Linear Algebra - GRE Subject Test: Math

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Question

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

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

Answer

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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Question

Evaluate the determinant of the following matrix.

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

Answer

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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Question

Evaluate the determinant of the following matrix.

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

Answer

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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Question

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

Answer

Step 1: We need to recall how to find the determinant of a matrix. To find the determinant of a $2\ast2$ matrix, we need to use the equation , where =Determinant, and 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.

. We see that there are two negative signs in the middle, which will become a plus sign.

. Simplify the right side.

. is the determinant.

The determinant of the matrix is .

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Question

Find the determinate of Matrix A.

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

Answer

Matrix A is given below.

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

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

Plugging in the values gives us:

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Question

Find the eigenvalues of the following matrix, if possible.

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

Answer

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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Question

Find the eigenvalues of the following matrix, if possible.

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

Answer

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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Question

Find the inverse of the following equation.

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

Answer

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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Question

Find the inverse of the following function.

Answer

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 in the resulting equation.

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

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

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

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

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Question

Find the inverse of the following function.

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

Answer

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 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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Question

Find the inverse of the function.

Answer

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

Switching the variables we get the following equation,

.

Now solve for y.

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

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Question

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

Answer

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

Next, we switch all of our and , 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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Question

Find the inverse of the function.

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

Answer

To find the inverse function, first replace 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 :

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

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

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

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Question

Find the inverse of the function .

Answer

To find the inverse of , interchange the and terms and solve for .

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

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

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Question

Find the inverse of the following matrix, if possible.

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

Answer

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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Question

Find the inverse of the following matrix, if possible.

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

Answer

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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Question

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

Answer

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

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.

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

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

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

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Question

Which of the following is the inverse of ?

Answer

Which of the following is the inverse of ?

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.

Next, swap x and y

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:

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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Question

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

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

Answer

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,

$B^{-1}=\frac{1}{28-54}\begin{bmatrix} 8&-4 \ -5& 2 \end{bmatrix}$

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

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Question

Find the inverse of the following matrix, if possible.

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

Answer

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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