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Linear Algebra : The Determinant

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Example Question #341 : Operations And Properties

Let be a four-by-four matrix.

Cofactor $C_{13}$ must be equal to:

Possible Answers:

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

Minor $M_{13}$

The reciprocal of Minor $M_{14}$

None of the other choices gives a correct response.

The additive inverse of Minor $M_{14}$

Correct answer:

Minor $M_{13}$

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

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

Set m = 1 , n= 3 ; the formula becomes

$C_{13}= ( -1 )^{1+3} M_{13}$

$C_{13}= ( -1 )^{ 4} M_{13}$

$C_{13}=1 \cdot M_{13}$

$C_{13}= M_{13}$ ,

making the quantities equal.

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Example Question #31 : The Determinant

True or False: If is a matrix obtained from (both being square matrices) by swapping any two rows, then det(A) = -det(B) .

Possible Answers:

True

False

Correct answer:

True

Explanation:

This is not easy to prove in a nutshell, but it can be believed by examining a few examples.

For instance

$\begin{vmatrix} 3& 1& 2& -1\\ 8& 0&2 & 0\\ -2&-1 & 2& 2\\ 2& 5& 6& 7 \end{vmatrix} = -366$ .

And swapping the 2nd and the 4th row gives

$\begin{vmatrix} 3& 1& 2& -1\\ -2& -1&2 & 2\\ 8&0 & 2& 0\\ 2& 5& 6& 7 \end{vmatrix} = 366$ .

So the determinants are negatives of each other. (Both determinants were evaluated using a calculator.)

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Example Question #32 : The Determinant

True or false: If det(cA) = 10 where is a constant, then det(2cA) = 20

Possible Answers:

True

False

Correct answer:

False

Explanation:

We cannot deduce anything about how the determinant of a multiple of changes without knowing the size of . In general det(cA) = c^ndet(A) , where is the size of the (square) matrix .

In our case $det(2cA) = 2^ndet(cA) = 2^n \times 10$ , which varies depending on the size of the matrix in question.

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Example Question #34 : The Determinant

and are nonsingular matrices with the same dimension.

$\det A = \sin \frac{5 \pi }{13}$

$\det B = \cos \frac{5 \pi }{13}$

Which of the following is equal to $\det (AB)$ ?

Possible Answers:

$\sin \frac{10 \pi}{13}$

$\frac{1}{2} \sin \frac{10 \pi}{13}$

$\tan \frac{10 \pi}{13}$

$\cos \frac{10 \pi}{13}$

$\frac{1}{2} \cos \frac{10 \pi}{13}$

Correct answer:

$\frac{1}{2} \sin \frac{10 \pi}{13}$

Explanation:

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

$\det (AB) = \det A \cdot \det B = \sin \frac{5 \pi }{13} \cdot \cos \frac{5 \pi }{13}$

By a trigonometric identity,

$2 \sin \theta \cos \theta = \sin 2 \theta$ , or $\sin \theta \cos \theta = \frac{1}{2} \sin 2 \theta$ , so, setting $\theta = \frac{5 \pi }{13}$ :

$\det (AB) =\frac{1}{2} \sin \frac{10 \pi }{13}$ .

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Example Question #31 : The Determinant

The equation of the plane that includes the points $(x_{1}, y_{1}, z_{1})$ , $(x_{2}, y_{2}, z_{2})$ , and $(x_{3}, y_{3}, z_{3})$ can be found by way of evaluating

$\begin{vmatrix} x & y & z & 1 \\ x_{1} & y_{1} & z_{1} & 1 \\ x_{2} & y_{2} & z_{2} & 1 \\ x_{3} & y_{3} & z_{3} & 1 \end{vmatrix}= 0$

Find the equation of the plane that passes through the points (4, -2, 0) , (6, -3, 5 ) , and (7, 0, -1) .

Possible Answers:

-9 x +17 y +7z +70= 0

11x+13y + z +142= 0

-9x+13y +7z +62= 0

-9 x +17 y + z +70= 0

11x+13y +7z +142= 0

Correct answer:

-9 x +17 y +7z +70= 0

Explanation:

A way to simplify this problem is to translate each of the points using the transformation

$(x,y,z) \rightarrow (x-4, y+2, z)$ .

This translates the points as follows:

$(4, -2, 0) \rightarrow (0,0,0)$

$(6, -3, 5 ) \rightarrow (2, -1, 5)$

$(7, 0, -1) \rightarrow (3, 2, -1)$

Replace (x-4, y+2, z) with (u, v, w) for the sake of simplicity.

