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GRE Subject Test: Math : Algebra

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Example Questions

Example Question #1 : Linear Algebra

Find the inverse of the following equation.

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

Possible Answers:

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

$y=\pm x^2 -2$

y=e^x+2

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

y=x-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 : Matrices

Find the inverse of the function.

y=x+5

Possible Answers:

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

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

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

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

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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Example Question #1 : Find The Inverse Of A Function

If $f(x)=-\frac{3x}{5}+11$ , what is its inverse function, $f^{-1}(x)$ ?

Possible Answers:

$f^{-1}(x)=\frac{5x}{3}+\frac{55}{3}$

$f^{-1}(x)=-\frac{5x}{3}+\frac{55}{3}$

$f^{-1}(x)=-\frac{5x}{3}-\frac{55}{3}$

$f^{-1}(x)=-\frac{3x}{5}+\frac{33}{5}$

$f^{-1}(x)=-\frac{3x}{5}-\frac{33}{5}$

Correct answer:

$f^{-1}(x)=-\frac{5x}{3}+\frac{55}{3}$

Explanation:

We begin by taking $f(x)=-\frac{3x}{5}+11$ and changing the f(x) to a , giving us $y=-\frac{3x}{5}+11$ .

Next, we switch all of our x's and y's , giving us $x=-\frac{3y}{5}+11$ .

Finally, we solve for by subtracting from each side, multiplying each side by , and dividing each side by , leaving us with,

$y=-\frac{5x}{3}+\frac{55}{3}$ .

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Example Question #4 : Inverses

Find $f^{-1}(x)$ for $f(x)=\frac{2x-7}{8x+3}$

Possible Answers:

$f^{-1}(x)=\frac{3x^{2}+4x}{8}$

$f^{-1}(x)=\sqrt{x^{2}+6}$

$f^{-1}(x)=\frac{x^{2}}{x+5}$

$f^{-1}(x)=\frac{2x+7}{4-3x}$

$f^{-1}(x)=\frac{3x+7}{2-8x}$

Correct answer:

$f^{-1}(x)=\frac{3x+7}{2-8x}$

Explanation:

$f(x)=\frac{2x-7}{8x+3}$

To find the inverse of a function, first swap the x and y in the given function.

$x=\frac{2y-7}{8y+3}$

Solve for y in this re-written form.

x(8y+3)=2y-7

8xy+3x=2y-7

8xy-2y=-7-3x

y(8x-2)=-7-3x

$y=\frac{-7-3x}{8x-2}$

$y=\frac{-(7+3x)}{-(-8x+2)}$

$y=\frac{3x+7}{2-8x}$

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Example Question #371 : Gre Subject Test: Math

Which of the following is the inverse of h(x) ?

h(x)=3lnx

Possible Answers:

$y=\frac{e^x}{3}$

$y=e^{\frac{x}{3}}$

$y=3e^{x}$

$y=9e^{x}$

Correct answer:

$y=e^{\frac{x}{3}}$

Explanation:

Which of the following is the inverse of h(x) ?

h(x)=3lnx

To find the inverse of a function, we need to swap x and y, and then rearrange to solve for y. The inverse of a function is basically the function we get if we swap the x and y coordinates for every point on the original function.

So, to begin, we can replace the h(x) with y.

y=3lnx

Next, swap x and y

x=3lny

Now, we need to get y all by itself; we can to begin by dividng the three over.

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

Now, recall that

And that we can rewrite any log as an exponent as follows:

log_b(a)=x

b^x=a

So with that in mind, we can rearrange our function to get y by itself:

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

Becomes our final answer:

$y=e^{\frac{x}{3}}$

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Example Question #181 : Algebra

Find the Inverse of Matrix B $(B^{-1})$ where

$B=\begin{bmatrix} 2&4 \\ 5&8 \end{bmatrix}$ .

Possible Answers:

$B^{-1}=\frac{-1}{4}\begin{bmatrix} 8&4 \\ 5& 2 \end{bmatrix}$

$B^{-1}=\frac{1}{4}\begin{bmatrix} 2&4 \\ 5& 8 \end{bmatrix}$

$B^{-1}=\frac{1}{4}\begin{bmatrix} 8&-4 \\ -5& 2 \end{bmatrix}$

$B^{-1}=\frac{-1}{4}\begin{bmatrix} 2&-4 \\ -5& 8 \end{bmatrix}$

$B^{-1}=\frac{-1}{4}\begin{bmatrix} 8&-4 \\ -5& 2 \end{bmatrix}$

Correct answer:

$B^{-1}=\frac{-1}{4}\begin{bmatrix} 8&-4 \\ -5& 2 \end{bmatrix}$

Explanation:

To find the inverse matrix of B use the following formula,

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

Since the matrix B is given as,

$B=\begin{bmatrix} 2&4 \\ 5&8 \end{bmatrix}$

the inverse becomes,

$B^{-1}=\frac{1}{2*8-5*4}\begin{bmatrix} 8&-4 \\ -5& 2 \end{bmatrix}$

$B^{-1}=\frac{-1}{4}\begin{bmatrix} 8&-4 \\ -5& 2 \end{bmatrix}$ .

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Example Question #11 : 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.

$-\frac{1}{32}$

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

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

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 #11 : Matrices

Given the following matrix, find the determinant, if possible.

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

Possible Answers:

There is no determinant.

Correct answer:

Explanation:

Write the formula to find the determinant given a 2 by 2 matrix.

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

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

$\begin{bmatrix} -4&4 \\ 4&-4 \end{bmatrix}=(-4)(-4)-(4)(4)=16-16=0$

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GRE Subject Test: Math : 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 #4 : Matrices

Example Question #181 : Algebra

Example Question #3 : Matrices

Example Question #1 : Find The Inverse Of A Function

Example Question #4 : Inverses

Example Question #371 : Gre Subject Test: Math

Example Question #181 : Algebra

Example Question #11 : Linear Algebra

Example Question #11 : Matrices

All GRE Subject Test: Math Resources

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