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Linear Algebra : Linear Algebra

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Example Questions

Example Question #41 : The Determinant

True or false: Square matrix is nilpotent if and only if $\det A = 0$ .

Possible Answers:

True

False

Correct answer:

False

Explanation:

A square matrix is nilpotent if, for some whole number , $A^{k} = O$ , the zero matrix of the same dimension as .

The determinant of the product of matrices is equal to the product of their determinants. It follows that if $A^{k} = O$ , $\left ( \det A \right )^{k} = \det O = 0$ , so $\det A = 0$ . Therefore, any nilpotent matrix must have determinant 0.

However, not all matrices with determinant 0 are nilpotent, as is proved by counterexample. Let

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

This matrix is diagonal, as its only nonzero entry is alone its main diagonal. To raise this to a power , simply raise all of the diagonal elements to the power of , and preserve the off-diagonal zeroes. It follows that for all $k \in \mathbb{N}$ ,

$A^{k} = \begin{bmatrix} 1^{k} & 0 \\ 0 & 0 ^{k} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \ne O$

Also, the determinant of is $\det A = 1(0)- 0(0) = 0$ .

Since a non-nilpotent matrix with determinant zero exists, the biconditonal is false.

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Example Question #351 : Operations And Properties

, , and are each $4 \times 4$ real matrices.

$\det A = 0$ ; $\det B = 1$ ; $\det C = 2$ .

True or false: It follows that ABC is a singular matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

The determinant of the product of matrices is equal to the product of their determinants; thus,

$\det (ABC) = (\det A)( \det B)( \det C) = 0 \cdot 1 \cdot 2 = 0$ .

A matrix has determinant 0 if and only if it is singular - that is, without an inverse. It therefore follows that ABC is singular.

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Example Question #354 : Operations And Properties

Let $(x_{1}, y_{1}, z_{1})$ , $(x_{2}, y_{2}, z_{2})$ , and $(x_{3}, y_{3}, z_{3})$ be three noncollinear points in Cartesian three-space. The equation of the plane through all three points is

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

Give the equation of the plane that includes the points (0,0,0) , (1, 3, 2) , and (-1, 2, -4) .

Possible Answers:

6x + y + z = 0

8x + 6y +z = 0

8x + 6y -5 z = 0

16x-2y+z = 0

16x-2y-5z = 0

Correct answer:

16x-2y-5z = 0

Explanation:

If the three known points (0,0,0) , (1, 3, 2) , and (-1, 2, -4) , and an unknown point (x, y, z) are coplanar, then

$\begin{vmatrix} 0& 0 & 0 & 1 \\1 & 3 & 2 & 1 \\ -1 & 2 & -4 & 1 \\ x&y&z & 1 \end{vmatrix} =0$

The variable equation can be formed from this determinant equation.

Using the first row, this can be rewritten as

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

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

where $M_{14}$ , the minor, is the determinant of the matrix formed by striking out Row 1 and Column 4:

$\begin{vmatrix} \times & \times & \times & \times\\1 & 3 & 2 & \times \\ -1 & 2 & -4 & \times \\x& y & z & \times\end{vmatrix}$

$M_{14} = \begin{vmatrix} 1 & 3 & 2 \\ -1 & 2 & -4 \\x & y & z \end{vmatrix}$

Adding the upper-left to lower-right products and subtracting the upper-right to lower left products:

$M_{14}= 1(2)(z)+ 3(-4)(x)+ 2(-1)(y)- 2(2)(x)- 1(-4)(y)- 3(-1)(z)$

=2z-12x -2y -4x +4y +3z

= -16x +2y +5z

$\begin{vmatrix} 0& 0 & 0 & 1 \\1 & 3 & 2 & 1 \\ -1 & 2 & -4 & 1 \\ x&y&z & 1 \end{vmatrix} =0 = C_{14} = -M_{14}= 16x-2y-5z$

Set this equal to 0 to get the equation of the plane:

16x-2y-5z = 0

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Example Question #355 : Operations And Properties

, , and are each $4 \times 4$ real matrices.

