MTH120 College Algebra Chapter 7.8

 7.8 Solving Systems with Cramer's Rule

I know what you are thinking it must be a Seinfeld reference to Kramer but that isn't the case. Then again most of likely never saw Seinfeld great show though. Cramer's Rule is a method for solving a system of linear equations using determinants. It is applicable when the system is represented in matrix form $��=�$, where:

• $�$ is the coefficient matrix,
• $�$ is the column vector of variables, and
• $�$ is the column vector of constants on the right-hand side.

Cramer's Rule provides a formula for finding the solution vector ${�}_{�}$ for each variable ${�}_{�}$:

${�}_{�}=\frac{\text{det}({�}_{�})}{\text{det}(�)}$

Here, ${�}_{�}$ is obtained by replacing the $�$-th column of $�$ with the column vector $�$, and $\text{det}(�)$ is the determinant of matrix $�$.

It's important to note that Cramer's Rule is only applicable when the determinant of the coefficient matrix $�$ is non-zero. If $\text{det}(�)=0$, Cramer's Rule cannot be used.

Example: Solving a System of Equations Using Cramer's Rule

Consider the following system of equations:

1. $2�+3�=8$
2. $4�-3�=5$

This system can be represented in matrix form as $��=�$, where:

$�=\left[\begin{array}{cc}2& 3\\ 4& -3\end{array}\right]$ $�=\left[\begin{array}{c}�\\ �\end{array}\right]$ $�=\left[\begin{array}{c}8\\ 5\end{array}\right]$

Step 1: Calculate the Determinant of $�$

$\text{det}(�)=(2\cdot (-3))-(3\cdot 4)=-18-12=-30$

Since $\text{det}(�)\mathrm{\ne }0$, Cramer's Rule can be applied.

Step 2: Calculate $�$ and $�$

Use Cramer's Rule to find $�$ and $�$:

$�=\frac{\text{det}({�}_{�})}{\text{det}(�)}$

$�=\frac{\text{det}({�}_{�})}{\text{det}(�)}$

Where ${�}_{�}$ is obtained by replacing the first column of $�$ with the column vector $�$, and ${�}_{�}$ is obtained by replacing the second column of $�$ with $�$.

${�}_{�}=\left[\begin{array}{cc}8& 3\\ 5& -3\end{array}\right]$

${�}_{�}=\left[\begin{array}{cc}2& 8\\ 4& 5\end{array}\right]$

Now calculate the determinants:

$\text{det}({�}_{�})=(8\cdot (-3))-(3\cdot 5)=-24-15=-39$

$\text{det}({�}_{�})=(2\cdot 5)-(8\cdot 4)=10-32=-22$

Now, calculate $�$ and $�$:

$�=\frac{\text{det}({�}_{�})}{\text{det}(�)}=\frac{-39}{-30}=\frac{13}{10}$

$�=\frac{\text{det}({�}_{�})}{\text{det}(�)}=\frac{-22}{-30}=\frac{11}{15}$

So, the solution to the system of equations is $�=\frac{13}{10}$ and $$�=1115

Evaluating the determinant of a 2x2 matrix is straightforward and can be done using a simple formula. For a 2x2 matrix:

$�=\left[\begin{array}{cc}�& �\\ �& �\end{array}\right]$

The determinant $\text{det}(�)$ is calculated as:

$\text{det}(�)=(�\cdot �)-(�\cdot �)$

Let's go through a few examples:

Example 1: Determinant of a 2x2 Matrix

Consider the matrix:

$�=\left[\begin{array}{cc}3& 4\\ 2& 1\end{array}\right]$

Using the formula, calculate the determinant:

$\text{det}(�)=(3\cdot 1)-(4\cdot 2)=3-8=-5$

So, the determinant of matrix $�$ is -5.

