of an "inverse" is hardly new, and we are probably already familiar with the inverse of a number. In fact, we can find the inverse of virtually anything, including trigonometric functions and even imaginary numbers. Finding the inverse of a matrix can be very useful in a range of operations. But what exactly is the inverse of a matrix? How can we find an inverse of a matrix, and why would we even want to do this in the first place? Let''s find out:

Inverse matrices explained

We write the inverse of matrix $A$ as $A^{-1}$ . The inverse of a matrix functions in a similar way to a reciprocal of a number. We may recall that when we find the reciprocal of a number, we simply "flip" the number over its division sign. For example, the reciprocal of "7" (or $\frac{7}{1}$ ) is $\frac{1}{7}$

Instead of flipping the matrix over a division symbol, we flip a few of its elements around (there are a few additional steps). Although this process is different, the end result is very similar.

Recall that when we multiply a number by its reciprocal, the result is always "1." For example, $\frac{7}{1}\times 7=1$ .

In the same way, $A\times A^{-1}$ (Written as $A{A}^{-1}$ ) $=I$ . Remember that we use " $=I$ " to represent an identity matrix, which is equivalent to the value "1" in linear algebra. It also doesn''t matter where we place our inverse matrix before carrying out our matrix multiplication. We can put our inverse matrix on the right of $A$ or the left of $A$ , and the result will still be the same. In other words, multiplying a given matrix by its inverse is commutative.

As a quick reminder, an identity matrix is a matrix with 1s as its main diagonal. The rest of its elements are 0s. It is always square. Here is an example of an identity matrix:

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

It''s also worth noting that we sometimes don''t have an inverse at all. There is no guarantee that every single matrix will have an inverse. The only way to check is by multiplying the matrix by its inverse and seeing whether we get "I." If we do not get I, then the matrix has no inverse. In other words, a matrix only has an inverse if:

$A{A}^{-1}=I$ Or: $A^{-1}A=I$

If a matrix has an inverse, we call it "nonsingular" or "invertible."

If the matrix has no inverse, we call it "singular."

How to find an inverse matrix

In order to find the inverse of a $2\times 2$ matrix, we need to follow a few steps:

$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

The inverse of this matrix becomes:

$\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

Let''s give that a try:

${\left[\begin{array}{cc}4& 7\\ 2& 6\end{array}\right]}^{-1}$

Can we solve this?

First, we flip the positions of a and d:

$\left[\begin{array}{cc}6& 7\\ 2& 4\end{array}\right]$

Then we put negatives in front of b and c:

$\left[\begin{array}{cc}6& -7\\ -2& 4\end{array}\right]$

Finally, we divide by $ad-bc$:

$\frac{1}{10}\left[\begin{array}{cc}6& -7\\ -2& 4\end{array}\right]$

We are left with:

$\left[\begin{array}{cc}0.6& -0.7\\ -0.2& 0.4\end{array}\right]$

Remember, we need to multiply the original matrix by this inverse matrix to find out whether the answer is I. If so, then we know that the matrix is nonsingular.

$\left[\begin{array}{cc}6& -7\\ -2& 4\end{array}\right]\left[\begin{array}{cc}0.6& -0.7\\ -0.2& 0.4\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Great! The result is an identity matrix, so we know that this inverse is correct. We also know that our original matrix is nonsingular or invertible.

In the above operation, $ad-bc$ is a very important value. It is so important, in fact, that we give it its own special name: The "determinant." In order to find the determinant, we must first check to see whether the matrix is square. If the matrix is not square, then we cannot find the determinant. We represent the determinant of a matrix $A$ as $\left|A\right|$ -- just like the absolute value of a number. Finding the determinant of a matrix becomes a little more complicated if we''re dealing with matrices bigger than $2\times 2$ , although it is possible.

We can also follow these steps to find our inverse of a square matrix:

1. Write the doubly augmented matrix $\left[A|I_n\right]$ , where "I" is the identity matrix of A.

2. Apply elementary matrix row operations to put the matrix in reduced row-echelon form

3. Find out whether the matrix is invertible

We can write this in the following formula:

$A^{-1}=\frac{1}{\left|A\right|}\left(\text{adj}\left(A\right)\right)$

Remember that in this formula, $\left|A\right|$ represents the determinant of the matrix.

What is a "doubly augmented matrix?" Basically, this is when we write two matrices as one. And what about "reduced echelon form?" This is when a matrix satisfies the following:

• In each row, its leftmost nonzero entry is 1 and the column that contains this 1 has zeros for all other elements. We call this a "leading 1."
• In the second row, the leading 1 is to the right of the leading 1 in the first row. This can also apply to any rows further down, including the third row or the fourth row, etc. The leading 1 must simply be to the right of the leading 1 above it.
• If a row contains only zeros, it must be placed at the bottom.

Finding the inverse of a matrix: Practice

Now that we have covered the basics of finding inverse matrices, it''s time to put our skills to good use and try it for ourselves.

Consider the following matrix:

$A=\left[\begin{array}{cc}1& 2\\ 1& 1\end{array}\right]$

Let''s say that this matrix is $A$ . Can we find $A^{-1}$

Remember, our first step is to write the doubly augmented matrix $\left[A|I_n\right]$ . This means we have to combine the identity matrix with the original matrix in one matrix:

$\left[A|I\right]=\left[\begin{array}{cccc}1& 2& 1& 0\\ 1& 1& 0& 1\end{array}\right]$

Our next step is to write our matrix in reduced row-echelon form. Right now, the matrix is not in this format. We need to make sure that each row has a 1 as its leftmost nonzero element. We also need to make sure the leading 1 in the second row is to the right of the leading 1 in the first row.

Let''s begin our matrix row operations:

$\left[\begin{array}{cccc}1& 2& 1& 0\\ 1& 1& 0& 1\end{array}\right]$

We can start by subtracting row 2 from row 1, and then replacing row 2 with the result $R_2=R_1-R_2$ . We are left with:

$\left[\begin{array}{cccc}1& 2& 1& 0\\ 0& 1& 1& -1\end{array}\right]$

Next, we can replace row 1 with $-2R_2+R_1$

$\left[\begin{array}{cccc}1& 0& -1& 2\\ 0& 1& 1& -1\end{array}\right]$

Great! Our doubly augmented matrix is now in reduced row-echelon form. In other words:

$\left[\begin{array}{cccc}1& 0& -1& 2\\ 0& 1& 1& -1\end{array}\right]=\left[I|A^{-1}\right]$

This tells us that the system has a solution and that A is invertible.

To find the inverse matrix, all we need to do is look at the right side of the matrix:

$A^{-1}=\left[\begin{array}{cc}-1& 2\\ 1& -1\end{array}\right]$

When do we use the inverse of matrices?

But why go through all this trouble to find the inverse? Basically, it comes down to one simple fact:

We cannot divide when we work with matrices. Because of this, we are left with no choice but to multiply by an inverse. This provides us with the same results. Even though it''s more complicated than simply dividing, it''s our only real option.

We also use inverse matrices to solve systems of linear equations, 3D transformations, and much more.

Topics related to the Inverse Matrices

Matrix Multiplication

Identity Matrix

Augmented Matrices

Flashcards covering the Inverse Matrices

Precalculus Flashcards

CLEP Precalculus Flashcards

Practice tests covering the Inverse Matrices

Precalculus Diagnostic Tests

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