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Linear Algebra : Operations and Properties

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Example Questions

Example Question #251 : Operations And Properties

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

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 #24 : The Inverse

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

Calculate $A^{-1}$ .

Possible Answers:

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

$A^{-1}$ is undefined.

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

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

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

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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Example Question #31 : The Inverse

is a singular matrix; is a nonsingular matrix.

Which is true of A B ?

Possible Answers:

A B must be a nonsingular matrix.

A B must be a singular matrix.

A B may be singular or nonsingular.

Correct answer:

A B must be a singular matrix.

Explanation:

A matrix is singular - that is, without an inverse - if and only if its determinant is 0. Since is a singular matrix and is a nonsingular matrix, $\det A = 0$ . The product of two matrices has as its determinant the product of the individual determinants, so

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

must be singular.

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Example Question #251 : Operations And Properties

A matrix has as its determinant $\cos \frac{2\pi}{17}$ .

True or false: the determinant of $A^{-1}$ is $\csc \frac{2\pi}{17}$ .

Possible Answers:

False

True

Correct answer:

False

Explanation:

The determinant of the inverse $A^{-1}$ of matrix is the reciprocal of the determinant of , so

$\det A^{-1} = \frac{1}{\det A } = \frac{1}{\cos \frac{2 \pi}{17} } = \sec \frac{2 \pi}{17}$ .

The statement is false.

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Example Question #255 : Operations And Properties

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

For which of the following values of is a singular matrix?

Possible Answers:

is singular regardless of the value of .

$a = \ln 2$

a = 1

a= 0

$a = \log 2$

Correct answer:

is singular regardless of the value of .

Explanation:

is singular - that is, it has no inverse - if and only if its determinant is equal to 0. the determinant of a $2 \times 2$ matrix

$\begin{bmatrix} a_{11} & a_{12 } \\ a_{21} & a_{22 } \end{bmatrix}$

is $\det A = a_{11} a_{22} - a_{12} a_{21}$

Substitute appropriately:

$\det A = 2^{a} \cdot 2^{-a}- 1 \cdot 1$

$= 2^{a -a}- 1$

$= 2^{0}- 1$

= 1 - 1

= 0

Regardless of the value of , has determinant 0 and is a singular matrix.

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Example Question #33 : The Inverse

$A = \begin{bmatrix} a & 3 \\ 9 & -6 \end{bmatrix}$

Give so that is a singular matrix.

Possible Answers:

a = -4.5

a=0

a=18

a=-18

a = 4.5

Correct answer:

a = -4.5

Explanation:

$A = \begin{bmatrix} a & 3 \\ 9 & -6 \end{bmatrix}$

is a singular matrix - one without an inverse - if and only if its determinant is equal to 0. The determinant of a $2 \times 2$ matrix

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

is equal to

$\det A = ad - bc$

Substituting appropriately and setting this quantity equal to 0, solve for :

a(-6)-3(9) = 0

-6a -27 = 0

-6a = 27

a = -4.5

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Example Question #252 : Operations And Properties

A matrix has as its determinant $\cot \frac{13\pi}{15}$ .

True or false: $A^{-1 }$ has as its determinant $\tan \frac{13\pi}{15}$ .

Possible Answers:

False

True

Correct answer:

True

Explanation:

The determinant of the inverse $A^{-1}$ of matrix is the reciprocal of the determinant of , so

$\det A^{-1} = \frac{1}{\det A } = \frac{1}{\cot \frac{13 \pi}{15} } = \tan \frac{13 \pi}{15}$ .

The statement is true.

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Example Question #33 : The Inverse

$A = \begin{bmatrix} a &b \\ c & d \end{bmatrix}$ such that $\det A = 4$ .

Which of the following is equal to $A^{-1}$ ?