Example 2: Determinant of a 2x2 Matrix

Let's take another matrix:

$�=\left[\begin{array}{cc}-2& 5\\ 7& 0\end{array}\right]$

Calculate the determinant:

$\text{det}(�)=(-2\cdot 0)-(5\cdot 7)=0-35=-35$

The determinant of matrix $�$ is -35.

In both examples, we used the formula to find the determinant of 2x2 matrices. The process is relatively simple and extends to any 2x2 matrix.

Cramer's Rule is a method for solving a system of two equations in two variables using determinants. It's applicable when the system can be represented as:

1. $��+��=�$
2. $��+��=�$

To use Cramer's Rule, you need to calculate the following determinants:

1. $\text{det}(�)$ (the determinant of the coefficient matrix)
2. $\text{det}({�}_{�})$ (the determinant of $�$ with the $�$-constants replaced)
3. $\text{det}({�}_{�})$ (the determinant of $�$ with the $�$-constants replaced)

Then, you can find $�$ and $�$ using the following formulas:

$�=\frac{\text{det}({�}_{�})}{\text{det}(�)}$ $�=\frac{\text{det}({�}_{�})}{\text{det}(�)}$

Let's go through an example:

Example: Solving a System of Two Equations Using Cramer's Rule

Consider the following system of equations:

1. $2�+3�=8$
2. $4�-3�=5$

We can represent this system as:

• $�=2$
• $�=3$
• $�=4$
• $�=-3$
• $�=8$
• $�=5$

Step 1: Calculate the Determinants

1. Calculate the determinant of the coefficient matrix $�$:

   $�=\mid \begin{array}{cc}2& 3\\ 4& -3\end{array}\mid =(2\cdot (-3))-(3\cdot 4)=-6-12=-18$

2. Calculate the determinant ${�}_{�}$ by replacing the first column of $�$ with the constants:

   ${�}_{�}=\mid \begin{array}{cc}8& 3\\ 5& -3\end{array}\mid =(8\cdot (-3))-(3\cdot 5)=-24-15=-39$

3. Calculate the determinant ${�}_{�}$ by replacing the second column of $�$ with the constants:

   ${�}_{�}=\mid \begin{array}{cc}2& 8\\ 4& 5\end{array}\mid =(2\cdot 5)-(8\cdot 4)=10-32=-22$

Step 2: Calculate $�$ and $�$

Now, use Cramer's Rule to find $�$ and (y):

$�=\frac{\text{det}({�}_{�})}{\text{det}(�)}=\frac{-39}{-18}=\frac{39}{18}=\frac{13}{6}$

$�=\frac{\text{det}({�}_{�})}{\text{det}(�)}=\frac{-22}{-18}=\frac{22}{18}=\frac{11}{9}$

So, the solution to the system of equations is $�=\frac{13}{6}$ and $�=\frac{11}{9}$.

Cramer's Rule for 2x2 systems of linear equations is a straightforward method for solving equations with two variables using determinants. For a system of equations in the form:

1. $��+��=�$
2. $��+��=�$

You can use Cramer's Rule to find the solutions for $�$ and $�$ as follows:

Step 1: Calculate the Determinant of the Coefficient Matrix $�$

The coefficient matrix $�$ is formed by the coefficients of the variables:

$�=\mid \begin{array}{cc}�& �\\ �& �\end{array}\mid =��-��$

Step 2: Calculate $�$ and $�$

Now, you can find $�$ and $�$ using the following formulas:

$�=\frac{\mid \begin{array}{cc}�& �\\ �& �\end{array}\mid }{\mid \begin{array}{cc}�& �\\ �& �\end{array}\mid }=\frac{��-��}{��-��}$

$�=\frac{\mid \begin{array}{cc}�& �\\ �& �\end{array}\mid }{\mid \begin{array}{cc}�& �\\ �& �\end{array}\mid }=\frac{��-��}{��-��}$

Here are a couple of examples:

Example 1:

Solve the system of equations:

1. $3�+2�=8$
2. $4�-3�=5$

In this case, $�=3$, $�=2$, $�=4$, $�=-3$, $�=8$, and $�=5$.

