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Linear Algebra : Matrices

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Example Questions

Example Question #91 : Matrix Matrix Product

is a $4 \times 3$ matrix.

is a $5 \times 6$ matrix.

is a $3 \times 8$ matrix.

is a $6 \times 4$ matrix.

is an $8 \times 5$ matrix.

Which of the following is a square matrix?

Possible Answers:

EBDAC

$A^{T} D^{T}B^{T}E^{T}C^{T}$

$B^{T}E^{T}C^{T}A^{T}D^{T}$

All four of the expressions given in the other choices are defined.

ACEBD

Correct answer:

All four of the expressions given in the other choices are defined.

Explanation:

Two matrices can be multiplied if and only if the number of columns in the first is equal to the number of columns in the second. The number of rows in the first matrix and the number of columns in the second are the number of rows and columns in the product, respectively.

For example, the following matrices can be put together in order:

is a $4 \times 3$ matrix; is a $3 \times 8$ matrix; is defined and is a $4 \times 8$ matrix.

is a $3 \times 8$ matrix; is an $8 \times 5$ matrix; is defined and is a $3 \times 5$ matrix.

By extension, matrices can be linked in a product of three or more such that, if two matrices appear together, this same relation must hold. For example, since is a $4 \times 3$ matrix, is a $3 \times 8$ matrix, and is an $8 \times 5$ matrix, ACE is defined and is a $4 \times 5$ matrix.

It follows by further extension that ACEBD is defined and is a $4 \times 4$ matrix, and that EBDAC is defined and is an $8 \times 8$ matrix.

DACEB is defined to be a $6 \times 6$ matrix. The transpose of the product of matrices is equal to the product of transposes in reverse, so

$B^{T}E^{T}C^{T}A^{T}D^{T} = (DACEB)^{T}$ , a $6 \times 6$ matrix.

Similarly, $A^{T} D^{T}B^{T}E^{T}C^{T} = (CEBDA)^{T}$ , a $3 \times 3$ matrix.

The correct response is that all four given matrices are square.

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Example Question #91 : Matrices

is a singular $4 \times 4$ matrix.

is a nonsingular $5 \times 5$ matrix.

is a $4 \times 5$ matrix.

All of the following are undefined except:

Possible Answers:

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

$ABC^{T}$

$BC^{-1}A$

$A^{-1} C B$

None of the expressions given in the other choices are defined.

Correct answer:

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

Explanation:

$BC^{-1}A$ can be eliminated as a choice; is not a square matrix, so the inverse of , $C^{-1}$ , does not exist.

$A^{-1} C B$ can also be eliminated; is a singular matrix, so, by definition, $A^{-1}$ does not exist.

Two matrices can be multiplied if and only if the number of columns in the first is equal to the number of columns in the second. $ABC^{T}$ can be eliminated, since has four columns and has five rows.

The remaining choice is $B^{-1}C^{T}A$ . is nonsingular, so $B^{-1}$ is defined. $B^{-1}$ , like , is a $5 \times 5$ matrix; is a $4 \times 5$ matrix, so its transpose $C^{T}$ is $5 \times 4$ ; thus, $B^{-1}C^{T}$ is $5 \times 4$ . is $4 \times 4$ , so $B^{-1}C^{T}A = (B^{-1}C^{T})A$ is $5 \times 4$ . $B^{-1}C^{T}A$ is defined and is the correct choice.

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Example Question #91 : Matrices

Let $A = \begin{bmatrix} 2\sqrt{2} & \sqrt{3} \end{bmatrix}$ and $B = \begin{bmatrix} \sqrt{2} & \sqrt{6} \\-2 \sqrt{6} & \sqrt{3} \end{bmatrix}$ .

Find .

Possible Answers:

$\begin{bmatrix} 7 & 6 \sqrt{2}+ 4\sqrt{3} \end{bmatrix}$

$\begin{bmatrix} 4-6 \sqrt{2} & 3+ 4 \sqrt{3}\end{bmatrix}$

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

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

is undefined.

Correct answer:

$\begin{bmatrix} 4-6 \sqrt{2} & 3+ 4 \sqrt{3}\end{bmatrix}$

Explanation:

First, it must be established that is defined. This is the case if and only if has as many columns as has rows. Since has two columns and has two rows, is defined.

Matrices are multiplied by multiplying each row of the first matrix by each column of the second - that is, by adding the products of the entries in corresponding positions. Thus,

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

$= \begin{bmatrix} 2\sqrt{2} (\sqrt{2} )+ \sqrt{3}(-2 \sqrt{6} ) & 2\sqrt{2} (\sqrt{6} )+ \sqrt{3}( \sqrt{3} )\end{bmatrix}$

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

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Example Question #92 : Matrices

Let $A = \begin{bmatrix} \frac{2}{5} & \frac{1}{4}\end{bmatrix}$ and $B = \begin{bmatrix} \frac{3}{2} & \frac{1}{4} \\ \\ \frac{3}{4} &- \frac{1}{5}\end{bmatrix}$ .

