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Linear Algebra : Operations and Properties

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Example Questions

Example Question #14 : The Determinant

is a five-by-five matrix with determinant 100.

Give the determinant of $S^{T}$ .

Possible Answers:

100

More information is needed to find the determinant of .

$\frac{1}{100}$

$-\frac{1}{100}$

Correct answer:

100

Explanation:

The transpose of a square matrix has the same determinant as the original matrix, so

$\det S ^{T}= \det S = 100$ .

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Example Question #11 : The Determinant

Given: a matrix such that $\det K = \frac{3}{4}$ .

Give $\det( K^{T} ) ^{-1}$ .

Possible Answers:

More information is needed to answer the question.

$-\frac{4}{3}$

$\frac{3}{4}$

$\frac{4}{3}$

$-\frac{3}{4}$

Correct answer:

$\frac{4}{3}$

Explanation:

The determinant of the transpose of a matrix is equal to that of the original matrix; the determinant of the inverse of a matrix is equal to the reciprocal of that of the original matrix. Therefore,

$\det( K^{T} ) ^{-1} = (\det K^{T} ) ^{-1} = (\det K ) ^{-1} =\left ( \frac{3}{4} \right )^{-1}= \frac{4}{3}$ .

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Example Question #21 : The Determinant

Consider the matrix

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

Calculate the cofactor $C_{2,1}$ of this matrix.

Possible Answers:

$C_{2,1} = -5$

$C_{2,1} = 7$

$C_{2,1} = -7$

$C_{2,1} =5$

$C_{2,1} =0$

Correct answer:

$C_{2,1} = -7$

Explanation:

The cofactor $C_{m, n}$ of a matrix , by definition, is equal to

$C_{m, n} = (-1)^{m + n} M_{m, n}$ ,

where $M_{m, n}$ is the minor of the matrix - the determinant of the matrix formed when Row and Column of are struck out. Therefore, we first find the minor $M_{2,1}$ of the matrix by striking out Row 2 and Column 1 of , as shown in the diagram below:

Minor

The minor is therefore equal to

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

This is equal to the product of the upper-left and lower-right elements minus the product of the upper-right to lower-right elements, which is

$M_{2,1} = 3(2) - (-1)(1) = 6 - (-1) = 7$

Setting m = 2 and n = 1 in the definition of the cofactor, the formula becomes

$C_{2,1} = (-1)^{2 + 1} M_{2,1} = (-1)^{3} M_{2,1} = -1 \cdot M_{2,1}$ ,

$C_{2,1} = -1 \cdot 7 = -7$ .

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Example Question #22 : The Determinant

Consider the system of linear equations:

- 3x+ 7y = 22

2x+ 5z = 29

-y + 8z= 36

By Cramer's rule, which of the following expressions is equal to ?

Possible Answers:

$\frac{\begin{vmatrix} -3& 7& 0\\ 2& 0& 5\\ 0&-1 & 8 \end{vmatrix}}{\begin{vmatrix} 22& 7& 0\\ 29& 0& 5\\ 36 &-1 & 8 \end{vmatrix}}$

$\frac{\begin{vmatrix} -3& 7& 0\\ 2& 0& 5\\ 0&-1 & 8 \end{vmatrix}}{\begin{vmatrix} -3& 7& 22\\ 2& 29& 5\\ 36&-1 & 8 \end{vmatrix}}$

$\frac{\begin{vmatrix} -3& 7& 22\\ 2& 29& 5\\ 36&-1 & 8 \end{vmatrix}}{\begin{vmatrix} -3& 7& 0\\ 2& 0& 5\\ 0&-1 & 8 \end{vmatrix}}$

$\frac{\begin{vmatrix} 22& 7& 0\\ 29& 0& 5\\ 36 &-1 & 8 \end{vmatrix}}{\begin{vmatrix} -3& 7& 0\\ 2& 0& 5\\ 0&-1 & 8 \end{vmatrix}}$

$\frac{\begin{vmatrix} -3& 7& 22\\ 2& 5& 29\\ -1 & 8 & 36\end{vmatrix}}{\begin{vmatrix} -3& 7& 0\\ 2& 0& 5\\ 0&-1 & 8 \end{vmatrix}}$

Correct answer:

$\frac{\begin{vmatrix} 22& 7& 0\\ 29& 0& 5\\ 36 &-1 & 8 \end{vmatrix}}{\begin{vmatrix} -3& 7& 0\\ 2& 0& 5\\ 0&-1 & 8 \end{vmatrix}}$

Explanation:

It will help if we rewrite the system so that the "missing" term in each equation is replaced with a term with a coefficient of zero:

- 3x+ 7y + 0z= 22

2x+ 0y+ 5z = 29

0x-y + 8z= 36

By Cramer's Rule, is equal to the fraction of the determinants of two matrices. The denominator is the determinant of the matrix of coefficients:

$\begin{vmatrix} -3& 7& 0\\ 2& 0& 5\\ 0&-1 & 8 \end{vmatrix}$

The numerator is the determinant of the matrix formed by replacing the first column (the -coefficients) by the constants at the right of the equality symbols:

$\begin{vmatrix} 22& 7& 0\\ 29& 0& 5\\ 36 &-1 & 8 \end{vmatrix}$

The correct choice is therefore the expression

$\frac{\begin{vmatrix} 22& 7& 0\\ 29& 0& 5\\ 36 &-1 & 8 \end{vmatrix}}{\begin{vmatrix} -3& 7& 0\\ 2& 0& 5\\ 0&-1 & 8 \end{vmatrix}}$ .

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Example Question #23 : The Determinant

and are both matrices with nonzero determinant .