Step 1: Calculate $�$

$�=(3\cdot (-3))-(2\cdot 4)=-9-8=-17$

Step 2: Calculate $�$ and $�$

$�=\frac{(8\cdot (-3))-(2\cdot 5)}{-17}=\frac{(-24-10)}{-17}=\frac{-34}{-17}=2$

$�=\frac{(3\cdot 5)-(8\cdot 4)}{-17}=\frac{(15-32)}{-17}=\frac{-17}{-17}=1$

So, the solution is $�=2$ and $�=1$.

Example 2:

Solve the system of equations:

1. $2�-�=3$
2. $4�+3�=7$

In this case, $�=2$, $�=-1$, $�=4$, $�=3$, $�=3$, and $�=7$.

Step 1: Calculate $�$

$�=(2\cdot 3)-(-1\cdot 4)=6+4=10$

Step 2: Calculate $�$ and $�$

$�=\frac{(3\cdot 3)-(-1\cdot 7)}{10}=\frac{(9+7)}{10}=\frac{16}{10}=\frac{8}{5}$

$�=\frac{(2\cdot 7)-(4\cdot 3)}{10}=\frac{(14-12)}{10}=\frac{2}{10}=\frac{1}{5}$

So, the solution is $�=\frac{8}{5}$ and $�=\frac{1}{5}$.

Evaluating the determinant of a 3x3 matrix can be done using a more general formula for a 3x3 matrix. For a 3x3 matrix:

$�=\left[\begin{array}{ccc}�& �& �\\ �& �& �\\ �& ℎ& �\end{array}\right]$

The determinant $\text{det}(�)$ is calculated as:

$\text{det}(�)=�(��-�ℎ)-�(��-��)+�(�ℎ-��)$

Let's go through an example:

Example: Determinant of a 3x3 Matrix

Consider the matrix:

$�=\left[\begin{array}{ccc}2& -1& 3\\ 0& 4& 5\\ 1& 2& -2\end{array}\right]$

Using the formula, calculate the determinant:

$\text{det}(�)=2(4(-2)-5(2))-(-1)(0(-2)-5(1))+3(0(2)-4(1))$

$\text{det}(�)=2(-8-10)+(1)(0-5)+3(0-4)$

$\text{det}(�)=-36+0-12$

$\text{det}(�)=-48$

So, the determinant of the matrix $�$ is -48.

This general formula for a 3x3 matrix determinant can be applied to any 3x3 matrix to calculate its determinant efficiently.

Cramer's Rule can be extended to solve systems of three equations in three variables. The system can be represented as:

1. $��+��+��=�$
2. $��+��+��=ℎ$
3. $��+��+��=�$

To use Cramer's Rule for a 3x3 system, you need to calculate the following determinants:

1. $\text{det}(�)$ (the determinant of the coefficient matrix)
2. $\text{det}({�}_{�})$ (the determinant of $�$ with the $�$-constants replaced)
3. $\text{det}({�}_{�})$ (the determinant of $�$ with the $�$-constants replaced)
4. $\text{det}({�}_{�})$ (the determinant of $�$ with the $�$-constants replaced)

Then, you can find $�$, $�$, and $�$ using the following formulas:

$�=\frac{\text{det}({�}_{�})}{\text{det}(�)}$ $�=\frac{\text{det}({�}_{�})}{\text{det}(�)}$ $�=\frac{\text{det}({�}_{�})}{\text{det}(�)}$

Let's go through an example:

Example: Solving a System of Three Equations Using Cramer's Rule

Consider the system of equations:

1. $2�-�+3�=8$
2. $4�+5�+2�=10$
3. $6�+7�+3�=12$

Step 1: Set Up the Coefficient Matrix $�$

$�=\mid \begin{array}{ccc}2& -1& 3\\ 4& 5& 2\\ 6& 7& 3\end{array}\mid$

Step 2: Calculate $\text{det}(�)$

$\text{det}(�)=(2\cdot 5\cdot 3)+(1\cdot 2\cdot 6)+(3\cdot 4\cdot 7)-(3\cdot 5\cdot 6)-(2\cdot 4\cdot 7)-(1\cdot 2\cdot 3)$

$\text{det}(�)=30+12+84-90-56-6$

$\text{det}(�)=74$

Step 3: Calculate $\text{det}({�}_{�})$, $\text{det}({�}_{�})$, and $\text{det}({�}_{�})$

Replace the columns of $�$ with the constants:

${�}_{�}=\mid \begin{array}{ccc}8& -1& 3\\ 10& 5& 2\\ 12& 7& 3\end{array}\mid$

${�}_{�}=\mid \begin{array}{ccc}2& 8& 3\\ 4& 10& 2\\ 6& 12& 3\end{array}\mid$

${�}_{�}=\mid \begin{array}{ccc}2& -1& 8\\ 4& 5& 10\\ 6& 7& 12\end{array}\mid$

Now, calculate each determinant:

$\text{det}({�}_{�})=(8\cdot 5\cdot 3)+(-1\cdot 2\cdot 12)+(3\cdot 10\cdot 7)-(3\cdot 5\cdot 12)-(2\cdot 10\cdot 7)-(-1\cdot 2\cdot 6)$

$\text{det}({�}_{�})=120+24+210-180-140+12$

$\text{det}({�}_{�})=46$

Similarly, calculate $\text{det}({�}_{�})=-36$ and $\text{det}({�}_{�})=96$.

Step 4: Calculate $�$, $�$, and $�$

$�=\frac{\text{det}({�}_{�})}{\text{det}(�)}=\frac{46}{74}$

$�=\frac{\text{det}({�}_{�})}{\text{det}(�)}=\frac{-36}{74}$

$�=\frac{\text{det}({�}_{�})}{\text{det}(�)}=\frac{96}{74}$

So, the solution is $�=\frac{23}{37}$, $�=-\frac{18}{37}$, and $�=\frac{48}{37}$.

Determinants are important mathematical quantities associated with square matrices. They have several properties that are useful when performing matrix operations. Here are some key properties of determinants with examples:

1. Determinant of the Identity Matrix:

   • The determinant of the identity matrix $�$ is equal to 1, regardless of its size.

   Example: For a 2x2 identity matrix, $\text{det}(�)=1$.

2. Scalar Multiplication:

   • If you multiply a matrix by a scalar $�$, the determinant is multiplied by $�$.

   Example: If $�=\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]$, then $\text{det}(2�)=2\text{det}(�)=2(2\cdot 3)=12$.

3. Row or Column Multiplication:

   • If you multiply a single row or column of a matrix by a scalar $�$, the determinant is multiplied by $�$.

   Example: If $�=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$, and you multiply the first column by 5, you get $�=\left[\begin{array}{cc}5& 2\\ 15& 4\end{array}\right]$. The determinant of $�$ is $\text{det}(�)=5\cdot 4-2\cdot 15=20-30=-10$, which is 5 times the determinant of $�$.

4. Row or Column Addition:

   • If you add a multiple of one row (or column) to another row (or column), the determinant remains unchanged.

   Example: If $�=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$, and you replace the second row with the sum of the first row and twice the second row, you get $�=\left[\begin{array}{cc}1& 2\\ 7& 10\end{array}\right]$. The determinant of $�$ is the same as the determinant of $�$.

5. Transpose of a Matrix:

   • The determinant of a matrix and its transpose are the same.

   Example: If $�=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$, and ${�}^{�}$ is the transpose of $�$, then $\text{det}(�)=\text{det}({�}^{�})$.

6. Determinant of a Product:

   • The determinant of a product of matrices is the product of the determinants of the individual matrices.

