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Linear Algebra : The Inverse

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Example Question #21 : The Inverse

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

True or false: Both and have inverses.

Possible Answers:

False

True

Correct answer:

True

Explanation:

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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Example Question #22 : The Inverse

and are both nonsingular two-by-two matrices.

True or false: A + B must also be nonsingular.

Possible Answers:

True

False

Correct answer:

False

Explanation:

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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Example Question #23 : The Inverse

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

Possible Answers:

True

False

Correct answer:

True

Explanation:

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 c = 5 , n = 4 , $\det R = 0$ :

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

$\det (5R) = 0$

Therefore, , having determinant 0, is also singular.

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Example Question #21 : The Inverse

is a nonsingular matrix.

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

Possible Answers:

True

False

Correct answer:

True

Explanation:

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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Example Question #25 : The Inverse

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

Possible Answers:

True

False

Correct answer:

True

Explanation:

To prove is invertible, we need to find another square matrix such that D(AB)=(AB)D = I .

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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Example Question #26 : The Inverse

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

Possible Answers:

A^2

$A^{-1}$

A^3

Correct answer:

A^2

Explanation:

To simplify

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

we used the identities:

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

AI=IA=A

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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Example Question #27 : The Inverse

Suppose that A,B,C are all invertible. What is the inverse of ABC ?

Possible Answers:

$A^{-1}B^{-1}C^{-1}$

$C^{-1}A^{-1}A^{-1}$

$C^{-1}B^{-1}A^{-1}$

$C^{-1}A^{-1}B^{-1}$

Correct answer:

$C^{-1}B^{-1}A^{-1}$

Explanation:

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

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Example Question #21 : The Inverse

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

Find $A^{-1}$ .

Possible Answers:

$A ^{-1}= \begin{bmatrix} \frac{1}{2} &\frac{1}{2} \\ \\ \frac{1}{4} - \frac{1}{4} & -\frac{1}{4} - \frac{1}{4} i\end{bmatrix}$

does not have an inverse.

$A ^{-1}= \begin{bmatrix} \frac{1}{2} & \frac{1}{4} - \frac{1}{4} i \\ \\ \frac{1}{2} & -\frac{1}{4} - \frac{1}{4} i\end{bmatrix}$

$A ^{-1}= \begin{bmatrix} -\frac{1}{2}i & \frac{1}{4} + \frac{1}{4} i \\ \\- \frac{1}{2} i& \frac{1}{4} - \frac{1}{4} i\end{bmatrix}$

$A ^{-1}= \begin{bmatrix} -\frac{1}{2}i & - \frac{1}{2} \\ \\ \frac{1}{4} + \frac{1}{4} i & \frac{1}{4} - \frac{1}{4} i\end{bmatrix}$

Correct answer:

$A ^{-1}= \begin{bmatrix} \frac{1}{2} & \frac{1}{4} - \frac{1}{4} i \\ \\ \frac{1}{2} & -\frac{1}{4} - \frac{1}{4} i\end{bmatrix}$

Explanation:

The inverse of a two-by-two matrix

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

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

Substituting the entries in the matrix for the variables:

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

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

$A ^{-1}=\frac{1}{-4} \begin{bmatrix} -2 & -1 + i \\- 2 & 1 + i \end{bmatrix}$

$A ^{-1}= \begin{bmatrix} \frac{1}{2} & \frac{1}{4} - \frac{1}{4} i \\ \\ \frac{1}{2} & -\frac{1}{4} - \frac{1}{4} i\end{bmatrix}$

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Example Question #29 : The Inverse

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

Find $A^{-1}$ .

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To find the inverse of a matrix , set up an augmented matrix [A | I ] , 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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Example Question #30 : The Inverse

$A = \begin{bmatrix} 2+ i & 3 - i & i \\ -4 & 1+ i & 3 \end{bmatrix}$ .

Calculate $A^{-1}$ .

Possible Answers:

$\begin{bmatrix} \frac{2}{7}- \frac{1}{7}i &\frac{ 3}{7} +\frac{1}{7} i & -\frac{1}{7}i \\ \\ \frac{4}{7} & \frac{1}{7}- \frac{1}{7}i &- \frac{3}{7} \end{bmatrix}$

$\begin{bmatrix} \frac{2}{11}- \frac{1}{11}i & \frac{4}{11} \\ \\ \frac{ 3}{11} +\frac{1}{11} i & \frac{1}{11}- \frac{1}{11}i \\ \\ -\frac{1}{11}i & - \frac{3}{11} \end{bmatrix}$

$A^{-1}$ is undefined.

$\begin{bmatrix} \frac{2}{11}- \frac{1}{11}i &\frac{ 3}{11} +\frac{1}{11} i & -\frac{1}{11}i \\ \\ \frac{4}{11} & \frac{1}{11}- \frac{1}{11}i &- \frac{3}{11} \end{bmatrix}$

$\begin{bmatrix} \frac{2}{7}- \frac{1}{7}i & \frac{4}{7} \\ \\ \frac{ 3}{7} +\frac{1}{7} i & \frac{1}{7}- \frac{1}{7}i \\ \\ -\frac{1}{7}i & - \frac{3}{7} \end{bmatrix}$

Correct answer:

$A^{-1}$ is undefined.

Explanation:

The matrix is not a square matrix - it has two rows and three columns - so it does not have an inverse.

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Linear Algebra : The Inverse

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #21 : The Inverse

Example Question #22 : The Inverse

Example Question #23 : The Inverse

Example Question #21 : The Inverse

Example Question #25 : The Inverse

Example Question #26 : The Inverse

Example Question #27 : The Inverse

Example Question #21 : The Inverse

Example Question #29 : The Inverse

Example Question #30 : The Inverse

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