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# Inverse Matrices

## Use a formula to identify the inverse of 2 X 2 matrices

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Finding the Inverse of a Matrix

At your school book fair, you buy 3 paperback and 2 hardcovers. Your best friend buys 2 paperbacks and 4 hardcovers. What is the inverse of the matrix represented by this situation?

### Inverse of a Matrix

Recall that the multiplicative inverse of a real number is the reciprocal of the number and that the product of a number and its multiplicative inverse is the multiplicative identity, or 1. For example: \begin{align*}\frac{3}{7} \times \frac{7}{3}=1\end{align*}. Now we need to define a multiplicative identity and a multiplicative inverse for a square matrix. For real numbers, 1 is considered the identity because we can multiply any number, \begin{align*}a\end{align*}, by 1 and the result is \begin{align*}a\end{align*}. In other words, the value of the number does not change. For matrices, the multiplicative inverse of a square matrix will be a square matrix in which the values of the main diagonal are 1 and the remaining values are all zero. The following are examples of identity matrices.

\begin{align*}I= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \qquad I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \qquad I= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}\end{align*}

The products below illustrate how we can multiply a matrix by the identity and the result will be the original matrix.

Given:

\begin{align*}A= \begin{bmatrix} 2 & -1\\ -3 & 5 \end{bmatrix}, \ AI=\begin{bmatrix} 2 & -1\\ -3 & 5 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}= \begin{bmatrix} 2(1)+(-1)(0) & 2(0)+(-1)(1)\\ (-3)(1)+5(0) & (-3)(0)+5(1) \end{bmatrix} = \begin{bmatrix} 2 & -1\\ -3 & 5 \end{bmatrix}\end{align*}

In fact, it does not matter which order we multiply by the identity matrix. In other words, \begin{align*}AI=IA=A\end{align*}.

Now that we have defined an identity matrix, we can determine an inverse matrix such that \begin{align*}A(A^{-1})=(A^{-1})A=I\end{align*}.

The formula for finding the Inverse of a \begin{align*}2 \times 2\end{align*} matrix is:

Given:

\begin{align*}A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}, \ A^{-1}= \frac{1}{det[A]}\begin{bmatrix} d & -b\\ -c & a \end{bmatrix}=\frac{1}{ad-bc}\begin{bmatrix} d & -b\\ -c & a \end{bmatrix},\end{align*}

where \begin{align*}ad-bc \ne 0\end{align*}

\begin{align*}^*\end{align*}Note: If \begin{align*}ad-bc = 0\end{align*} or \begin{align*}det[A]=0\end{align*}, the matrix \begin{align*}A\end{align*} is called singular. The inverse of a singular matrix cannot be determined.

Let's find the inverse of the following matrices and verify our solution is correct (if the inverse exists).

1.      \begin{align*}\begin{bmatrix} 1 & 2\\ -3 & 7 \end{bmatrix}\end{align*}

First, use the formula above to find the inverse.

\begin{align*}\begin{bmatrix} 1 & 2\\ -3 & 7 \end{bmatrix}^{-1} = \frac{1}{(1)(7)-(2)(-3)} \begin{bmatrix} 7 & -2\\ 3 & 1 \end{bmatrix} = \frac{1}{13}\begin{bmatrix} 7 & -2\\ 3 & 1 \end{bmatrix} = \begin{bmatrix} \frac{7}{13} & -\frac{2}{13}\\ \frac{3}{13} & \frac{1}{13} \end{bmatrix}\end{align*}

To verify that this is indeed the inverse, we must show that the product of the inverse and the original matrix is the identity matrix for a \begin{align*}2 \times 2\end{align*} matrix. It will be easier to find this product using the form of the inverse where the reciprocal of the determinant has not been distributed inside the matrix as shown below:

\begin{align*}\frac{1}{13} \begin{bmatrix} 7 & 2\\ -3 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2\\ 3 & 7 \end{bmatrix} = \frac{1}{13}\begin{bmatrix} (7)(1)+(-2)(-3) & (7)(-2)+(2)(7)\\ (-3)(1)+(1)(3) & (-3)(-2)+(1)(7) \end{bmatrix} = \frac{1}{13}\begin{bmatrix} 13 & 0\\ 0 & 13 \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\end{align*}

