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GRE Subject Test: Math : Linear Algebra

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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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Example Question #13 : Linear Algebra

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

h(x)=3lnx

Possible Answers:

$y=3e^{x}$

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

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

$y=9e^{x}$

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 #14 : Linear Algebra

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

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

Possible Answers:

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

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

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

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

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

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 #15 : 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.

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

$-\frac{1}{32}$

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

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 #181 : Algebra

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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Example Question #17 : Linear Algebra

Evaluate the determinant of the following matrix.

$\begin{bmatrix} 5& -3\\ 7& 8 \end{bmatrix}$

Possible Answers:

Correct answer:

Explanation:

Remember, to evaluate the determinant of a matrix use the following:

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

The first step would be to write the determinant of the matrix:

$\begin{vmatrix} 5& -3\\ 7&8 \end{vmatrix}$

Now we can evaluate:

$\begin{vmatrix} 5& -3\\ 7&8 \end{vmatrix}=(5)(8)- (-3)(7)=40+21=61$

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Example Question #18 : Linear Algebra

Evaluate the determinant of the following matrix.

$\begin{bmatrix} 2& 4& 3\\ 3& 0& 1\\ 1& 2& 5 \end{bmatrix}$

Possible Answers:

Correct answer:

Explanation:

To find the determinant of a 3 x 3 matrix, we must use the following:

$\begin{vmatrix} a& b& c\\ d& e& f\\ g& h&k \end{vmatrix}=a\begin{vmatrix} e& f\\ h& k \end{vmatrix} -b\begin{vmatrix} d& f\\ g& k \end{vmatrix} + c\begin{vmatrix} d& e\\ g& h \end{vmatrix}$

The first thing we must do is write the determinant:

$\begin{vmatrix} 2& 4& 3\\ 3& 0& 1\\ 1& 2& 5 \end{vmatrix}$

Now we can proceed to evaluate the determinant

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

Notice that the numbers 2, 4, and 3 are being multiplied by the determinants of the 2x2 matrices so we have:

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Example Question #1 : Determinants

Find the determinant of the matrix: $\begin{vmatrix} -4 & -7 \\ 1 & 2 \end{vmatrix}$

Possible Answers:

Correct answer:

Explanation:

Step 1: We need to recall how to find the determinant of a 2*2 matrix. To find the determinant of a $2\ast2$ matrix, we need to use the equation D=ad-bc , where =Determinant, and a,b,c,d are from the matrix.

Step 2: Identify a,b,c, and d in the original matrix.
a=first number on top row, b=second number on top row (next to a), c=first number on the bottom row, and d is the second number on the bottom row (next to c).

In this matrix, a=, b=, c=, and d=.

Step 3: Substitute the values of a,b,c, and d into the equation to find the determinant of the matrix.

$ad-bc=(-4\cdot 2)-(-7\cdot 1)$ We will simplify the right side.

ad-bc=(-8)-(-7) . We see that there are two negative signs in the middle, which will become a plus sign.

ad-bc=-8+7 . Simplify the right side.

ad-bc=-1 . is the determinant.

The determinant of the matrix is .

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Example Question #1 : Determinants

Find the determinate of Matrix A.

$A=\begin{bmatrix} 8&2 \\ 5& 3 \end{bmatrix}$

Possible Answers:

Correct answer:

Explanation:

Matrix A is given below.

$A=\begin{bmatrix} a&b \\ c& d \end{bmatrix}$

The formula for the determinate of a 2x2 matrix is:

det(A)=ad-bc

Plugging in the values gives us:

det(A)=8*3-5*2=24-10=14

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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 #11 : Linear Algebra

Example Question #12 : Linear Algebra

Example Question #13 : Linear Algebra

Example Question #14 : Linear Algebra

Example Question #15 : Linear Algebra

Example Question #181 : Algebra

Example Question #17 : Linear Algebra

Example Question #18 : Linear Algebra

Example Question #1 : Determinants

Example Question #1 : Determinants

All GRE Subject Test: Math Resources

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