$\det A = 0$ ; $\det B = 1$ ; $\det C = -1$ .

True or false: It follows that A+B+ C is a nonsingular matrix.

Possible Answers:

False

True

Correct answer:

False

Explanation:

The statement can be proved false through counterexample.

Let

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

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

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

Each of the matrices has only zero elements on its off-diagonal elements, so each is a diagonal matrix. Consequently, the determinant of each matrix is the product of its diagonal elements.

is a zero matrix, so $\det A = 0$ ; is an identity matrix, so $\det B = 1$ ; and

$\det C = \begin{vmatrix}{\color{DarkGreen}\textbf{-1}} & 0 & 0 & 0 \\ 0 & {\color{DarkGreen}\textbf{1}} & 0 & 0 \\ 0 & 0 & {\color{DarkGreen}\textbf{1}} & 0 \\ 0 & 0 & 0 & {\color{DarkGreen}\textbf{1}} \end{vmatrix} = (-1)(1)(1)(1) =-1$

, , and meet the conditions of the problem. Now, add the matrices by adding corresponding entries:

A+ B+ C

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

$= \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 &2 & 0 \\ 0 & 0 & 0 &2 \end{bmatrix}$

This matrix is also diagonal, so

$\det (A+ B+ C) = \begin{bmatrix} \textbf{{\color{DarkGreen} 0}} & 0 & 0 & 0 \\ 0 & \textbf{{\color{DarkGreen} 2}} & 0 & 0 \\ 0 & 0 & \textbf{{\color{DarkGreen} 2}} & 0 \\ 0 & 0 & 0 & \textbf{{\color{DarkGreen} 2}} \end{bmatrix}= (0)(2)(2)(2) = 0$

A matrix has determinant 0 if and only if it is singular - that is, without an inverse. It therefore follows that A+B+ C is singular. Thus, we have proved through counterexample that A+B+ C need not be nonsingular.

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Example Question #42 : The Determinant

, , and are each $4 \times 4$ real matrices.

$\det A = 0$ ; $\det B = 1$ ; $\det C = -1$ .

True or false: It follows that A+B+ C is a singular matrix.

Possible Answers:

False

True

Correct answer:

False

Explanation:

The statement can be proved false through counterexample.

Let

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

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

$C = \begin{vmatrix} -0.5 & 0 & 0 & 0 \\ 0 & 0.5 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 &2 \end{vmatrix}$

Each of the matrices has only zero elements on its off-diagonal elements, so each is a diagonal matrix. Consequently, the determinant of each matrix is the product of its diagonal elements.

is a zero matrix, so $\det A = 0$ ; is an identity matrix, so $\det B = 1$ ; and

$\det C = \begin{vmatrix} {\color{DarkGreen}\textbf{-0.5}} & 0 & 0 & 0 \\ 0 & {\color{DarkGreen}\textbf{0.5}} & 0 & 0 \\ 0 & 0 & {\color{DarkGreen}\textbf{2}} & 0 \\ 0 & 0 & 0 & {\color{DarkGreen}\textbf{2}} \end{vmatrix} = (-0.5)(0.5)(2)(2) = -1$

, , and meet the conditions of the problem. Now, add the matrices by adding corresponding entries:

A+ B+ C

$= \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}+ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 &1 \end{bmatrix}+ \begin{bmatrix} -0.5 & 0 & 0 & 0 \\ 0 & 0.5 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 &2 \end{bmatrix}$

$= \begin{bmatrix} 0.5 & 0 & 0 & 0 \\ 0 & 1.5 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix}$

This matrix is also diagonal, so

$\det (A+ B+ C) = \begin{vmatrix} \textbf{{\color{DarkGreen} 0.5}} & 0 & 0 & 0 \\ 0 & \textbf{{\color{DarkGreen} 1.5}} & 0 & 0 \\ 0 & 0 & \textbf{{\color{DarkGreen} 3}} & 0 \\ 0 & 0 & 0 & \textbf{{\color{DarkGreen} 3}} \end{vmatrix}= (0.5)(1.5)(3)(3) =6.75$

A matrix has a nonzero determinant if and only if it is nonsingular - that is, with an inverse. It therefore follows that A+B+ C is nonsingular. Thus, we have proved through counterexample that A+B+ C need not be singular.

