Inverses of Matrices

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Pre-Calculus › Inverses of Matrices

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Find the inverse of the matrix.

$A=\begin{pmatrix} 6& 2\ 1& 5 \end{pmatrix}$

$A^{-1}=\begin{pmatrix} \frac{5}{28}&\frac{-1}{14} \ \ \frac{-1}{28}& \frac{3}{14} \end{pmatrix}$

$A^{-1}=\begin{pmatrix} 5&-2 \ -1&6 \end{pmatrix}$

$A^{-1}=\begin{pmatrix} \frac{3}{14}&\frac{1}{14} \ \ \frac{1}{28}&\frac{5}{28} \end{pmatrix}$

$A^{-1}=\begin{pmatrix} \frac{3}{14}&\frac{-1}{14} \ \ \frac{-1}{28}&\frac{5}{28} \end{pmatrix}$

Explanation

We use the inverse of a 2x2 matrix formula to determine the answer. Given a matrix

$A=\begin{pmatrix} a&b \ c& d \end{pmatrix}$ it's inverse is given by the formula:

$A^{-1}=\frac{1}{\det A}\begin{pmatrix} d&-b \ -c&a \end{pmatrix}$

First we define the determinant of our matrix:

$\det A=ad-bc=6\cdot5-2\cdot1=30-2=28$

Then,

$A^{-1}=\frac{1}{28}\begin{pmatrix} 5&-2 \ -1& 6 \end{pmatrix}=\begin{pmatrix} \frac{5}{28}&\frac{-1}{14} \ \ \frac{-1}{28}&\frac{3}{14} \end{pmatrix}$

Find the inverse of the matrix.

$A=\begin{pmatrix} 6& 2\ 1& 5 \end{pmatrix}$

$A^{-1}=\begin{pmatrix} \frac{5}{28}&\frac{-1}{14} \ \ \frac{-1}{28}& \frac{3}{14} \end{pmatrix}$

$A^{-1}=\begin{pmatrix} 5&-2 \ -1&6 \end{pmatrix}$

$A^{-1}=\begin{pmatrix} \frac{3}{14}&\frac{1}{14} \ \ \frac{1}{28}&\frac{5}{28} \end{pmatrix}$

$A^{-1}=\begin{pmatrix} \frac{3}{14}&\frac{-1}{14} \ \ \frac{-1}{28}&\frac{5}{28} \end{pmatrix}$

Explanation

We use the inverse of a 2x2 matrix formula to determine the answer. Given a matrix

$A=\begin{pmatrix} a&b \ c& d \end{pmatrix}$ it's inverse is given by the formula:

$A^{-1}=\frac{1}{\det A}\begin{pmatrix} d&-b \ -c&a \end{pmatrix}$

First we define the determinant of our matrix:

$\det A=ad-bc=6\cdot5-2\cdot1=30-2=28$

Then,

$A^{-1}=\frac{1}{28}\begin{pmatrix} 5&-2 \ -1& 6 \end{pmatrix}=\begin{pmatrix} \frac{5}{28}&\frac{-1}{14} \ \ \frac{-1}{28}&\frac{3}{14} \end{pmatrix}$

Find the multiplicative inverse of the following matrix:

$\begin{bmatrix} 2&3 \ 1&1 \end{bmatrix}$

$\begin{bmatrix} -1&3 \ 1&-2 \end{bmatrix}$

$\begin{bmatrix} \frac{1}{2}&\frac{1}{3} \ 1& 1 \end{bmatrix}$

$\begin{bmatrix} -2&-3 \ -1&-1 \end{bmatrix}$

$\begin{bmatrix} 1&1 \ 3& 2 \end{bmatrix}$

This matrix has no inverse.

Explanation

By writing the augmented matrix $\begin{bmatrix} 2& 3&| 1 &0 \ 1& 1& |0&1 \end{bmatrix}$ , and reducing the left side to the identity matrix, we can implement the same operations onto the right side, and we arrive at $\begin{bmatrix} 1&0 |& -1& 3\ 0& 1|& 1& -2 \end{bmatrix}$ , with the right side representing the inverse of the original matrix.

Find the multiplicative inverse of the following matrix:

$\begin{bmatrix} 2&3 \ 1&1 \end{bmatrix}$

$\begin{bmatrix} -1&3 \ 1&-2 \end{bmatrix}$

$\begin{bmatrix} \frac{1}{2}&\frac{1}{3} \ 1& 1 \end{bmatrix}$

$\begin{bmatrix} -2&-3 \ -1&-1 \end{bmatrix}$

$\begin{bmatrix} 1&1 \ 3& 2 \end{bmatrix}$

This matrix has no inverse.

Explanation

Find the inverse of the matrix

$A = \begin{pmatrix} 2 & -1 \ 1 & -5 \end{pmatrix}$

$\begin{pmatrix} \frac{5}{9} & -\frac{1}{9} \ \frac{1}{9} & -\frac{2}{9} \end{pmatrix}$

$\begin{pmatrix} 5 & -1 \ 1 & -2 \end{pmatrix}$

$\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}$

$\begin{pmatrix} \frac{5}{9} & \frac{1}{9} \ \frac{1}{9} & \frac{2}{9} \end{pmatrix}$

None of the other answers.

