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GRE Subject Test: Math : Inverses

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Example Question #3 : Linear Algebra

Find the inverse of the following matrix, if possible.

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

Possible Answers:

$-\frac{1}{32}$

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

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

The inverse does not exist.

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 #4 : Linear Algebra

Find the inverse of the following matrix, if possible.

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

Possible Answers:

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

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

The inverse does not exist.

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

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 #5 : Linear Algebra

Find the inverse of the function.

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

Possible Answers:

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

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

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

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

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 #6 : Linear Algebra

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

Possible Answers:

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

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

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 #3 : 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=x-2

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

y=e^x+2

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

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 : Find The Inverse Of A Function

Find the inverse of the following function.

y=x^3+e^0

Possible Answers:

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

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

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

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

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 #5 : Find The Inverse Of A Function

Find the inverse of the following function.

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

Possible Answers:

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

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

Does not exist

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

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 #1 : Inverses

Find the inverse of the function.

y=x+5

Possible Answers:

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

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

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

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

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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Example Question #11 : Linear Algebra

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

Possible Answers:

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

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

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

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

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

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 #12 : Linear Algebra

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

Possible Answers:

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

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

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

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

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

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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GRE Subject Test: Math : Inverses

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

All GRE Subject Test: Math Resources

Example Questions

Example Question #3 : Linear Algebra

Example Question #4 : Linear Algebra

Example Question #5 : Linear Algebra

Example Question #6 : Linear Algebra

Example Question #3 : Find The Inverse Of A Function

Example Question #4 : Find The Inverse Of A Function

Example Question #5 : Find The Inverse Of A Function

Example Question #1 : Inverses

Example Question #11 : Linear Algebra

Example Question #12 : Linear Algebra

All GRE Subject Test: Math Resources

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