   Example: If $�$ and $�$ are two matrices, then $\text{det}(��)=\text{det}(�)\cdot \text{det}(�)$.

These properties are important for simplifying calculations involving determinants, especially when dealing with larger matrices and when using determinants to solve systems of linear equations or calculate the inverse of a matrix.

Cramer's Rule, a method for solving systems of linear equations using determinants, has applications in various real-world scenarios. Here are some examples:

1. Electrical Circuits: In electrical engineering, Cramer's Rule can be used to solve complex circuits with multiple unknown currents. Each junction or loop in the circuit can be represented by a linear equation, and Cramer's Rule can help find the current in each branch.

2. Equations: ${�}_1-{�}_2-{�}_3=0$, $2{�}_1+3{�}_2-{�}_3=5$, ${�}_1+{�}_2-2{�}_3=-3$

3. Chemical Reactions: In chemistry, chemical reactions can be described by a system of linear equations. The coefficients of reactants and products in chemical reactions can be used as coefficients in these equations. Cramer's Rule can help determine the quantities of reactants and products involved in chemical reactions.

4. Equations: $2�+3�=4$, $4�-2�=10$

5. Structural Analysis: Engineers use Cramer's Rule in structural analysis to calculate the forces and stresses in various parts of a structure. For example, it can be applied to determine the forces acting on different elements of a bridge or a building to ensure structural integrity.

6. Finance and Investment: Cramer's Rule can be used in finance to optimize investment portfolios. It helps find the optimal allocation of assets in a portfolio to maximize returns while minimizing risk. The relationships between different investment options can be represented as a system of equations.

7. Chemical Engineering: Chemical engineers often encounter systems of equations that represent mass and energy balances in chemical processes. Cramer's Rule can be applied to determine the concentrations of various substances or the energy flows within a chemical reactor.

8. Optics and Lens Design: In optics, the design of lenses and optical systems can involve solving systems of equations related to the behavior of light rays. Cramer's Rule can help determine the properties of lenses and optical components to achieve specific optical outcomes.

9. Traffic Flow and Network Analysis: Transportation engineers use Cramer's Rule to model and analyze traffic flow in transportation networks. The equations represent traffic volumes, travel times, and congestion, and Cramer's Rule can be applied to optimize traffic management strategies.

10. Game Theory: In game theory, Cramer's Rule can be used to find solutions to two-player games with simultaneous moves. The payoffs and strategies of players can be represented as a system of equations, and Cramer's Rule can determine the equilibrium strategies.

11. Biological Modeling: In biology, mathematical models are often used to describe biological processes, such as population growth, disease spread, or genetic inheritance. Cramer's Rule can help solve these models by finding the values of parameters.

These examples demonstrate the versatility of Cramer's Rule in solving systems of equations in various fields, making it a valuable tool for engineers, scientists, and analysts.

College Algebra Chapter 7 Quiz

Problem 1: The system of equations is:

1. $-5�-�=12$
2. $12�+4�=9$

The ordered pair is $(-3,3)$. To check if it's a solution, substitute these values into the equations:

For Equation 1: $-5(-3)-3=15-3=12$

For Equation 2: $12(-3)+4(3)=-36+12=-24$

So, the ordered pair $(-3,3)$ is not a solution to the system of equations.

Problem 2: The system of equations is:

1. $12�-13�=43$
2. $2�-�=0$

Let's solve this system using elimination:

First, multiply Equation 2 by 13 to make the coefficients of $�$ in both equations equal:

1. $12�-13�=43$
2. $26�-13�=0$

Now, subtract Equation 2 from Equation 1:

$(12�-13�)-(26�-13�)=43-0$

Simplify:

$-14�=43$

Now, solve for $�$:

$�=-\frac{43}{14}$

Substitute this value of $�$ into Equation 2 to find $�$:

$2(-\frac{43}{14})-�=0$

Simplify:

$-\frac{43}{7}-�=0$

Now, solve for $�$:

$�=-\frac{43}{7}$

So, the solution to the system is $�=-\frac{43}{14}$ and $�=-\frac{43}{7}$.

