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GRE Subject Test: Math : Linear Algebra

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Example Questions

Example Question #1 : Linear Algebra

Perform the following operation.

$\begin{bmatrix} 2& 4& 1\\ 5& 3& -2\\ -3& -1&0 \end{bmatrix}$ $\cdot \begin{bmatrix} 2\\ 4\\ 1 \end{bmatrix}$ $+\begin{bmatrix} -7\\ 4\\ 11 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 14& 24& 1 \end{bmatrix}$

$\begin{bmatrix} 7\\ 20\\ 7 \end{bmatrix}$

$\begin{bmatrix} 14\\ 24 \\ 1 \end{bmatrix}$

$\begin{bmatrix} 34\\ -25\\ 7 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 14\\ 24 \\ 1 \end{bmatrix}$

Explanation:

The first step to solving this operation is to do the multiplication:

$\begin{bmatrix} 2& 4& 1\\ 5& 3& -2\\ -3& -1&0 \end{bmatrix}$ $\cdot \begin{bmatrix} 2\\ 4\\ 1 \end{bmatrix}$ $=\begin{bmatrix} 2(2)& 4(4)& 1(1)\\ 5(2)& 3(4)& -2(1)\\ -3(2)& -1(4)&0(1) \end{bmatrix}$ $=\begin{bmatrix} 4+16+1\\ 10+12-2\\ -6-4+0 \end{bmatrix}$ $=\begin{bmatrix} 21\\ 20\\ -10 \end{bmatrix}$

Once we have multiplied the matrices, we can perform the addition portion:

$\begin{bmatrix} 21\\ 20\\ -10 \end{bmatrix} + \begin{bmatrix} -7\\ 4\\ 11 \end{bmatrix} = \begin{bmatrix} 14\\ 24\\ 1 \end{bmatrix}$

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Example Question #2 : Matrices

Perform the following operation.

$3\left(\begin{bmatrix} 7\\ -3\\ 2 \end{bmatrix}+\begin{bmatrix} -1\\ 4\\ 6 \end{bmatrix}\right)$

Possible Answers:

$\begin{bmatrix} 20\\ -5\\ 12 \end{bmatrix}$

$\begin{bmatrix} 4\\ 9\\ 20 \end{bmatrix}$

$\begin{bmatrix} 18\\ 3\\ 24 \end{bmatrix}$

$\begin{bmatrix} 18& 3& 24 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 18\\ 3\\ 24 \end{bmatrix}$

Explanation:

The first step is to solve whatever is in the parentheses, in this case it is addition:

$\begin{bmatrix} 7\\ -3\\ 2 \end{bmatrix}+\begin{bmatrix} -1\\ 4\\ 6 \end{bmatrix}=\begin{bmatrix} 6\\ 1\\ 8 \end{bmatrix}$

We then substitute our solution into the parentheses:

$3\left(\begin{bmatrix} 6\\ 1\\ 8 \end{bmatrix}\right)$

Our next, and final step in this problem, is to carry out the multiplication:

$3\left(\begin{bmatrix} 6\\ 1\\ 8 \end{bmatrix}\right)=\begin{bmatrix} 18\\ 3\\ 24 \end{bmatrix}$

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Example Question #2 : Linear Algebra

Find the inverse of the following matrix, if possible.

$A=\begin{bmatrix} 4&-4 \\ -4&-4 \end{bmatrix}$

Possible Answers:

The inverse does not exist.

$\begin{bmatrix} \frac{1}{4}&-\frac{1}{4} \\ -\frac{1}{4}& -\frac{1}{4} \end{bmatrix}$

$-\frac{1}{32}$

$\begin{bmatrix} \frac{1}{8}&-\frac{1}{8} \\ -\frac{1}{8}& -\frac{1}{8} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \frac{1}{8}&-\frac{1}{8} \\ -\frac{1}{8}& -\frac{1}{8} \end{bmatrix}$

Explanation:

Write the formula to find the inverse of a matrix.

$\begin{bmatrix} a&b\\ c&d \end{bmatrix}^-^1=\frac{1}{ad-bc}\begin{bmatrix} d& -b\\ -c& a \end{bmatrix}$

Substituting in the given matrix we are able to find the inverse matrix.

$\begin{bmatrix} 4&-4 \\ -4&-4 \end{bmatrix}^-^1=\frac{1}{(4)(-4)-(-4)(-4)}\begin{bmatrix} -4&4 \\ 4&4 \end{bmatrix}$

$\frac{1}{-32}\begin{bmatrix} -4&4 \\ 4&4 \end{bmatrix}=\begin{bmatrix} \frac{1}{8}&-\frac{1}{8} \\ -\frac{1}{8}& -\frac{1}{8} \end{bmatrix}$

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Example Question #1 : Matrices

Find the inverse of the following matrix, if possible.

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

Possible Answers:

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

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

The inverse does not exist.

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

Correct answer:

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

Explanation:

Write the formula to find the inverse of a matrix.

$\begin{bmatrix} a&b\\ c&d \end{bmatrix}^-^1=\frac{1}{ad-bc}\begin{bmatrix} d& -b\\ -c& a \end{bmatrix}$

Using the given information we are able to find the inverse matrix.

$\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}^-^1=\frac{1}{(0)(0)-(1)(1)}\begin{bmatrix} 0&-1 \\-1&0 \end{bmatrix}$

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

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Example Question #2 : Matrices

Find the inverse of the function.