Are \begin{align*}\begin{bmatrix} 4 & -3\\ -3 & 2 \end{bmatrix}\end{align*} and \begin{align*}\begin{bmatrix} -2 & -3\\ -3 & -4 \end{bmatrix}\end{align*} inverses?

If the matrices are inverses then their product will be the identity matrix.

\begin{align*}\begin{bmatrix} 4 & -3\\ -3 & 2 \end{bmatrix} \begin{bmatrix} -2 & -3\\ -3 & -4 \end{bmatrix}= \begin{bmatrix} (4)(-2)+(-3)(-3) & (4)(-3)+(-3)(-4)\\ (-3)(-2)+(2)(-3) & (-3)(-3)+(2)(-4) \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\end{align*}

Since the product is the identity matrix, the matrices are inverses of one another.

1.

\begin{align*}\begin{bmatrix} 4 & 6\\ -2 & -3 \end{bmatrix}.\end{align*}

Use the formula above to find the inverse.

\begin{align*}\begin{bmatrix} 4 & 6\\ -2 & -3 \end{bmatrix}^{-1} = \frac{1}{(4)(-3)-(6)(-2)} \begin{bmatrix} -3 & -6\\ 2 & 4 \end{bmatrix} = \frac{1}{0} \begin{bmatrix} -3 & -6\\ 2 & 4 \end{bmatrix} \Rightarrow \ & \text{the inverse does not exist.}\\ & \text{This matrix is singular.}\end{align*}

### Examples

#### Example 1

Earlier, you were asked to find the inverse of the matrix that represents the book fair purchases.

The matrix that represents this situation is: \begin{align*}\begin{bmatrix} 3 & 2\\ 2 & 4 \end{bmatrix}\end{align*}

Use the formula you learned in this lesson to find the inverse.

\begin{align*}\begin{bmatrix} 3 & 2\\ 2 & 4 \end{bmatrix}^{-1} = \frac{1}{[(3)(4)-(2)(2)]} \begin{bmatrix} 4 & -2\\ -2 & 3 \end{bmatrix} = \frac{1}{8}\begin{bmatrix} 4 & -2\\ -2 & 3 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{4}\\ -\frac{1}{4} & \frac{3}{8} \end{bmatrix}\end{align*}

#### Example 2

Are matrices \begin{align*}\begin{bmatrix} -\frac{1}{3} & 1\\ -\frac{1}{2} & 2 \end{bmatrix}\end{align*} and \begin{align*}\begin{bmatrix} -12 & 6\\ -3 & 2 \end{bmatrix}\end{align*} inverses of each other?

\begin{align*}\begin{bmatrix} -\frac{1}{3} & 1\\ -\frac{1}{2} & 2 \end{bmatrix} \begin{bmatrix} -12 & 6\\ -3 & 2 \end{bmatrix} = \begin{bmatrix} \left(-\frac{1}{3}\right)(-12)+(1)(-3) & \left(-\frac{1}{3}\right)(6)+(1)(2) \\ \left(-\frac{1}{2}\right)(-12)+(2)(-3) & \left(-\frac{1}{2}\right)(6)+(2)(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.\end{align*}

Yes, they are inverses.