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Example Question #357 : Operations And Properties

Let $(x_{1}, y_{1}, z_{1})$ , $(x_{2}, y_{2}, z_{2})$ , $(x_{3}, y_{3}, z_{3})$ , and $(x_{4}, y_{4}, z_{4})$ be three points in Cartesian three-space. These points are coplanar if and only if

$\begin{vmatrix} x_{1}& y_{1} & z_{1} & 1 \\ x_{2}& y_{2} & z_{2} & 1 \\ x_{3}& y_{3} & z_{3} & 1 \\ x_{4}& y_{4} & z_{4} & 1 \end{vmatrix} =0$

Three points are given: (0,0,0) , (1, 3, 2) , and (-1, 2, -4) . Which point along the line of the equation

x = y = 6

is on the plane that includes these points?

Possible Answers:

None of the other choices gives the correct response.

$\left ( 6,6, \frac{84}{5}\right )$

$\left ( 6,6, 84\right )$

$\left ( 6,6,- \frac{84}{5}\right )$

$\left ( 6,6, -84\right )$

Correct answer:

$\left ( 6,6, \frac{84}{5}\right )$

Explanation:

The three points (0,0,0) , (1, 3, 2) , and (-1, 2, -4) , and a point of the form (6,6,t)

for some real are on the same plane. Therefore, it follows that we must find so that

$\begin{vmatrix} 0& 0 & 0 & 1 \\1 & 3 & 2 & 1 \\ -1 & 2 & -4 & 1 \\ 6&6& t & 1 \end{vmatrix} =0$

Using the first row, this can be rewritten as

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

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

where $M_{14}$ , the minor, is the determinant of the matrix formed by striking out Row 1 and Column 4:

$\begin{vmatrix} \times & \times & \times & \times\\1 & 3 & 2 & \times \\ -1 & 2 & -4 & \times \\6 & 6 & t & \times\end{vmatrix}$

$M_{14} = \begin{vmatrix} 1 & 3 & 2 \\ -1 & 2 & -4 \\6 & 6 & t \end{vmatrix}$

Adding the upper-left to lower-right products and subtracting the upper-right to lower left products:

$M_{14} = 1(2)(t)+ 3(-4)(6)+2(-1)(6) - 2(2)(6)- (-4)(6)(1) - 3(-1)(t)$

= 2t -72 -12 - 24 +24 +3t

= 5t -84

Thus,

$C_{14}= -5t + 84$ ,

and

$\begin{vmatrix} 0& 0 & 0 & 1 \\1 & 3 & 2 & 1 \\ -1 & 2 & -4 & 1 \\ 6&6& t & 1 \end{vmatrix} = -5t + 84$

et this equal to 0 and solve for :

-5t + 84 = 0

-5t = -84

$t = \frac{84}{5}$

The desired point is $\left ( 6,6, \frac{84}{5}\right )$ .

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Example Question #47 : The Determinant

Let $(x_{1}, y_{1}, z_{1})$ , $(x_{2}, y_{2}, z_{2})$ , $(x_{3}, y_{3}, z_{3})$ , and $(x_{4}, y_{4}, z_{4})$ be three points in Cartesian three-space. These points are coplanar if and only if

$\begin{vmatrix} x_{1}& y_{1} & z_{1} & 1 \\ x_{2}& y_{2} & z_{2} & 1 \\ x_{3}& y_{3} & z_{3} & 1 \\ x_{4}& y_{4} & z_{4} & 1 \end{vmatrix} =0$

Three points are given: (0,0,0) , (1, 3, 2) , and (-1, 2, -4) . Which point along the line of the equation

x+ 2 = y + 1 = z

is on the plane that includes these points?

