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Linear Algebra : The Determinant

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Example Question #411 : Linear Algebra

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:

$C_{2,1} =0$

$C_{2,1} =5$

$C_{2,1} = 7$

$C_{2,1} = -7$

$C_{2,1} = -5$

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:

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

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

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

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

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

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:

$n^{2}$

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

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:

$C_{23} = 0$

$C_{23} = 10$

$C_{23} = -2$

$C_{23} = -10$

$C_{23} = 2$

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:

$C_{33}= -11$

$C_{33}= 0$

$C_{33}= 11$

$C_{33}=- 5$

$C_{33}= 5$

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:

$\sqrt{3}$

$\frac{1}{9}$

$\sqrt[3]{3}$

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 #21 : 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 #21 : 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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Example Question #21 : The Determinant

$A = \begin{bmatrix} 4&0 &0\\ a& 1& 2\\ 0& 3& 7 \end{bmatrix}$

Evaluate so that the matrix has 4 as its determinant.

Possible Answers:

A = 1

has 4 as its determinant regardless of the value of .

a = -2

a = -1

a = 2

Correct answer:

has 4 as its determinant regardless of the value of .

Explanation:

One way to calculate the determinant of a three-by-three matrix is as follows:

Choose a row or column of the matrix, multiply each element of the row or column $a_{mn}$ by its corresponding cofactor $C_{mn}$ , and add the products. The best row or column to choose is Row 1, which has two zeroes; the determinant will be

$\det A = a_{11} C_{11}+ a_{12}C_{12}+ a_{13}C_{13}$

Examining $A = \begin{bmatrix} 4&0 &0\\ a& 1& 2\\ 0& 3& 7 \end{bmatrix}$ , we see that $a_{11}= 4, a_{12}= a_{13}= 0$ , so the expression simplifies to

$\det A =4 C_{11}$ ,

that is, 4 times the cofactor $C_{11}$ .

To find cofactor $C_{11}$ , first find the minor $M_{11}$ , the determinant of the matrix formed by striking out Row 1 and Column 1, as seen below:

Minor

$M_{11} = \begin{vmatrix} 1& 2\\ 3 & 7 \end{vmatrix}$ , which can be calculated by taking the product of the upper-left and lower-right elements and subtracting that of the other two:

$M_{11} = 1 \cdot 7 - 2 \cdot 3 = 7 - 6 = 1$

Cofactor $C_{mn}$ can be calculated from a minor $M_{mn}$ using the formula

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

so, setting m = n = 1 :

$C_{11}= ( -1 )^{1+1} M_{11} = (-1)^{2} \cdot 1 = 1 \cdot 1 = 1$

The determinant of is

$\det A =4 C_{11} = 4 \cdot 1 = 4$

regardless of the value of .

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Example Question #30 : The Determinant

Let be a five-by-five matrix.

Cofactor $C_{14}$ must be equal to:

Possible Answers:

Minor $M_{14}$

The additive inverse of the reciprocal of Minor $M_{14}$

None of the other choices gives a correct response.

The reciprocal of Minor $M_{14}$

The additive inverse of Minor $M_{14}$

Correct answer:

The additive inverse of Minor $M_{14}$

Explanation:

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

Set m = 1 , n= 4 ; the formula becomes

$C_{14}= ( -1 )^{1+4} M_{14}$

$C_{14}= ( -1 )^{5} M_{14}$

$C_{14}=-1 \cdot M_{14}$

$C_{14}=- M_{14}$ .

Therefore, the cofactor $C_{14}$ must be equal to the opposite of the minor $M_{14}$ .

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Linear Algebra : The Determinant

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #411 : Linear Algebra

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 #21 : The Determinant

Example Question #21 : The Determinant

Example Question #21 : The Determinant

Example Question #30 : The Determinant

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