The Inverse - Linear Algebra

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Question

Calculate $A^{-1}$ , where

$A=\begin{bmatrix} 1 & 5\ 2 & 4 \end{bmatrix}$

Answer

The first step, is to create an augmented matrix with the identity Matrix.

$A=\left[ \begin{matrix} 1 & 5 \ 2 &4 \ \end{matrix} \right | \left \begin{matrix} 1 & 0 \ 0 & 1 \ \end{matrix} \right ]$

To find the inverse, all we need to do is get the Identity Matrix on the left hand side.

$2R_1-R_2\rightarrow R_2$

$A=\left[ \begin{matrix} 1 & 5 \ 0 &6 \ \end{matrix} \right | \left \begin{matrix} 1 & 0 \ 2 & -1 \ \end{matrix} \right ]$

$\frac{1}{6}R_2\rightarrow R_2$

$A=\left[\begin{matrix} 1 & 5\ 0 & 1 \end{matrix}\right| \left.\begin{matrix} 1& 0\ \frac{1}{3}& -\frac{1}{6} \end{matrix}\right]$

$-5R_2+R_1\rightarrow R_1$

$A=\left[\begin{matrix} 1 & 0\ 0 & 1 \end{matrix}\right| \left.\begin{matrix} -\frac{2}{3}& \frac{5}{6}\ \frac{1}{3}& -\frac{1}{6} \end{matrix}\right]$

Since we have the Identity Matrix on the left hand side, we are done solving for the inverse.

$A^{-1}=\left[\begin{matrix} -\frac{2}{3}& \frac{5}{6}\ \frac{1}{3}& -\frac{1}{6} \end{matrix}\right]$

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Question

Find the inverse of the matrix $\begin{bmatrix} 1 & -2\ -3& 3\end{bmatrix}$

Answer

To find the inverse, first find the determinant. In this case, the determinant is $1\cdot3 - (-3)(-2) = 3 - 6 = -3$

The inverse is found by multiplying $-\frac{1}{3} \begin{bmatrix} 3 & 2\ 3& 1\end{bmatrix} = \begin{bmatrix} -1 &-\frac{2}{3} \ -1& -\frac{1}{3}\end{bmatrix}$

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Question

Find the inverse of the matrix $\begin{bmatrix} 5 & 0\ -1& 2\end{bmatrix}$

Answer

First, find the determinant:

Now multiply $\frac{1}{10}$ by the matrix $\begin{bmatrix} 2 &0 \1 & 5\end{bmatrix}$ , the original matrix with 2 and 5 switched and the signs changed on -1 and 0.

$0.1 \begin{bmatrix} 2&0 \1 & 5\end{bmatrix} = \begin{bmatrix} 0.2 &0 \ 0.1 & 0.5\end{bmatrix}$

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Question

Find the inverse of matrix A.

$A = \begin{bmatrix} 2&3 \ 4&-6 \end{bmatrix}$

Answer

For any 2x2 matrix, to determine if it is invertible, we must first calculate its determinant. If the determinant is equal to 0, then the matrix is not invertible. If it isn’t equal to 0, then its inverse can be found using this formula:

$Let A=\begin{bmatrix} a&b \ c&d \end{bmatrix}$

$A^{-1}= \frac{1}{ad-bc}\begin{bmatrix} d& -b\ -c& a\end{bmatrix}$

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Question

Find the inverse of matrix A.

$A=\begin{bmatrix} 4&3 \ 6&5 \end{bmatrix}$

Answer

For any 2x2 matrix, to determine if it is invertible, we must first calculate its determinant. If the determinant is equal to 0, then the matrix is not invertible. If it isn’t equal to 0, then its inverse can be found using this formula:

$Let A=\begin{bmatrix} a&b \ c&d \end{bmatrix}$

$A^{-1}= \frac{1}{ad-bc}\begin{bmatrix} d& -b\ -c& a\end{bmatrix}$

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Question

Determine the inverse of matrix A where

$A=\begin{bmatrix} 2&-8 \ 1&4 \end{bmatrix}$

Answer

To find the inverse of a matrix first look to verify that the matrix is square. If it is not square, it does not have an inverse. Next, you must find the determinant. If the determinant is 0, then the matrix does not have an inverse. The determinant for this matrix is ad-bc = 16, therefore it has an inverse. To find the inverse of a 2x2 matrix we first write it in augmented form.