Evaluate $\det (x^{T}Y^{-1})$ . If applicable, give the determinant in terms of .

Possible Answers:

$\frac{1}{n^{2}}$

$n^{2}$

Correct answer:

Explanation:

The determinant of the transpose of a matrix is equal to that of the original matrix; therefore, $\det X^{T} = \det X = n$ .

The determinant of the inverse of a matrix is equal to the reciprocal of that of the determinant of the original matrix; therefore, $\det Y^{-1} = \frac{1}{\det Y} = \frac{1}{n}$ .

The determinant of the product of two matrices is equal to the product of the determinants of the individual matrices, so

$\det (x^{T}Y^{-1})= \det X^{T} \cdot \det Y^{-1} = n \cdot \frac{1}{n} = 1$ .

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Example Question #24 : The Determinant

Consider the matrix

$A = \begin{bmatrix} -1&4 &1 \\ -2& 3& 1\\ 1&6 & 2 \end{bmatrix}$ .

Give cofactor $C_{2 3}$ of this matrix.

Possible Answers:

$C_{23} = 0$

$C_{23} = -10$

$C_{23} = 10$

$C_{23} = 2$

$C_{23} = -2$

Correct answer:

$C_{23} = 10$

Explanation:

The minor of a matrix $M_{mn}$ is the determinant of the matrix formed by striking out Row and Column . By definition, the corresponding cofactor $C_{mn}$ can be calculated from this minor using the formula

$C_{mn}= ( -1 )^{m+n} M_{mn}$

To find cofactor $C_{2 3}$ , we first find minor $M_{23}$ by striking out Row 2 and Column 3, as follows:

Minor

$M_{23}$ is equal to the determinant

$\begin{vmatrix} -1& 4\\ 1& 6 \end{vmatrix}$

which is found by taking the product of the upper left and lower right entries and subtracting that of the other two entries:

$M_{23} = -1(6)- 1(4) = -6 - 4 = -10$

In the cofactor equation, set m = 2, n = 3 :

$C_{23}= ( -1 )^{2+3} M_{23}= ( -1 )^{5} (-10) = (-1)(-10) = 10$

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Example Question #25 : The Determinant

Consider the matrix

$A = \begin{bmatrix} -1&4 &1 \\ -2& 3& 1\\ 1&6 & 2 \end{bmatrix}$ .

Give cofactor $C_{33}$ of this matrix.

Possible Answers:

$C_{33}=- 5$

$C_{33}= 11$

$C_{33}= 5$

$C_{33}= -11$

$C_{33}= 0$

Correct answer:

$C_{33}= 5$

Explanation:

$C_{mn}= ( -1 )^{m+n} M_{mn}$

To find cofactor $C_{33}$ , we first find minor $M_{33}$ by striking out Row 3 and Column 3, as follows:

Minor

$M_{33}$ is equal to the determinant

$\begin{vmatrix} -1& 4\\-2& 3 \end{vmatrix}$

which is found by taking the product of the upper left and lower right entries and subtracting that of the other two entries:

$M_{33} = (-1)(3)- 4 (-2) = -3 -(-8)= 5$

In the cofactor equation, set m = 3, n = 3 :

$C_{33}= ( -1 )^{3+3 } M_{33} = ( -1 )^{6 } \cdot 5 = 1 \cdot 5 = 5$

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Example Question #26 : The Determinant

The determinant of a three-by-three matrix is 3. Give the determinant of $\frac{1}{3}Z$ .

Possible Answers:

$\frac{1}{9}$

$\sqrt[3]{3}$

$\sqrt{3}$

Correct answer:

$\frac{1}{9}$

Explanation:

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

$\det (cZ) = c^{n}(\det Z)$ .

Set $\det Z = 3$ , $c = \frac{1}{3}$ , n = 3 ;

$\det \left ( \frac{1}{3}Z \right )= \left ( \frac{1}{3} \right )^{3} (3) = \frac{1}{27} \cdot 3 = \frac{1}{9}$

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Example Question #27 : The Determinant

$A = \begin{bmatrix} 1& 3&-2 \\ 2& -1&4 \\ a& b& c \end{bmatrix}$ . $B = \begin{bmatrix} 1& 3&-2 \\ 2& -1&4 \\ 2a& 2b& 2c \end{bmatrix}$

The determinant of is 2. Give the determinant of .

Possible Answers:

Correct answer:

Explanation:

is the same matrix as , except each entry in one row (the third row) of has been multiplied by the same scalar, 2. This has the effect of multiplying that determinant by that scalar. Therefore,

$\det B = 2 \det A = 2 \cdot 2 = 4$ .

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Example Question #28 : The Determinant

$A = \begin{bmatrix} 4& 5& -1\\ 2&-2 &0 \\ a& b& c \end{bmatrix}$ . $B = \begin{bmatrix} 4& 5& -1 \\ a& b& c \\ 2&-2 &0 \end{bmatrix}$ .

The determinant of is equal to 5. True or false: the determinant of is also 5.

Possible Answers:

False

True

Correct answer:

False

Explanation:

is obtained from by switching one of its rows with another. This makes the determinant of equal to the additive inverse of that of . Since has determinant 5, has determinant .

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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 #14 : The Determinant

Example Question #11 : The Determinant

Example Question #21 : The Determinant

Example Question #22 : The Determinant

Example Question #23 : The Determinant

Example Question #24 : The Determinant

Example Question #25 : The Determinant

Example Question #26 : The Determinant

Example Question #27 : The Determinant

Example Question #28 : The Determinant

All Linear Algebra Resources

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