$g(x)= \frac{x+3}{2x-4}$

Possible Answers:

$g^{-1}(x)= \frac{x+3}{-4x+1}$

$g^{-1}(x)= \frac{4x+3}{2x-1}$

$g^{-1}(x)= \frac{3x+4}{x-2}$

$g^{-1}(x)= \frac{x+4}{2-x}$

Correct answer:

$g^{-1}(x)= \frac{4x+3}{2x-1}$

Explanation:

To find the inverse function, first replace g(x) with :

$y= \frac{x+3}{2x-4}$

Now replace each with an and each with a :

$x= \frac{y+3}{2y-4}$

Solve the above equation for :

x(2y-4)=y+3

2xy-4x=y+3

2xy-y=4x+3

(2x-1)y=4x+3

$y= \frac{4x+3}{2x-1}$

Replace with $g^{-1}(x)$ . This is the inverse function:

$g^{-1}(x)= \frac{4x+3}{2x-1}$

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Example Question #3 : Find The Inverse Of A Relation

Find the inverse of the function y=2x^2+1 .

Possible Answers:

$f^-^1(x)=\pm \sqrt{\frac{x-1}{2}}$

$f^-^1(x)=\pm \sqrt{x-1}$

f^-^1(x)=x+1

$f^-^1(x)=\pm \sqrt{\frac{x+1}{2}}$

f^-^1(x)=-x+1

Correct answer:

$f^-^1(x)=\pm \sqrt{\frac{x-1}{2}}$

Explanation:

To find the inverse of y=2x^2+1 , interchange the and terms and solve for .

y=2x^2+1

x=2y^2+1

x-1=2y^2

$\frac{x-1}{2}=y^2$

$f^-^1(x)=\pm \sqrt{\frac{x-1}{2}}$

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Example Question #1 : Find The Inverse Of A Function

Find the inverse of the following equation.

$y=\sqrt{\ln(x)+2}$ .

Possible Answers:

$y=\pm x^2 -2$

y=e^x+2

y=x-2

$y=e^\pm ^{^{x^2}-2}$

$y=e^x^{^{2}-2}$

Correct answer:

$y=e^x^{^{2}-2}$

Explanation:

To find the inverse in this case, we need to switch our x and y variables and then solve for y.

Therefore,

$y=\sqrt{\ln(x)+2}$ becomes,

$x=\sqrt{\ln(y)+2}$

To solve for y we square both sides to get rid of the sqaure root.

$x^2=\ln(y)+2$

We then subtract 2 from both sides and take the exponenetial of each side, leaving us with the final answer.

$x^2-2=\ln(y)$

$e^{x^2-2}=e^{\ln(y)}$

$e^x^{^{2}-2}=y$

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Example Question #4 : Matrices

Find the inverse of the following function.

y=x^3+e^0

Possible Answers:

$y^{-1}=\sqrt[3]{x+e}$

$y^{-1}=1-x^{3}}$

$y^{-1}=\sqrt[{3}]{x-1}$

$y^{-1}=e^{0}}$

Correct answer:

$y^{-1}=\sqrt[{3}]{x-1}$

Explanation:

To find the inverse of y, or

$y^{-1}$

first switch your variables x and y in the equation.

$x_{i}=y_i^3+e^0$

Second, solve for the variable y_i in the resulting equation.

x_i-e^0=y_i^3

$(x_i-e^0)^\frac{1}{3}=(y_i^{3})^\frac{1}{3}$

$y_i=\sqrt[3]{x_i-e^0}$

Simplifying a number with 0 as the power, the inverse is

$y^{-1}=\sqrt[{3}]{x-1}$

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Example Question #181 : Algebra

Find the inverse of the following function.

$y=\ln\, x\: +\: \pi$

Possible Answers:

$y^{-1}=\pi-\ln(x)$

$y^{-1}=e^{x-\pi}$

$y^{-1}=\ln(y)-\pi$

Does not exist

Correct answer:

$y^{-1}=e^{x-\pi}$

Explanation:

To find the inverse of y, or

$y^{-1}$

first switch your variables x and y in the equation.

$x_i=\ln(y_i)+\pi$

Second, solve for the variable y_i in the resulting equation.

$x_i-\pi=\ln\,y_i$

And by setting each side of the equation as powers of base e,

$y^{-1}=e^{x-\pi}$

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Example Question #3 : Linear Algebra

Find the inverse of the function.

y=x+5

Possible Answers:

$y^{-1}=5+x$

$y^{-1}=5-x$

$y^{-1}=-x-5$

$y^{-1}=x-5$

Correct answer:

$y^{-1}=x-5$

Explanation:

To find the inverse we need to switch the variables and then solve for y.

y=x+5

Switching the variables we get the following equation,

x=y+5 .

Now solve for y.

$y^{-1}=x-5$

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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 #1 : Linear Algebra

Example Question #2 : Matrices

Example Question #2 : Linear Algebra

Example Question #1 : Matrices

Example Question #2 : Matrices

Example Question #3 : Find The Inverse Of A Relation

Example Question #1 : Find The Inverse Of A Function

Example Question #4 : Matrices

Example Question #181 : Algebra

Example Question #3 : Linear Algebra

All GRE Subject Test: Math Resources

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