$\begin{bmatrix} 2&-8 &1 &0 \ 1&4 &0 & 1 \end{bmatrix}$ First we will swap rows 1 and 2 $\begin{bmatrix} 1&4 &0 &1 \ 2&-8 &1& 0 \end{bmatrix}$ next we will eliminate the first column by taking R2-2R1 $\begin{bmatrix} 1&4 &0 &1 \ 0&-16 &1& -2 \end{bmatrix}$ , next we will divide R2/16 to set the second pivot. $\begin{bmatrix} 1&4 &0 &1 \ 0&1 &-1/16& 1/8 \end{bmatrix}$ . Next we will eliminate the second column by taking R1-4R2. $\begin{bmatrix} 1&0 &1/4 &1/2 \ 0&1 &-1/16& 1/8 \end{bmatrix}$ . Now that we have the identity matrix on the left, our answer is on the right. There is, however, an easier way to determine the inverse of a 2x2 matrix. The trick is to swap the numbers in spots a and d, put negatives in front of the numbers in spots b and c and then divide everything by the determinant. For this example, $\begin{bmatrix} 4&8 \ -1& 2 \end{bmatrix}$ then divide by the determinant which is 16 and simplify.

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Question

Determine the inverse of matrix A where

$A=\begin{bmatrix} 32&1 \ 5&8 \ -1&6 \end{bmatrix}$

Answer

The matrix is not square so it does not have an inverse.

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Question

Determine the inverse of matrix A where

$A = \begin{bmatrix} 1&5 \ 4&20 \end{bmatrix}$

Answer

The matrix is square, so it could have an inverse. Next we find the determinant. This matrix has a determinant of 0, so it does not have an inverse.

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Question

Determine the inverse of matrix A where

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

Answer

To find the inverse of a matrix first look to verify that the matrix is square. If it is not square, it does not have an inverse. Next, you must find the determinant. If the determinant is 0, then the matrix does not have an inverse. The determinant for this matrix is ad-bc = 9, therefore it has an inverse. To find the inverse of a 2x2 matrix we first write it in augmented form.

$\begin{bmatrix} 2&-1 &1 &0 \ 5&2 &0 & 1 \end{bmatrix}$ First we will divide R1/2 $\begin{bmatrix} 1&-1/2 &1/2 &0 \ 5&2 &0& 1\end{bmatrix}$ next we will eliminate the first column by taking R2-5R1 $\begin{bmatrix} 1&-1/2 &1/2 &0 \ 0&9/2 &-5/2& 1 \end{bmatrix}$ , next we will divide 9R2/2 to set the second pivot. $\begin{bmatrix} 1&-1/2 &1/2 &0 \ 0&1 &-5/9& 2/9 \end{bmatrix}$ . Next we will eliminate the second column by taking R1+1/2R2. $\begin{bmatrix} 1&0 &2/9 &1/9 \ 0&1 &-5/9& 2/9 \end{bmatrix}$ . Now that we have the identity matrix on the left, our answer is on the right. There is, however, an easier way to determine the inverse of a 2x2 matrix. The trick is to swap the numbers in spots a and d, put negatives in front of the numbers in spots b and c and then divide everything by the determinant. For this example, $\begin{bmatrix} 2&1 \ -5& 2 \end{bmatrix}$ then divide by the determinant which is 9 and simplify.

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Question

Determine the inverse of matrix A where

$A=\begin{bmatrix} 5&2 \ 1&0 \end{bmatrix}$

Answer

To determine the inverse of a matrix, you must first verify that the matrix is square. Next calculate the determinant. The determinant for this matrix is -2. Now you can swap the numbers in a and d and put a negative in front of the numbers in b and c. $\begin{bmatrix} 0&-2 \ -1& 5 \end{bmatrix}$ once you have done that, then you divide each number by the determinant.

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Question

Determine the inverse of matrix A where

$A = \begin{bmatrix} 3&2 & 6\ 7&0 &1 \end{bmatrix}$

Answer

The matrix is not square, so it does not have an inverse.

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Question

Determine the inverse of matrix A where

$A=\begin{bmatrix} 2& 3\ 4& 6 \end{bmatrix}$

Answer

To determine the inverse of a matrix, you must first verify that the matrix is square. Next calculate the determinant. The determinant for this matrix is 0, so it does not have an inverse.

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Question

Determine the inverse of matrix A where

$\begin{bmatrix} 8&-15 \ 22& 4 \end{bmatrix}$

Answer

The matrix is square, so it could have an inverse. Next you need to find the determinant which is 362. Swap the numbers in spots a and d and put a negative in front of the numbers in spots b and c.

$\begin{bmatrix} 4&15 \ -22& 8 \end{bmatrix}$ then divide each number by the determinant and simplify.

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Question

Determine the inverse of matrix A where

$A=\begin{bmatrix} 15&3 \ 12&2 \end{bmatrix}$

Answer

The matrix is square, so it could have an inverse. Next you need to find the determinant which is -6. Swap the numbers in spots a and d and put a negative in front of the numbers in spots b and c.

$\begin{bmatrix} 2&-3 \ -12& 15 \end{bmatrix}$ then divide each number by the determinant and simplify.