Possible Answers:

$\left ( \frac{4}{3}, \frac{7}{3}, \frac{10}{3} \right )$

$\left (- \frac{4}{5}, \frac{1}{5}, \frac{6}{5} \right )$

$\left ( \frac{4}{5}, \frac{9}{5}, \frac{14}{5} \right )$

$\left (- \frac{4}{3},- \frac{1}{3}, \frac{2}{3} \right )$

None of the other choices gives the correct response.

Correct answer:

$\left ( \frac{4}{3}, \frac{7}{3}, \frac{10}{3} \right )$

Explanation:

The three points (0,0,0) , (1, 3, 2) , and (-1, 2, -4) , and a point of the form (t, t+ 1, t+ 2 ) for some real are on the same plane. Therefore, it follows that we must find so that

$\begin{vmatrix} 0& 0 & 0 & 1 \\1 & 3 & 2 & 1 \\ -1 & 2 & -4 & 1 \\ t & t+ 1 & t+2 & 1 \end{vmatrix} =0$

Using the first row, this can be rewritten as

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

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

where $M_{14}$ , the minor, is the determinant of the matrix formed by striking out Row 1 and Column 4:

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

$M_{14} = \begin{vmatrix} 1 & 3 & 2 \\ -1 & 2 & -4 & \\ t & t+ 1 & t+2 \end{vmatrix}$

Adding the upper-left to lower-right products and subtracting the upper-right to lower left products:

$\begin{matrix} M_{14}= 1(2)(t+2) + 3(-4)(t)+ 2(-1)(t+1) - 2(2)(t) - (1)(-4)(t+1) -(3)(-1)(t+2) \end{matrix}$

= 2t+4 -12t-2t-2 -4t +4t+4 +3t+6

= -9t +12

It follows that

$C_{14} = 9t - 12$ ,

and

$\begin{vmatrix} 0& 0 & 0 & 1 \\1 & 3 & 2 & 1 \\ -1 & 2 & -4 & 1 \\ t & t+ 1 & t+2 & 1 \end{vmatrix} =9t-12$

Set this equal to 0 to find :

9t-12 = 0

9t =12

$t= \frac{4}{3}$

The desired point is $\left ( \frac{4}{3}, \frac{7}{3}, \frac{10}{3} \right )$ .

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Example Question #46 : The Determinant

, , and are each $4 \times 4$ real matrices.

$\det A = 0$ ; $\det B = 1$ ; $\det C = 2$ .

True or false: It follows that A+B+ C is a nonsingular matrix.

Possible Answers:

False

True

Correct answer:

False

Explanation:

The statement can be proved false through counterexample.

Let

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

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

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

Each of the matrices has only zero elements on its off-diagonal elements, so each is a diagonal matrix. Consequently, the determinant of each matrix is the product of its diagonal elements.

is a zero matrix, so $\det A = 0$ ; is an identity matrix, so $\det B = 1$ ; and

$\det C = \begin{vmatrix}{\color{DarkGreen}\textbf{-1}} & 0 & 0 & 0 \\ 0 & {\color{DarkGreen}\textbf{-1}} & 0 & 0 \\ 0 & 0 & {\color{DarkGreen}\textbf{1}} & 0 \\ 0 & 0 & 0 & {\color{DarkGreen}\textbf{2}} \end{vmatrix} = (-1)(-1)(1)(2) = 2$

, , and meet the conditions of the problem. Now, add the matrices by adding corresponding entries:

A+ B+ C

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

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

This matrix is also diagonal, so

$\det (A+ B+ C) = \begin{bmatrix} \textbf{{\color{DarkGreen} 0}} & 0 & 0 & 0 \\ 0 & \textbf{{\color{DarkGreen} 0}} & 0 & 0 \\ 0 & 0 & \textbf{{\color{DarkGreen} 2}} & 0 \\ 0 & 0 & 0 & \textbf{{\color{DarkGreen} 3}} \end{bmatrix}= (0)(0)(2)(3) = 0$

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Example Question #41 : The Determinant

$A = \begin{bmatrix}6 & -3 \\ 8 & x \end{bmatrix}$ for some real such that has an inverse.

Give $A^{-1}$ in terms of .

