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Linear Algebra : The Determinant

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Example Questions

Example Question #11 : The Determinant

Calculate the determinant of matrix A.

$A=\begin{bmatrix} 2&5 &4 &7 &1 \\ 8& 6&11 & 3&9 \\4 & 1&3 &0 &6 \\ 7& 0&2 &9 &1 \end{bmatrix}$

Possible Answers:

Not possible

504

Correct answer:

Not possible

Explanation:

It is not possible to calculate the determinant of this matrix because only square matrices (nxn) have determinants.

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Example Question #12 : The Determinant

Calculate the determinant of matrix A.

$A=\begin{bmatrix} 1&5 &3 \\ 0&-2 &7 \\ 0&0 &-9 \end{bmatrix}$

Possible Answers:

None of the other answers

Correct answer:

Explanation:

To find the determinant of a lower or upper triangular matrix, simply find the product of the diagonal entries of matrix A:

$\det(A) = (1)(-2)(-9) = 18$

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Example Question #13 : The Determinant

Calculate the determinant of matrix A.

$A= \begin{bmatrix} 3& 0& 0&0 \\- 6&2 & 0&0 \\4 & 1& -1 &0 \\ 7&0 & 8 &5 \end{bmatrix}$

Possible Answers:

None of the other answers

Correct answer:

Explanation:

To find the determinant of a lower or upper triangular matrix, simply find the product of the diagonal entries of matrix A:

$\det(A) = (3)(2)(-1)(5) = -30$

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Example Question #12 : The Determinant

$A = \begin{bmatrix} 3&1 &0 \\ 0&a &12 \\ 1& 0 & a \end{bmatrix}$

is a singular matrix for what values of ?

Possible Answers:

a = 2 or a = 2i

$a = \pm 2$

$a=\pm 2i$

a = 0 or a = 2i

a = 0 or a = 2

Correct answer:

$a=\pm 2i$

Explanation:

A matrix is singular - without an inverse - if and only if its determinant is equal to 0.

One way to calculate the determinant of a three-by-three matrix is to add the products of the three diagonals going from upper-left to lower-right, then subtract the products of the three diagonals going from upper-right to lower left. From the diagram below:

Determinant

we see that the products of the three upper-left to lower-right diagonals are:

$3 \cdot a \cdot a = 3a^{2}$

$1 \cdot 12 \cdot 1 = 12$

$0 \cdot 0 \cdot 0 = 0$

From the diagram below:

Determinant

we see that the products of the three upper-right to lower-left diagonals are:

$0 \cdot a \cdot 1 = 0$

$3 \cdot 12 \cdot 0 = 0$

$1 \cdot 0 \cdot a = 0$

Add the first three and subtract the last three:

$3a^{2}+ 12+ 0 - 0 - 0 -0 = 3a^{2}+ 12$

This must be equal to 0, so set it as such, and solve for :

$3a^{2}+ 12 = 0$

$3a^{2} = -12$

$a^{2} =-4$

$a = \pm \sqrt{-4} = \pm 2i$

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Example Question #11 : The Determinant

Given: matrix such that $\det C = \frac{3}{5}$ .

Evaluate $\det C^{T}C$ .

Possible Answers:

$-\frac{9}{25}$

$\frac{9}{25}$

Correct answer:

$\frac{9}{25}$

Explanation:

The determinant of $C^{T}$ is equal to that of the transpose of ; also, the determinant of the matrix product of two matrices is equal to the product of the determinants. Therefore,

$\det C^{T}C$

$= \det C^{T} \cdot \det C$

$= \det C \cdot \det C$

$= \frac{3}{5} \cdot \frac{3}{5}$

$= \frac{9}{25}$

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Example Question #15 : The Determinant

Consider the matrix

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

where is a real number.

Evaluate so that the minor $M_{31}$ of this matrix is equal to 12.

Possible Answers:

y = -8

y = -3

$M_{31}$ cannot be equal to 12 regardless of the value of .

y = 8

y = 3

Correct answer:

$M_{31}$ cannot be equal to 12 regardless of the value of .

Explanation:

The minor $M_{31}$ of the matrix is the determinant of the matrix formed when Row 3 and Column 1 of are struck out. This is shown below:

Minor

The minor is therefore equal to

$\begin{vmatrix} 2 & 0\\ y& -2 \end{vmatrix}$

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

$M_{31} = 2(-2) - 0 \cdot y = -4 - 0 = -4$

Therefore, regardless of the value of , the minor $M_{31}$ cannot be equal to 12.

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Example Question #12 : The Determinant

Consider the matrix

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

where is a real number.

Evaluate so that the minor $M_{13}$ of this matrix is equal to 9.

Possible Answers:

y = 3

$M_{13}$ cannot be equal to 9 regardless of the value of .

y = -3

y = -12

y = 12

Correct answer:

y = -12

Explanation:

The minor $M_{13}$ of the matrix is the determinant of the matrix formed when Row 1 and Column 3 of are struck out. This is shown below:

Minor

The minor is therefore equal to

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

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

$M_{13} = 3 (-5 ) - y (2) =-15 - 2y$

Set this equal to 9 and solve for :

-15 - 2y = 9

-2y = 24

y = -12

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Example Question #11 : The Determinant

Consider the matrix

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

where is a real number.

Evaluate so that the minor $M_{33}$ of this matrix is equal to 77.

Possible Answers:

x = 77

x =-77

x = -11

x = 11

$M_{33}$ cannot be equal to 77 regardless of the value of .

Correct answer:

x = 11

Explanation:

The minor $M_{33}$ of the matrix is the determinant of the matrix formed when Row 3 and Column 3 of are struck out. This is shown below:

Minor

The minor is therefore equal to

$\begin{vmatrix} x& 3\\ -x& 4 \end{vmatrix}$

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

$M_{33} = x (4) -x (-3) = 4x - (-3x ) = 7x$

Set this equal to 77 and solve for :

7x = 77

x = 11

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Example Question #14 : The Determinant

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

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

Possible Answers:

$-\frac{1}{100}$

100

$\frac{1}{100}$

More information is needed to find the determinant of .

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 #14 : The Determinant

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

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

Possible Answers:

$-\frac{3}{4}$

$\frac{3}{4}$

$\frac{4}{3}$

$-\frac{4}{3}$

More information is needed to answer the question.

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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Linear Algebra : The Determinant

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #11 : The Determinant

Example Question #12 : The Determinant

Example Question #13 : The Determinant

Example Question #12 : The Determinant

Example Question #11 : The Determinant

Example Question #15 : The Determinant

Example Question #12 : The Determinant

Example Question #11 : The Determinant

Example Question #14 : The Determinant

Example Question #14 : The Determinant

All Linear Algebra Resources

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