#### Example 3

Find the inverse of

\begin{align*}\begin{bmatrix} 4 & 2\\ 10 & 5 \end{bmatrix}.\end{align*}

\begin{align*} \begin{bmatrix} 4 & 2\\ 10 & 5 \end{bmatrix}^{-1} = \frac{1}{(4)(5)-(2)(10)} \begin{bmatrix} 5 & -2\\ -10 & 4 \end{bmatrix} = \frac{1}{0} \begin{bmatrix} 5 & -2\\ -10 & 4 \end{bmatrix} \Rightarrow & \ \text{inverse does not exist.}\\ & \text{This matrix is singular.}\end{align*}

#### Example 4

Find the inverse of

\begin{align*}\begin{bmatrix} 3 & -4\\ 6 & -7 \end{bmatrix}.\end{align*}

\begin{align*}\begin{bmatrix} 3 & -4\\ 6 & -7 \end{bmatrix}^{-1} = \frac{1}{(3)(-7)-(-4)(6)} \begin{bmatrix} -7 & 4\\ -6 & 3 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} -7 & 4\\ -6 & 3 \end{bmatrix} = \begin{bmatrix} -\frac{7}{3} & \frac{4}{3}\\ -2 & 1 \end{bmatrix}\end{align*}

### Review

Determine whether the following pairs of matrices are inverses of one another.

1. .
\begin{align*}\begin{bmatrix} 5 & -15\\ 3 & -10 \end{bmatrix}\end{align*} and \begin{align*}\begin{bmatrix} 2 & -3\\ \frac{3}{5} & -1 \end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} 3 & 7\\ 1 & 2 \end{bmatrix}\end{align*} and \begin{align*}\begin{bmatrix} -2 & 7\\ 1 & -3 \end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} -5 & 8\\ 1 & 3 \end{bmatrix}\end{align*} and \begin{align*}\begin{bmatrix} 3 & -8\\ -1 & -5 \end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} -5 & 4\\ -9 & 7 \end{bmatrix}\end{align*} and \begin{align*}\begin{bmatrix} 7 & -4\\ 9 & -5 \end{bmatrix}\end{align*}

Find the inverse of each matrix below, if it exists.

1. .
\begin{align*}\begin{bmatrix} -11 & 7\\ -3 & 2 \end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} 6 & -3\\ 8 & -5 \end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} 1 & 2\\ 4 & 9 \end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} -2 & -3\\ 6 & 9 \end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} -2 & 4\\ 5 & 3 \end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} 0 & 1\\ 3 & 2 \end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} -4 & 7\\ 0 & 2 \end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} 2 & -6\\ -6 & 18 \end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} 7 & 5\\ 14 & 10 \end{bmatrix}\end{align*}
1. .
\begin{align*}\begin{bmatrix} -2 & 5\\ -2 & 6 \end{bmatrix}\end{align*}
1. For two 2x2 matrices, A and B, to be inverses of each other, what must be true of AB and BA?

To see the Review answers, open this PDF file and look for section 4.9.

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### Vocabulary Language: English

TermDefinition
determinant The determinant is a single number descriptor of a square matrix. The determinant is computed from the entries of the matrix, and has many properties and interpretations explored in linear algebra.
identity matrix An identity matrix is a matrix with zeros everywhere except along the diagonal where there are ones.
invertible A square matrix is invertible if it has an inverse. If the determinant of a square matrix is zero then the matrix is not invertible.
linearly independent Three equations are linearly independent if each equation cannot be produced by a linear combination of the other two. Systems that are linearly independent will have just one solution.
Multiplication Rule States that for 2 events (A and B), the probability of A and B is given by: P(A and B) = P(A) x P(B).
Multiplicative Identity for Matrices The multiplicative identity for matrices is the square matrix that can be multiplied by any square matrix of the same size without changing the values of the original matrix. This matrix will contain all 1’s in the main diagonal and zeros elsewhere.
Multiplicative Inverse The multiplicative inverse of the square matrix $A$ is the matrix $A^{-1}$, such that $A^{-1}A=AA^{-1}=I$.
Square matrices Square matrices are matrices in which the number of rows equals the number of columns.
Square matrix A square matrix is a matrix in which the number of rows equals the number of columns.