Find .

Possible Answers:

$\begin{bmatrix} \frac{53}{80} & \frac{1}{4} \end{bmatrix}$

is undefined.

$\begin{bmatrix} \frac{63}{80} & \frac{1}{20} \end{bmatrix}$

$\begin{bmatrix} \frac{53}{80} \\ \\ \frac{1}{4} \end{bmatrix}$

$\begin{bmatrix} \frac{63}{80} \\ \\ \frac{1}{20} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \frac{63}{80} & \frac{1}{20} \end{bmatrix}$

Explanation:

First, it must be established that is defined. This is the case if and only if has as many columns as has rows. Since has two columns and has two rows, is defined.

Matrices are multiplied by multiplying each row of the first matrix by each column of the second - that is, by adding the products of the entries in corresponding positions. Thus,

$AB= \begin{bmatrix} \frac{2}{5} & \frac{1}{4}\end{bmatrix} \begin{bmatrix} \frac{3}{2} & \frac{1}{4} \\ \\ \frac{3}{4} &- \frac{1}{5}\end{bmatrix}$ .

$= \begin{bmatrix} \frac{2}{5}\left ( \frac{3}{2} \right ) + \frac{1}{4} \left ( \frac{3}{4} \right )& \frac{2}{5}\left ( \frac{1}{4} \right ) + \frac{1}{4} \left ( -\frac{1}{5} \right ) \end{bmatrix}$

$= \begin{bmatrix} \frac{3}{5} + \frac{3}{16} & \frac{1}{10} - \frac{1}{20} \end{bmatrix}$

$= \begin{bmatrix} \frac{63}{80} & \frac{1}{20} \end{bmatrix}$

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Example Question #93 : Matrices

Let $A= \begin{bmatrix} \frac{3}{2} & \frac{1}{4} \\ \\ \frac{3}{4} &- \frac{1}{5}\end{bmatrix}$ and $B = \begin{bmatrix} \frac{2}{5} & \frac{1}{4}\end{bmatrix}$ .

Find .

Possible Answers:

$\begin{bmatrix} \frac{53}{80} & \frac{1}{4} \end{bmatrix}$

is undefined.

$\begin{bmatrix} \frac{63}{80} & \frac{1}{20} \end{bmatrix}$

$\begin{bmatrix} \frac{63}{80} \\ \\ \frac{1}{20} \end{bmatrix}$

$\begin{bmatrix} \frac{53}{80} \\ \\ \frac{1}{4} \end{bmatrix}$

Correct answer:

is undefined.

Explanation:

First, it must be established that is defined. This is the case if and only if has as many columns as has rows. Since has two columns and has one row, is not defined.

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Example Question #94 : Matrices

$A=\begin{bmatrix} a & 1 \\ 1 & 1\end{bmatrix}$ and $B = \begin{bmatrix} b & 1 \\ 1 & 1 \end{bmatrix}$ , where and stand for real quantities.

Which of the following must be a true statement?

Possible Answers:

$BA = -(AB)^{T}$

BA = AB

None of the statements given in the other choices are correct.

BA = -(AB)

$BA = (AB)^{T}$

Correct answer:

$BA = (AB)^{T}$

Explanation:

Matrices are multiplied by multiplying each row of the first matrix by each column of the second - that is, by adding the products of the entries in corresponding positions. Thus,

$AB=\begin{bmatrix} a & 1 \\ 1 & 1\end{bmatrix}\begin{bmatrix} b & 1 \\ 1 & 1 \end{bmatrix}$

$= \begin{bmatrix} a(b)+ 1(1) & a(1)+1(1) \\ 1(b)+ 1(1) & 1(1)+1(1) \end{bmatrix}$

$= \begin{bmatrix} ab+1 &a+1 \\ b+1 & 2 \end{bmatrix}$

$BA=\begin{bmatrix} b & 1 \\ 1 & 1\end{bmatrix}\begin{bmatrix}a & 1 \\ 1 & 1 \end{bmatrix}$

$= \begin{bmatrix} b(a)+ 1(1) & b(1)+1(1) \\ 1(a)+ 1(1) & 1(1)+1(1) \end{bmatrix}$

$= \begin{bmatrix} ab+1 &b+1 \\ a+1 & 2 \end{bmatrix}$

If rows and columns are transposed in , it can be seen that

$(AB)^{T}= \begin{bmatrix} ab+1 &b+1 \\ a+1 & 2 \end{bmatrix} = BA$ .

