The Inverse - Linear Algebra

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Question

True or False: If , are square and invertible matrices then is also invertible.

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Answer

To prove is invertible, we need to find another square matrix such that .

Since $A^{-1},B^{-1}$ exist, take $D = B^{-1}A^{-1}$ , then we have

$D(AB)=(B^{-1}A^{-1})AB = B^{-1}(A^{-1}A)B = B^{-1}B = I$ ,

and

$(AB)D = (AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AA^{-1} = I$ .

Hence is invertible.

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Question

Suppose that is an invertible matrix. Simplify $AAA^{-1}AAA^{-1}A^{-1}A$ .

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Answer

To simplify

$AAA^{-1}AAA^{-1}A^{-1}A$

we used the identities:

$AA^{-1}=A^{-1}A=I$

so we get

$AAA^{-1}AAA^{-1}A^{-1}A=A(AA^{-1})A(AA^{-1})(A^{-1}A)=AIAII=AA=A^2$

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Question

Suppose that are all invertible. What is the inverse of ?

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Answer

The inverse of is $C^{-1}B^{-1}A^{-1}$ since we can multiply it by to get:

$C^{-1}B^{-1}A^{-1}ABC=C^{-1}B^{-1}(A^{-1}A)BC=C^{-1}B^{-1}IBC$

$=C^{-1}B^{-1}BC=C^{-1}(B^{-1}B)C=C^{-1}C=I$

Therefore $C^{-1}B^{-1}A^{-1}$ is the inverse of

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Question

Determine the inverse of matrix A where

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

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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

and are both singular two-by-two matrices.

True or false: must also be singular.

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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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Question

and are both two-by-two matrices. has an inverse.

True or false: Both and have inverses.

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Answer

A matrix is nonsingular - that is, it has an inverse - if and only if its determinant is nonzero. Also, the determinant of the product of two matrices is equal to the product of their individual determinants. Combining these ideas:

$\det AB = \det A \cdot \det B$

If either $\det A = 0$ or $\det B = 0$ , then it must hold that

$\det AB = 0 \cdot 0 = 0$ .

Equivalently, if either or has no inverse, then has no inverse. Contrapositively, if has an inverse, it must hold that each of and has an inverse.

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Question

and are both nonsingular two-by-two matrices.

True or false: must also be nonsingular.

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Answer

We can prove that the sum of two nonsingular matrices need not be nonsingular by counterexample.

Let $A = \begin{bmatrix} 1& 0\ 0& 1 \end{bmatrix}$ , $B= \begin{bmatrix} -1& 0\ 0& -1 \end{bmatrix}$ .

A matrix is nonsingular - that is, with an inverse - if and only if its determinant is nonzero. 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:

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

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

Both and are nonsingular.

Now add the matrices by adding them term by term.

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

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

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

the zero matrix, whose determinant is 0 and which is therefore not nonsingular.

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Question

is a singular four-by-four matrix. True or false: must also be a singular matrix.

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Answer

A matrix is singular - that is, it has no inverse - if and only if its determinant is equal to 0. is singular, so

$\det R = 0$ .

The determinant of the scalar product of and an $n \times n$ matrix is

$\det (cR) = c^{n}(\det R)$ ;

setting , , $\det R = 0$ :

$\det (5R) = 5^{4 } \cdot 0$

$\det (5R) = 0$

Therefore, , having determinant 0, is also singular.

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Question

is a nonsingular matrix.

True or false: the inverse of the matrix is $\frac{1}{3}R^{-1}$ .

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Answer

By definition,

$R R^{-1} = I$ and $R^{-1}R = I$ .

Multiply:

$3R \cdot \frac{1}{3}R^{-1} = 3 \cdot \frac{1}{3} \cdot RR^{-1} = 1 \cdot I = I$

Similarly,

$\frac{1}{3}R^{-1}\cdot 3R = \frac{1}{3} \cdot 3 \cdot R^{-1} R= 1 \cdot I = I$

Therefore, $\frac{1}{3}R^{-1}$ is the inverse of .

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Question

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

Find $A^{-1}$ .

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Answer

To find the inverse of a matrix , set up an augmented matrix , as shown below:

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

Perform row operations on this matrix until it is in reduced row-echelon form.

The following operations are arguably the easiest:

$-R1+ R2 \rightarrow R2$

$\begin{bmatrix} 1 & 1 + i &0 &1 & 0 & 0 \ 0 & -2i & 0 & -1 & 1 & 0 \ 0 & 1 & i & 0 & 0 & 1 \end{bmatrix}$

$R2 \leftrightarrow R3$

$\begin{bmatrix} 1 & 1 + i &0 &1 & 0 & 0\ 0 & 1 & i & 0 & 0 & 1 \ 0 & -2i & 0 & -1 & 1 & 0 \end{bmatrix}$

$(-1 - i )R2 + R1 \rightarrow R1$

$(2i)R2 + R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0&1-i &1 & 0 & -1-i\ 0 & 1 & i & 0 & 0 & 1 \ 0 & 0 & -2 & -1 & 1 & 2i \end{bmatrix}$

$-\frac{1}{2} R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0&1-i &1 & 0 & -1-i\ 0 & 1 & i & 0 & 0 & 1 \ 0 & 0 &1 & \frac{1}{2} & - \frac{1}{2} & -i \end{bmatrix}$

$(-1+i)R3 + R1 \rightarrow R1$

$(- i)R3 + R2 \rightarrow R2$

$\begin{bmatrix} 1 & 0&0& \frac{1}{2}+\frac{1}{2}i & \frac{1}{2}-\frac{1}{2}i & 0\ 0 & 1 & 0& -\frac{1}{2} i & \frac{1}{2} i & 0 \ 0 & 0 &1 & \frac{1}{2} & - \frac{1}{2} & -i \end{bmatrix}$

The augmented matrix is in reduced row-echelon form $[I |A^{-1}]$ . The inverse is therefore

$A ^{-1} = \begin{bmatrix} \frac{1}{2}+\frac{1}{2}i & \frac{1}{2}-\frac{1}{2}i & 0\ -\frac{1}{2} i & \frac{1}{2} i & 0 \ \frac{1}{2} & - \frac{1}{2} & -i \end{bmatrix}$ .

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Question

Determine the inverse of matrix A where

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

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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

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

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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

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

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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

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

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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

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

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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.

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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.

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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

Calculate $A^{-1}$ , where

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

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

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

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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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