Possible Answers:

$\frac{1}{ 6x - 24}\begin{bmatrix}x & -3 \\ 8 & 6 \end{bmatrix}$

$\frac{1}{ 6x + 24}\begin{bmatrix}x & 3 \\ -8 & 6 \end{bmatrix}$

None of the other choices gives the correct response.

$\frac{1}{ 6x + 24}\begin{bmatrix}x & -3 \\ 8 & 6 \end{bmatrix}$

$\frac{1}{ 6x - 24}\begin{bmatrix}x & 3 \\ -8 & 6 \end{bmatrix}$

Correct answer:

$\frac{1}{ 6x + 24}\begin{bmatrix}x & 3 \\ -8 & 6 \end{bmatrix}$

Explanation:

The inverse of a two-by-two matrix

$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21}& a_{22} \end{bmatrix}$

is the matrix

$A^{-1} =\frac{1}{D} \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21}& a_{11} \end{bmatrix}$ .

is the determinant of the matrix, which is the product of the main diagonal elements minus the product of the other two:

D= 6(x)-(-3)(8)

= 6x + 24

The inverse is

$A^{-1} = \frac{1}{ 6x + 24}\begin{bmatrix}x & 3 \\ -8 & 6 \end{bmatrix}$ .

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Example Question #432 : Linear Algebra

$A = \begin{bmatrix} \sqrt{5} &2 \\2 \sqrt{2} & \sqrt{10} \end{bmatrix}$

Find $A^{-1}$ .

Possible Answers:

$\begin{bmatrix} \sqrt{5}& \sqrt{2}\\ -2 & -\frac{1}{2} \sqrt{10} \end{bmatrix}$

$\begin{bmatrix} \sqrt{5} &-2 \\-2 \sqrt{2} & \sqrt{10} \end{bmatrix}$

$\begin{bmatrix} -\sqrt{5} &2 \\2 \sqrt{2} & -\sqrt{10} \end{bmatrix}$

$\begin{bmatrix} \sqrt{5}& -\sqrt{2}\\ -2 & \frac{1}{2} \sqrt{10} \end{bmatrix}$

$\begin{bmatrix}- \sqrt{5}& \sqrt{2}\\ 2 &- \frac{1}{2} \sqrt{10} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \sqrt{5}& -\sqrt{2}\\ -2 & \frac{1}{2} \sqrt{10} \end{bmatrix}$

Explanation:

The inverse of a two-by-two matrix

$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21}& a_{22} \end{bmatrix}$

is the matrix

$A^{-1} =\frac{1}{D} \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21}& a_{11} \end{bmatrix}$ .

is the determinant of the matrix, which is the product of the main diagonal elements minus the product of the other two:

$D = a_{11}a_{22}- a_{12}a_{21}$

$= \sqrt{5}(\sqrt{10} )-2( 2 \sqrt{2} )$

$= 5\sqrt{2} -4 \sqrt{2}$

$=\sqrt{2}$

Therefore,

$A^{-1} =\frac{1}{\sqrt{2}} \begin{bmatrix} \sqrt{10} &-2 \\-2 \sqrt{2} & \sqrt{5} \end{bmatrix}$

$= \begin{bmatrix}\frac{1}{\sqrt{2}}( \sqrt{10}) &\frac{1}{\sqrt{2}}(-2) \\ \\ \frac{1}{\sqrt{2}} (-2 \sqrt{2}) &\frac{1}{\sqrt{2}}( \sqrt{5} )\end{bmatrix}$

$= \begin{bmatrix} \sqrt{5}& -\sqrt{2}\\ -2 & \frac{1}{2} \sqrt{10} \end{bmatrix}$

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #41 : The Determinant

Example Question #351 : Operations And Properties

Example Question #354 : Operations And Properties

Example Question #355 : Operations And Properties

Example Question #42 : The Determinant

Example Question #357 : Operations And Properties

Example Question #47 : The Determinant

Example Question #46 : The Determinant

Example Question #41 : The Determinant

Example Question #432 : Linear Algebra

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