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GRE Subject Test: Math : Linear Algebra

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Example Question #481 : Gre Subject Test: Math

Solve the systems of equations.

8x + 2y=4

4x + 2y = 8

Possible Answers:

(-1,6)

(-6,-1)

(6, -1)

(1,6)

Correct answer:

(-1,6)

Explanation:

The first step in solving this system of equations is to remove either the term or the term. This is done by multiplying the second equation by .

-1(4x + 2y) = 8 (-1)

-4x - 2y = -8

Now add these two equations together and the terms will be removed.

8x + 2y = 4

+ -4x - 2y =- 8

_____________________

4x = -4

Divide both sides of the equation by

x = -1

Put the value of , which is , into one of the equations to get the value of

-8 + 2y = 4

Add to both sides of the equation.

2y = 12

Divide both sides by

y = 6

(-1,6) is the correct answer.

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Example Question #11 : Solving Systems Of Equations

Solve this system of equations.

3x + y =11

-3x + y = 9

Possible Answers:

$(\frac{1}{3},10)$

$(-\frac{1}{3},-10)$

$(\frac{1}{3},-10)$

$(10, \frac{1}{3})$

Correct answer:

$(\frac{1}{3},10)$

Explanation:

To solve this system of equations, add. This will eliminate or remove the terms.

3x + y = 11

+ -3x + y = 9

___________________

2y = 20

Divide both sides by

y = 10

To get the value of replace the variable with the value of that variable in one of the equations.

3x + 10 = 11

Subtract from both sides of the equation.

3x + 10 - 10 = 11-10

3x = 1

Divide both sides of the equation by

$x = \frac{1}{3}$

$(\frac{1}{3},10)$ is the correct answer.

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Example Question #11 : Systems Of Equations

Solve this system of equations.

5x - 2y = 25

4x-2y = 24

Possible Answers:

(-9,-10)

(-10,9)

(-10,-9)

(1,-10)

Correct answer:

(1,-10)

Explanation:

5x -2y = 25

4x-2y=24

To solve this system of equations, subtract the second equation from the first.

5x - 2y = 25

-4x + 2y = -24

____________________________

x=1

Now, substitute in the value for x into one of the equations to solve for the value of .

$\\5x-2y = 25 \\5(1)-2y=25 \\5-2y=25$

Now subtract five from each side.

$\\5-5-2y=25-5 \\-2y=20$

Divide both sides by negative 2:

$\\ \frac{-2y}{-2}=\frac{20}{-2} \\ \\y=-10$

x = 1

(1,-10) is the correct answer.

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Example Question #131 : Linear Algebra

Solve this system of equations:

$y = \frac{1}{5}x +2$

$y = -\frac{1}{5}x -4$

Possible Answers:

(-15,-1)

(-15,1)

(15,-1)

(-1,-15)

Correct answer:

(-15,-1)

Explanation:

To solve this system of equations, set both equations equal to one another.

$\frac{1}{5}x +2 = -\frac{1}{5}x-4$

Add $\frac{1}{5}x$ to both sides of the equation.

$\frac{1}{5}x + \frac{1}{5}x + 2= -\frac{1}{5}x +\frac{1}{5}x -4$

$\frac{2}{5}x + 2 = -4$

Subtract from both sides of the equation.

$\frac{2}{5}x + 2 -2 = -4-2$

$\frac{2}{5}x =-6$

Multiply both sides of the equation by $\frac{5}{2}$ .

$\frac{2}{5}x \times \frac{5}{2} = \frac{-6}{1} \times \frac{5}{2}$

$x = \frac{-30}{2}$

x = -15

Plug the value of , which is into one of the equations to get the value of

$-\frac{1}{5 }\times \frac{-15}{1 } - 4 = y$

3-4 = y

y = -1

(-15,-1) is the correct answer for this system of equations.

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Example Question #11 : Solving Systems Of Equations

Find the value of and that satisfy the equations:

(1) x-y=5

and

(2) x+y=7 .

Possible Answers:

x=6,y=1

x=6,y=-1

x=-6,y=-1

x=-6,y=1

Correct answer:

x=6,y=1

Explanation:

Step 1: Write the two equations, one below another and line up the terms.

x-y=5
x+y=7
----------------

Step 2: We see that we have and . We can add these two equations up, which will isolate y and let us solve for x.