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Question

Determine the inverse of matrix A where

$A=\begin{bmatrix} 25&65 \ 40& 104 \end{bmatrix}$

Answer

The matrix is square, so it could have an inverse. Next, calculate the determinant. The determinant for this matrix is 0 so it does not have an inverse.

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Question

Determine the inverse of matrix A where

$A=\begin{bmatrix} 2&3 &1 \ 5&4 &8 \ 2& 1&5 \end{bmatrix}$

Answer

A matrix must be square to have an inverse. This matrix is square, so it could have an inverse. Next the determinant of the matrix must not be 0 to have an inverse. The determinant of this matrix is -6, so it has an inverse. To find the inverse of a 3x3 matrix, first write it in augmented form.

$\begin{bmatrix} 2&3 &1 &1 &0 &0 \ 5&4 &8 &0 &1 &0 \ 2&1 &5 &0 &0 &1 \end{bmatrix}$

Next find the pivot in the first column by dividing R1/2.

$\begin{bmatrix} 1&3/2 &1/2 &1/2 &0 &0 \ 5&4 &8 &0 &1 &0 \ 2&1 &5 &0 &0 &1 \end{bmatrix}$

Next, eliminate the first column by subtracting R2-5R1 and R3-2R1

$\begin{bmatrix} 1&3/2 &1/2 &1/2 &0 &0 \ 0&-7/2 &11/2 &-5/2 &1 &0 \ 0&-2 &4 &-1 &0 &1 \end{bmatrix}$

Next, find the pivot in column 2 by dividing -7R2/2

$\begin{bmatrix} 1&3/2 &1/2 &1/2 &0 &0 \ 0&1 &-11/7 &5/7 &-2/7 &0 \ 0&-2 &4 &-1 &0 &1 \end{bmatrix}$

Next, eliminate the second column by subtracting R1-3R2/2 and R3+2R2

$\begin{bmatrix} 1&0 &20/7 &-4/7 &3/7 &0 \ 0&1 &-11/7 &5/7 &-2/7 &0 \ 0&0 &6/7 &3/7 &-4/7 &1 \end{bmatrix}$

Next find the pivot in column 3 by dividing 6R3/7

$\begin{bmatrix} 1&0 &20/7 &-4/7 &3/7 &0 \ 0&1 &-11/7 &5/7 &-2/7 &0 \ 0&0 &1 &1/2 &-2/3 &7/6 \end{bmatrix}$

Finally, Eliminate the third column by subtracting R1-20R3/7 and R2-

$\begin{bmatrix} 1&0 &0 &-2 &7/3 &-10/3 \ 0&1 &0 &3/2 &-4/3 &11/6 \ 0&0 &1 &1/2 &-2/3 &7/6 \end{bmatrix}$

Now that we have the identity matrix on the left side, the right side is our answer.

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Question

Determine the inverse of matrix A where

$A=\begin{bmatrix} 1&45 &41 \ 0&0 &12 \ 0&0 &20 \end{bmatrix}$

Answer

The matrix must be square to have an inverse. This matrix is square so it could have an inverse. A matrix must also have a non-zero determinant. The determinant for this matrix is zero, so it does not have an inverse.

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Question

True or false: A matrix with five rows and four columns has as its inverse a matrix with four rows and five columns.

Answer

Only a square matrix - a matrix with an equal number of rows and columns - has an inverse. Therefore, a matrix with five rows and four columns cannot even have an inverse.

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Question

True or false:

If a matrix with four rows and four columns has an inverse, then the inverse also has four rows and four columns.

Answer

The inverse of a square matrix - that is, a matrix with an equal number of rows and columns - if it exists, is equal in dimension to that matrix. Therefore, any inverse of a four-by-four matrix is itself a four-by-four matrix.

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Question

and are both singular two-by-two matrices.

True or false: must also be singular.

Answer

To prove a statement false, it suffices to find one case in which the statement does not hold. We show that

$A = \begin{bmatrix} 1& 0\ 0& 0 \end{bmatrix}$ and $B= \begin{bmatrix} 0& 0\ 0& 1 \end{bmatrix}$

provide a counterexample.

A matrix is singular - that is, without an inverse - if and only if its determinant is equal to zero. The determinant of a two-by-two matrix is equal to the product of its upper left to lower right entries minus that of its upper right to lower left entries, so:

$\det A = 1 \cdot 0 - 0 \cdot 0 = 0 - 0 = 0$

$\det B = 0 \cdot 1 - 0 \cdot 0 = 0 - 0 = 0$

Both and are singular.

Now add the matrices by adding them term by term.

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

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

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

This is simply the two-by-two identity, which has an inverse - namely, itself.

The statement has been proved false by counterexample.

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