The determinant equation becomes

$\begin{vmatrix} u & v & w & 1 \\0 & 0 & 0 & 1 \\ 2 & -1 & 5 & 1 \\3 & 2 & -1 & 1 \end{vmatrix}=0$

The determinant can be calculated by multiplying each entry in one row or column by its corresponding cofactor, then adding the products.The easiest row or column to choose is the second row, which has only one nonzero entry. This is equal to

$0 C_{21}+ 0C_{22}+ 0 C_{23}+ 1C_{24}= C_{24}$

$C_{24}$ is equal to $(-1)^{2+4}M_{24} =( -1)^{6}M_{24}= M_{24}$ ,

Minor is the determinant of the matrix formed by the deletion of the second row and the fourth column; this determinant is

$\begin{vmatrix} u & v & w \\ 2 & -1 & 5 \\3 & 2 & -1 \end{vmatrix}=0$

The determinant of a three-by-three matrix can be found by adding the upper-left to lower-right products and subtracting the upper-right to lower-left products, as shown:

Determinant

u + 15v + 4w - (-3w) - 10u - (-2v) = -9u+17v +7w

Set this determinant equal to 0. The equation of the plane in terms of u,v,w is

-9u+17v +7w = 0

Substituting back for these variables, the equation becomes

-9 (x-4)+17(y+2) +7z = 0

-9 x+36+17 y+34+7z = 0

-9 x +17 y +7z +70= 0 ,

the correct response.

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Example Question #341 : Operations And Properties

The equation of the plane that includes the points $(x_{1}, y_{1}, z_{1})$ , $(x_{2}, y_{2}, z_{2})$ , and $(x_{3}, y_{3}, z_{3})$ can be found by way of evaluating

$\begin{vmatrix} x & y & z & 1 \\ x_{1} & y_{1} & z_{1} & 1 \\ x_{2} & y_{2} & z_{2} & 1 \\ x_{3} & y_{3} & z_{3} & 1 \end{vmatrix}= 0$

Find the equation of the plane that passes through the origin, (2, -1, 5) , and (3, 2, -1) .

Possible Answers:

11x +25y + 7z = 0

11x +17y + z = 0

-9x +17y +7z = 0

-9x +25y + 7z = 0

11x +17y +7z = 0

Correct answer:

-9x +17y +7z = 0

Explanation:

Replace the first three elements in each of the last three rows with the given coordinates of the points:

$\begin{vmatrix} x & y & z & 1 \\0 & 0 & 0 & 1 \\ 2 & -1 & 5 & 1 \\3 & 2 & -1 & 1 \end{vmatrix}=0$

$0 C_{21}+ 0C_{22}+ 0 C_{23}+ 1C_{24}= C_{24}$

$C_{24}$ is equal to $(-1)^{2+4}M_{24} =( -1)^{6}M_{24}= M_{24}$ ,

Minor $M_{24}$ is the determinant of the matrix formed by the deletion of the second row and the fourth column; this determinant is

$\begin{vmatrix} x & y & z \\ 2 & -1 & 5 \\3 & 2 & -1 \end{vmatrix}$

The determinant of a three-by-three matrix can be found by adding the upper-left to lower-right products and subtracting the upper-right to lower-left products, as shown:

Determinant

x + 15y + 4z - (-3z) - 10x - (-2y) = -9x +17y +7z

Set this determinant equal to 0. The equation of the plane is

-9x +17y +7z = 0 .

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Example Question #33 : The Determinant

and are nonsingular matrices of the same size.

True or false: $\det (B^{-1}AB ) = \det A$ must be a true statement.

Possible Answers:

True

False

Correct answer:

True

Explanation:

The determinant of the product of matrices is equal to the product of the individual determinants; also, the determinant of the inverse of a matrix is the reciprocal of the determinant of the original matrix. Therefore,

$\det (B^{-1}AB ) = \det B \cdot \det A \cdot \det B^{-1}$

$\det (B^{-1}AB ) = \det B \cdot \det A \cdot \frac{1}{\det B }$

$\det (B^{-1}AB ) = \det A$

The statement is true.

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Example Question #344 : Operations And Properties

and are square matrices of the same size.

True or false: $\det (A^{T}B A )= \det B$ must be a true statement.