$x{\color{Red} -y}=5$
$x+{\color{Blue} y}=7$ We add here.
----------------
2x=12

Step 3: We will isolate x by itself. We need to divide by 2 on both sides to get x.

$\frac {2x}{2}=\frac {12}{2}$
$\rightarrow x=6$

Step 4: We found x, so we can plug in that value into any one of the two equations and solve for y. Let's choose equation (1).

(1)... x-y=5
(6)-y=5 . Isolate y by itself. We are going to subtract 6 from both sides.
6-6-y=5-6 . Simplify the left hand side.
-y=-1

Step 5: We will divide by -1 to get the value of y.

$\frac {-y}{-1}=\frac {-1}{-1}$
$\rightarrow y=1$

The values that solve this system of equations is x=6 and y=1 .

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Example Question #11 : Systems Of Equations

4x+3y=11

8x+4y=12

Possible Answers:

Unsolvable

x=1

y=-5

x=-1

y=5

x=5

y=-1

x=-5

y=1

Correct answer:

x=-1

y=5

Explanation:

To answer this question you must first solve for one of the variables. This can be done with either variable with either equation. In this example of how to solve the problem we will solve for y using the second equation

8x+4y=12

subtract 8x from both sides

4y=12-8x

divide both sides by y

y=3-2x

Now we plug this into the first equation for the y variable

4x+3(3-2x)=11

Distribute the 3

4x+9-6x=11

Simplify

-2x+9=11

subtract 9 from both sides

-2x=2

divide by -2 on both sides

x=-1

Using this we solve for y in the second equation

8(-1) +4y=12

simplify

-8+4y=12

add 8 to both sides

4y=20

divide by 4 on both sides

y=5

Final answer x=-1 and y=5

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Example Question #1 : Eigenvalues

Find the eigenvalues of the following matrix, if possible.

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

Possible Answers:

$\lambda_{1,2}=\pm1$

The eigenvalues do not exist.

$\lambda_{1,2}=1\pm2i$

$\lambda=\frac{5}{2}$

$\lambda=\begin{bmatrix} 2i&4i \\ -2i&i \end{bmatrix}$

Correct answer:

$\lambda_{1,2}=1\pm2i$

Explanation:

In order to find the eigenvalues of a matrix, apply the following formula:

$det(A-\lambda I) =0$

is the identity matrix.

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

Compute the determinant and set it equal to zero.

$(1-\lambda)(1-\lambda)-(2)(-2)=0$

$1-2\lambda+\lambda^2+4=0$

$\lambda^2-2\lambda+5=0$

Solve for lambda by using the quadratic formula.

$\lambda_{1,2}=1\pm2i$

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Example Question #2 : Eigenvalues

Find the eigenvalues of the following matrix, if possible.

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

Possible Answers:

$\lambda=\frac{5}{2}$

$\lambda_{1,2}=1\pm2i$

$\lambda_{1,2}=\pm1$

$\lambda=\begin{bmatrix} 2i&4i \\ -2i&i \end{bmatrix}$

The eigenvalues do not exist.

Correct answer:

$\lambda_{1,2}=1\pm2i$

Explanation:

In order to find the eigenvalues of a matrix, apply the following formula:

$det(A-\lambda I) =0$

is the identity matrix.

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

Compute the determinant and set it equal to zero.

$(1-\lambda)(1-\lambda)-(2)(-2)=0$

$1-2\lambda+\lambda^2+4=0$

$\lambda^2-2\lambda+5=0$

Solve for lambda by using the quadratic formula.

$\lambda_{1,2}=1\pm2i$

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GRE Subject Test: Math : Linear Algebra

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #481 : Gre Subject Test: Math

Example Question #11 : Solving Systems Of Equations

Example Question #11 : Systems Of Equations

Example Question #131 : Linear Algebra

Example Question #11 : Solving Systems Of Equations

Example Question #11 : Systems Of Equations

Example Question #1 : Eigenvalues

Example Question #2 : Eigenvalues

All GRE Subject Test: Math Resources

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