Explanation

There are a couple of ways to do this. I will use the determinant method.

First we need to find the determinant of this matrix, which is

for a matrix in the form:

$\begin{pmatrix} a & b \ c & d \end{pmatrix}$ .

Substituting in our values we find the determinant to be:

Now one formula for finding the inverse of the matrix is

$\frac{1}{det[A]}adj[A]=\frac{1}{ad-bc}\begin{pmatrix} d& -b\ -c&a \end{pmatrix}$

$= \frac{1}{-9}\begin{pmatrix} -5 &1 \ -1 & 2 \end{pmatrix}$

$= \begin{pmatrix} \frac{5}{9} & -\frac{1}{9} \ \frac{1}{9} & -\frac{2}{9} \end{pmatrix}$ .

Find the inverse of the matrix

$A = \begin{pmatrix} 2 & -1 \ 1 & -5 \end{pmatrix}$

$\begin{pmatrix} \frac{5}{9} & -\frac{1}{9} \ \frac{1}{9} & -\frac{2}{9} \end{pmatrix}$

$\begin{pmatrix} 5 & -1 \ 1 & -2 \end{pmatrix}$

$\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}$

$\begin{pmatrix} \frac{5}{9} & \frac{1}{9} \ \frac{1}{9} & \frac{2}{9} \end{pmatrix}$

None of the other answers.

Explanation

There are a couple of ways to do this. I will use the determinant method.

First we need to find the determinant of this matrix, which is

for a matrix in the form:

$\begin{pmatrix} a & b \ c & d \end{pmatrix}$ .

Substituting in our values we find the determinant to be:

Now one formula for finding the inverse of the matrix is

$\frac{1}{det[A]}adj[A]=\frac{1}{ad-bc}\begin{pmatrix} d& -b\ -c&a \end{pmatrix}$

$= \frac{1}{-9}\begin{pmatrix} -5 &1 \ -1 & 2 \end{pmatrix}$

$= \begin{pmatrix} \frac{5}{9} & -\frac{1}{9} \ \frac{1}{9} & -\frac{2}{9} \end{pmatrix}$ .

Find the inverse of the following matrix.

$\begin{bmatrix} 1&\pi \ \pi & \pi^2 \end{bmatrix}$

This matrix has no inverse.

$\begin{bmatrix} 1&\pi \ -\pi & \pi^2 \end{bmatrix}$

$\begin{bmatrix} 1&\sqrt{\pi} \ \pi & \pi^2 \end{bmatrix}$

$\begin{bmatrix} 1+\pi &\sqrt{\pi} \ -\pi^2 & \pi^2-\pi \end{bmatrix}$

$\begin{bmatrix}1 & 0\0&0 \end{bmatrix}$

Explanation

This matrix has no inverse because the columns are not linearly independent. This means if you row reduce to try to compute the inverse, one of the rows will have only zeros, which means there is no inverse.

Find the inverse of the following matrix.

$\begin{bmatrix} 1&\pi \ \pi & \pi^2 \end{bmatrix}$

This matrix has no inverse.

$\begin{bmatrix} 1&\pi \ -\pi & \pi^2 \end{bmatrix}$

$\begin{bmatrix} 1&\sqrt{\pi} \ \pi & \pi^2 \end{bmatrix}$

$\begin{bmatrix} 1+\pi &\sqrt{\pi} \ -\pi^2 & \pi^2-\pi \end{bmatrix}$

$\begin{bmatrix}1 & 0\0&0 \end{bmatrix}$

Explanation

What is the inverse of the identiy matrix $\begin{bmatrix} 1& 0\ 0 &1 \end{bmatrix}$ ?

The identity matrix $\begin{bmatrix} 1&0 \0 & 1 \end{bmatrix}$

$\begin{bmatrix} 1& 1\ 1 & 1 \end{bmatrix}$

$\begin{bmatrix} 0&0 \0 & 0 \end{bmatrix}$

$\begin{bmatrix} 2& -1\-1 & 2 \end{bmatrix}$

Explanation

By definition, an inverse matrix is the matrix B that you would need to multiply matrix A by to get the identity. Since the identity matrix yields whatever matrix it is being multiplied by, the answer is the identity itself.

What is the inverse of the identiy matrix $\begin{bmatrix} 1& 0\ 0 &1 \end{bmatrix}$ ?

The identity matrix $\begin{bmatrix} 1&0 \0 & 1 \end{bmatrix}$

$\begin{bmatrix} 1& 1\ 1 & 1 \end{bmatrix}$

$\begin{bmatrix} 0&0 \0 & 0 \end{bmatrix}$

$\begin{bmatrix} 2& -1\-1 & 2 \end{bmatrix}$

Explanation

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