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

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

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

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

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

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

We then substitute our solution into the parentheses:

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

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

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

Write the formula to find the inverse of a matrix.

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

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

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

\(\displaystyle \frac{1}{-32}\begin{bmatrix} -4&4 \\ 4&4 \end{bmatrix}=\begin{bmatrix} \frac{1}{8}&-\frac{1}{8} \\ -\frac{1}{8}& -\frac{1}{8} \end{bmatrix}\)

Write the formula to find the inverse of a matrix.

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

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

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

\(\displaystyle -1\begin{bmatrix} 0&-1 \\ -1&0 \end{bmatrix}=\begin{bmatrix} 0& 1\\ 1&0 \end{bmatrix}\)

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

\(\displaystyle y= \frac{x+3}{2x-4}\)

Now replace each \(\displaystyle y\) with an \(\displaystyle x\) and each \(\displaystyle x\) with a \(\displaystyle y\):

\(\displaystyle x= \frac{y+3}{2y-4}\)

Solve the above equation for \(\displaystyle y\):

\(\displaystyle x(2y-4)=y+3\)

\(\displaystyle 2xy-4x=y+3\)

\(\displaystyle 2xy-y=4x+3\)

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

\(\displaystyle y= \frac{4x+3}{2x-1}\)

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

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

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

\(\displaystyle y=2x^2+1\)

\(\displaystyle x=2y^2+1\)

\(\displaystyle x-1=2y^2\)

\(\displaystyle \frac{x-1}{2}=y^2\)

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

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

Therefore,

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

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

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

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

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

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

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

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

To find the inverse of y, or 

\(\displaystyle y^{-1}\)

first switch your variables x and y in the equation. 

 \(\displaystyle x_{i}=y_i^3+e^0\)

Second, solve for the variable \(\displaystyle y_i\) in the resulting equation. 

\(\displaystyle x_i-e^0=y_i^3\)

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

\(\displaystyle y_i=\sqrt[3]{x_i-e^0}\)

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

\(\displaystyle y^{-1}=\sqrt[{3}]{x-1}\)

To find the inverse of y, or 

\(\displaystyle y^{-1}\)

first switch your variables x and y in the equation. 

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

Second, solve for the variable \(\displaystyle y_i\) in the resulting equation. 

\(\displaystyle x_i-\pi=\ln\,y_i\)

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

\(\displaystyle y^{-1}=e^{x-\pi}\)

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

\(\displaystyle y=x+5\)

Switching the variables we get the following equation,

\(\displaystyle x=y+5\).

Now solve for y.

\(\displaystyle y^{-1}=x-5\)