Find the Multiplicative Inverse of a Matrix - Pre-Calculus

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What is the inverse of the following nxn matrix

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

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Note the first and the last columns are equal.

Therefore, when we try to find the determinant using the following formula we get the determinant equaling 0:

$\begin{bmatrix} 1 & 0 & 1\1 & 0 & 1\1 & 0 & 1\end{bmatrix}=1\begin{bmatrix}0 &1\0&1 \end{bmatrix}-0\begin{bmatrix}1 & 1\1 &1 \end{bmatrix}+1\begin{bmatrix} 1 &0\1&0\end{bmatrix}$

This means simply, that the matrix does not have an inverse.

Find the inverse of the matrix

$\begin{pmatrix} 1&2 \ 2&4 \end{pmatrix}$ .

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For a 2x2 matrix

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

the inverse can be found by

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

Because the determinant is equal to zero in this problem, or

$1\cdot4-2\cdot2=4-4=0$ ,

the inverse does not exist.

Find the inverse of the matrix

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

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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.

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

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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:

First we define the determinant of our matrix:

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

Then,

Find the inverse of the following matrix.

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

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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 multiplicative inverse of the following matrix:

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

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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.

What is the inverse of the following nxn matrix

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

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Note the first and the last columns are equal.

Therefore, when we try to find the determinant using the following formula we get the determinant equaling 0:

$\begin{bmatrix} 1 & 0 & 1\1 & 0 & 1\1 & 0 & 1\end{bmatrix}=1\begin{bmatrix}0 &1\0&1 \end{bmatrix}-0\begin{bmatrix}1 & 1\1 &1 \end{bmatrix}+1\begin{bmatrix} 1 &0\1&0\end{bmatrix}$

This means simply, that the matrix does not have an inverse.

Find the inverse of the matrix

$\begin{pmatrix} 1&2 \ 2&4 \end{pmatrix}$ .

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For a 2x2 matrix

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

the inverse can be found by

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

Because the determinant is equal to zero in this problem, or

$1\cdot4-2\cdot2=4-4=0$ ,

the inverse does not exist.

Find the inverse of the matrix

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

Tap to see back →

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.

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

Tap to see back →

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:

First we define the determinant of our matrix:

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

Then,

Find the inverse of the following matrix.

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

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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 multiplicative inverse of the following matrix:

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

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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.

What is the inverse of the following nxn matrix

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

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Note the first and the last columns are equal.

Therefore, when we try to find the determinant using the following formula we get the determinant equaling 0:

$\begin{bmatrix} 1 & 0 & 1\1 & 0 & 1\1 & 0 & 1\end{bmatrix}=1\begin{bmatrix}0 &1\0&1 \end{bmatrix}-0\begin{bmatrix}1 & 1\1 &1 \end{bmatrix}+1\begin{bmatrix} 1 &0\1&0\end{bmatrix}$

This means simply, that the matrix does not have an inverse.

Find the inverse of the matrix

$\begin{pmatrix} 1&2 \ 2&4 \end{pmatrix}$ .

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For a 2x2 matrix

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

the inverse can be found by

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

Because the determinant is equal to zero in this problem, or

$1\cdot4-2\cdot2=4-4=0$ ,

the inverse does not exist.

Find the inverse of the matrix

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

Tap to see back →

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.

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

Tap to see back →

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:

First we define the determinant of our matrix:

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

Then,

Find the inverse of the following matrix.

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

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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 multiplicative inverse of the following matrix:

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

Tap to see back →

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.

What is the inverse of the following nxn matrix

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

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Note the first and the last columns are equal.

Therefore, when we try to find the determinant using the following formula we get the determinant equaling 0:

$\begin{bmatrix} 1 & 0 & 1\1 & 0 & 1\1 & 0 & 1\end{bmatrix}=1\begin{bmatrix}0 &1\0&1 \end{bmatrix}-0\begin{bmatrix}1 & 1\1 &1 \end{bmatrix}+1\begin{bmatrix} 1 &0\1&0\end{bmatrix}$

This means simply, that the matrix does not have an inverse.

Find the inverse of the matrix

$\begin{pmatrix} 1&2 \ 2&4 \end{pmatrix}$ .

Tap to see back →

For a 2x2 matrix

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

the inverse can be found by

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

Because the determinant is equal to zero in this problem, or

$1\cdot4-2\cdot2=4-4=0$ ,

the inverse does not exist.