Possible Answers:

$\begin{bmatrix} \frac{d }{4}&- \frac{c}{4} \\ \\ - \frac{b}{4} & \frac{a}{4} \end{bmatrix}$

$\begin{bmatrix} 4 d &-4 b \\ -4 c & 4a \end{bmatrix}$

$\begin{bmatrix} \frac{a }{4}&- \frac{b}{4} \\ \\ - \frac{d}{4} & \frac{c}{4} \end{bmatrix}$

$\begin{bmatrix} \frac{d }{4}&- \frac{b}{4} \\ \\ - \frac{c}{4} & \frac{a}{4} \end{bmatrix}$

$\begin{bmatrix} 4 a &-4 b \\ -4 c & 4d \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \frac{d }{4}&- \frac{b}{4} \\ \\ - \frac{c}{4} & \frac{a}{4} \end{bmatrix}$

Explanation:

The inverse of a matrix $A = \begin{bmatrix} a &b \\ c & d \end{bmatrix}$ , if it exists, is

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

Since $\det A = 4$ ,

$A^{-1} =\frac{1}{4} \begin{bmatrix} d&- b \\- c &a \end{bmatrix}$

$=\begin{bmatrix} \frac{d }{4}&- \frac{b}{4} \\ \\ - \frac{c}{4} & \frac{a}{4} \end{bmatrix}$

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Example Question #256 : Operations And Properties

$A = \begin{bmatrix} \cos \theta & \sin \theta \\ \cos (\frac{\pi}{2} - \theta )& \sin (\frac{\pi}{2}- \theta ) \end{bmatrix}$

For which of the following values of $\theta$ is a singular matrix?

Possible Answers:

$\theta = \frac{11 \pi}{3 }$

cannot be singular for any value of $\theta \;$ .

$\theta = \frac{11 \pi}{6}$

$\theta = \frac{11 \pi}{2 }$

$\theta = \frac{11 \pi}{4 }$

Correct answer:

$\theta = \frac{11 \pi}{4 }$

Explanation:

For any value of $\theta \;$ ,

$\cos\left ( \frac{\pi}{2} - \theta \right ) = \sin \theta$

and, equivalently,

$\sin \left ( \frac{\pi}{2} - \theta \right ) = \cos \theta$ .

can be rewritten as

$A = \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta &\cos \theta \end{bmatrix}$

is singular - that is, it has no inverse - if and only if its determinant is equal to 0. The determinant of a $2 \times 2$ matrix

$\begin{bmatrix} a_{11} & a_{12 } \\ a_{21} & a_{22 } \end{bmatrix}$

is $\det A = a_{11} a_{22} - a_{12} a_{21}$

Substitute appropriately:

$\det A = \cos \theta \cdot \cos \theta - \sin \theta \cdot \sin \theta$

$= \cos ^{2} \theta - \sin ^{2} \theta$

$= \cos 2 \theta$

by a trigonometric identity.

Therefore, is singular if and only if

$\cos 2 \theta = 0$ .

It follows that for to be singular,

$2 \theta = \frac{\pi}{2} + n \pi = \frac{\pi+ 2n \pi }{2}$ for some integer value .

or, equivalently,

$\theta = \frac{\pi+ 2n \pi }{4}$

This can be restated:

$\theta = \frac{ (2n +1) \pi }{4}$ , or

$\theta = \frac{N \pi }{4}$ for an odd positive or negative integer .

The only choice that fits this description is $\theta = \frac{11 \pi}{4 }$ , which is the correct choice.

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Example Question #333 : Linear Algebra

is an involutory matrix.

True or false: It follows that $A^{-1}$ is also an involutory matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

A matrix is involutory if $A = A^{-1}$ . Since and $A^{-1}$ are the same matrix, it immediately follows that $A^{-1}$ is involutory.

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Linear Algebra : Operations and Properties

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #251 : Operations And Properties

Example Question #24 : The Inverse

Example Question #31 : The Inverse

Example Question #251 : Operations And Properties

Example Question #255 : Operations And Properties

Example Question #33 : The Inverse

Example Question #252 : Operations And Properties

Example Question #33 : The Inverse

Example Question #256 : Operations And Properties

Example Question #333 : Linear Algebra

All Linear Algebra Resources

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