Possible Answers:

True

False

Correct answer:

False

Explanation:

The determinant of the product of matrices is equal to the product of the individual determinants; also, the determinant of the transpose of a matrix is equal to that of the original matrix. It follows that

$\det (A^{T}B A )= \det A^{T} \cdot \det B \cdot \det A$

$= \det A \cdot \det B \cdot \det A$

$= \det B \cdot( \det A)^{2}$

Therefore, $\det A^{T}B A = \det B$ is only true if $(\det A)^{2} = 1$ , or, equivalently, if $\det A = \pm 1$ . The statement is false.

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Example Question #39 : The Determinant

The volume of a tetrahedron in coordinate space with its vertices at the points $(x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}), (x_{3}, y_{3}, z_{3}), (x_{4}, y_{4}, z_{4})$ is given by the following equation:

$V = \frac{1}{6} |\det A|$

where

$A \begin{bmatrix} x_{1} & y_{1} & z_{1} & 1 \\ x_{2} & y_{2} & z_{2} & 1 \\ x_{3} & y_{3} & z_{3} & 1 \\ x_{4} & y_{4} & z_{4} & 1 \end{bmatrix}$

A tetrahedron has its vertices at (4, 3, 7) , (6, -3, 8 ) , (2, 9, 10 ) , and (5, -8, 2) . Give the volume of the tetrahedron.

Possible Answers:

$3\frac{2}{3}$

$42\frac{2}{3}$

$7\frac{2}{3}$

$44\frac{2}{3}$

$10\frac{2}{3}$

Correct answer:

$10\frac{2}{3}$

Explanation:

For reasons that will become clear, the problem is made easier if each point is translated similarly so that one of them translates to the origin (0,0,0) . Select the translation $(x, y, z)\rightarrow (x-4, y-3, y-7 )$ . Then the translations are:

$(4, 3, 7) \rightarrow (0,0,0)$

$(6, -3, 8 ) \rightarrow (2, -6, 1)$

$(2, 9, 10 ) \rightarrow (-2, 6, 3 )$

$(5, -8, 2) \rightarrow (1, -11, -5)$

Replace the variables in the matrix with these modified values:

$A= \begin{bmatrix}0 & 0& 0 & 1 \\ 2 & -6& 1 & 1 \\-2 & 6 & 3& 1 \\ 1 &-11 & -5 & 1 \end{bmatrix}$

The determinant can be found by choosing any row or column and adding the products of each value and its cofactor. Since the first row has three zeroes, thanks to our translations, choose this row:

$\det A= 0C_{11}+ 0C_{12}+ 0 C_{13}+ 1C_{14} = C_{14}$

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

where $M_{14}$ is the minor - the determinant of the matrix formed when Row 1 and Column 4 are removed:

$M_{14}= \begin{vmatrix} 2 & -6& 1 \\-2 & 6 & 3 \\ 1 &-11 & -5 \end{vmatrix}$

The determinant of a $3 \times 3$ matrix can be found by adding the products of the upper-left to lower-right diagonals and subtracting the products of the upper-right to lower-left diagonals:

Determinant

$M_{14} = -60 + (-18) + 22 - 6 - (-66) - (-60) = 64$ ,

and $\det A = -64$ .

$V = \frac{1}{6} |-64| = 10\frac{2}{3}$ ,

the volume of the tetrahedron.

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Example Question #40 : The Determinant

$A =\begin{bmatrix} 2+ 5i & 3 - 3i \\ 3+3i & 2 - 5i \end{bmatrix}$

Give the determinant of .

Possible Answers:

47i

11i

Correct answer:

Explanation:

The determinant of a $2 \times 2$ matrix can be calculated by subtracting the upper-right to lower-left product from the upper-left to lower-right product:

$\det A =\begin{vmatrix} \textbf{{\color{Red} 2 + 5i}}& \textbf{{\color{DarkGreen} 3 - 3i}} \\ \textbf{{\color{DarkGreen}3 + 3i}}& \textbf{{\color{Red} 2 - 5i}} \end{vmatrix}$

= (2+5i)(2-5i) - (3- 3i)(3+3i )

$= (2^{2}+ 5^{2} )- (3^{2} + 3^{2})$

= (4 + 25) - (9+9)

= 11

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Linear Algebra : The Determinant

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #341 : Operations And Properties

Example Question #31 : The Determinant

Example Question #32 : The Determinant

Example Question #34 : The Determinant

Example Question #31 : The Determinant

Example Question #341 : Operations And Properties

Example Question #33 : The Determinant

Example Question #344 : Operations And Properties

Example Question #39 : The Determinant

Example Question #40 : The Determinant

All Linear Algebra Resources

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