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Example Question #95 : Matrices

$A=\begin{bmatrix}1 & a \\-a & 1\end{bmatrix}$ and $B = \begin{bmatrix} 1 & b \\ -b& 1 \end{bmatrix}$ , where and stand for real quantities.

Which of the following must be a true statement?

Possible Answers:

None of the statements given in the other choices are correct.

$BA = -(AB)^{T}$

BA = -AB

$BA = (AB)^{T}$

BA = AB

Correct answer:

BA = AB

Explanation:

Matrices are multiplied by multiplying each row of the first matrix by each column of the second - that is, by adding the products of the entries in corresponding positions. Thus,

$AB=\begin{bmatrix}1 & a \\-a & 1\end{bmatrix} \begin{bmatrix} 1 & b \\ -b& 1 \end{bmatrix}$

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

$=\begin{bmatrix}-ab+1&a+b\\-a -b& -ab+1 \end{bmatrix}$

$BA=\begin{bmatrix}1 & b \\-b & 1\end{bmatrix} \begin{bmatrix} 1 & a \\ -a& 1 \end{bmatrix}$

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

$=\begin{bmatrix}-ab+1&a+b\\-a -b& -ab+1 \end{bmatrix}$

= AB

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Example Question #96 : Matrices

$A^{2}$ is a symmetric matrix.

True or false: It follows that is also a symmetric matrix.

Possible Answers:

False

True

Correct answer:

False

Explanation:

This statement can be proved by counterexample.

Let $A= \begin{bmatrix} 6 & -5\\ 7& -6 \end{bmatrix}$ .

is not symmetric, since its transpose,

$A^{T}= \begin{bmatrix} 6 & 7\\ -5& -6 \end{bmatrix}$

is not equal to .

Then

$A^{2}= \begin{bmatrix} 6 & -5\\ 7& -6 \end{bmatrix}\begin{bmatrix} 6 & -5\\ 7& -6 \end{bmatrix}$

Matrices are multiplied by multiplying each row of the first matrix by each column of the second - that is, by adding the products of the entries in corresponding positions. Thus,

$A^{2} = \begin{bmatrix} 6(6)+(-5)(7) &6(-5)+(-5)(-6) \\ 7(6)+(-6)(7) &7(-5)+(-6)(-6) \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ .

Therefore, $A^{2}= I$ , the two-by-two identity, which is symmetric.

Since a nonsymmetric matrix exists whose square is symmetric, then the given statement is false.

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Example Question #97 : Matrices

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

Calculate $A^{20 }$ .

Possible Answers:

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

$\begin{bmatrix} i & 0 & 0 \\ 0 &0 &i\end{bmatrix}$

$A^{20 }$ is undefined.

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

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

Correct answer:

$A^{20 }$ is undefined.

Explanation:

Only square matrices can be taken to any power. Since is not a square matrix, having two rows and three columns, $A^{20 }$ is undefined.

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Example Question #98 : Matrices

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

Calculate $(A^{T}A)^{77}$ .

Possible Answers:

None of the other choices gives the correct response.

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

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

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

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

Correct answer:

None of the other choices gives the correct response.

Explanation:

$A^{T}$ , the transpose of , can be found by transposing rows with columns.

$A= \begin{bmatrix} i & 0 & 0 \\ 0 &0 &-i\end{bmatrix}$ , so

$A^{T}= \begin{bmatrix} i & 0 \\0& 0 \\ 0 &-i\end{bmatrix}$

Matrices are multiplied by multiplying each row of the first matrix by each column of the second - that is, by adding the products of the entries in corresponding positions. Thus,

$A^{T}A= \begin{bmatrix} i & 0 \\0& 0 \\ 0 &-i\end{bmatrix} \begin{bmatrix} i & 0 & 0 \\ 0 &0 &-i\end{bmatrix}$

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

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

This is a square matrix, so it can be raised to a power. To raise a diagonal matrix to a power, simply raise each number in the main diagonal to that power:

$(A^{T}A)^{77} =\begin{bmatrix} (-1)^{77} & 0 & 0 \\ 0 &0^{77} & 0 \\ 0 & 0 & (-1)^{77} \end{bmatrix}=\begin{bmatrix} -1 & 0 & 0 \\ 0 &0 & 0 \\ 0 & 0 & -1 \end{bmatrix}$ .

This is not among the choices.

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Linear Algebra : Matrices

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #91 : Matrix Matrix Product

Example Question #91 : Matrices

Example Question #91 : Matrices

Example Question #92 : Matrices

Example Question #93 : Matrices

Example Question #94 : Matrices

Example Question #95 : Matrices

Example Question #96 : Matrices

Example Question #97 : Matrices

Example Question #98 : Matrices

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