Problem 3: The system of equations is:

1. $-12�-4�=4$
2. $2�+16�=2$

We'll solve this using elimination. First, let's simplify Equation 1:

1. $-3�-�=1$

Now, multiply Equation 1 by 4 and Equation 2 by 1 to make the coefficients of $�$ equal:

1. $-12�-4�=4$
2. $2�+16�=2$

Now, add Equation 1 and Equation 2:

$(-12�-4�)+(2�+16�)=4+2$

Simplify:

$-10�+12�=6$

Now, solve for $�$:

$-10�=6-12�$

$-10�=6-12�$

Simplify:

$�=-\frac{6-12�}{10}$

Simplify further:

$�=-\frac{3-6�}{5}$

Now, we have an expression for $�$ in terms of $�$. This system has infinitely many solutions because it represents a line.

Problem 4: The system of equations is:

1. $5�-�=1$
2. $-10�+2�=-2$

Let's solve this system using elimination. First, multiply Equation 1 by 2:

1. $10�-2�=2$

Now, add Equation 1 and Equation 2:

$(5�-�)+(10�-2�)=1+2$

Simplify:

$15�-3�=3$

Now, divide both sides by 3 to simplify further:

$5�-�=1$

This equation is equivalent to Equation 1. The two equations represent the same line, so there are infinitely many solutions.

Problem 5: The system of equations is:

1. $4�-6�-2�=1$
2. $10�-7�+5�=-14$
3. $3�+6�-9�=65$

Let's use elimination to solve this system. Start by adding Equation 1, Equation 2, and Equation 3:

$(4�-6�-2�)+(10�-7�+5�)+(3�+6�-9�)=1-14+65$

Simplify:

$17�-7�=52$

Now, we have an equation involving $�$ and $�$, while $�$ is eliminated. This system represents a plane in 3D space. There are infinitely many solutions along this plane.

Problem 6: The system of equations is:

1. $�+�=20$
2. $�+�+�=20$
3. $�+2�+�=10$

Let's solve this system using substitution. From Equation 1, we can express $�$ in terms of $�$:

z=20−x

Now, substitute this expression for $�$ into Equation 2 and Equation 3:

2. $�+�+(20-�)=20$
3. $�+2�+(20-�)=10$

Simplify these equations:

2. $�=0$
3. $�+2�+20-�=10$

Simplify further:

3. $2�+10=10$

Now, solve for $�$:

$2�=0$

$�=0$

So, the solution to this system is $�=10$, $�=0$, and $�=10$.

Problem 7: The system of equations is:

1. $5�-4�-3�=0$
2. $2�+�+2�=0$
3. $-6�-7�=0$

Let's solve this system using elimination. First, solve Equation 3 for $�$:

$�=-\frac{7�}{6}$

Now, substitute this expression for $�$ into Equation 1 and Equation 2:

1. $5(-\frac{7�}{6})-4�-3�=0$
2. $2(-\frac{7�}{6})+�+2�=0$

Simplify these equations:

1. $-\frac{35�}{6}-4�-3�=0$
2. $-\frac{7�}{3}+�+2�=0$

Now, add Equation 1 and Equation 2:

$-\frac{35�}{6}-4�-3�-\frac{7�}{3}+�+2�=0$

Simplify:

$-\frac{35�}{6}-\frac{7�}{3}-4�+�-3�+2�=0$

Further simplify:

$-\frac{49�}{6}-3�=0$

Now, solve for $�$:

$-3�=\frac{49�}{6}$

$�=-\frac{49�}{18}$

This system has infinitely many solutions, and the solutions lie along a line in 3D space.

Problem 8: The system of equations is:

1. $�={�}^2+2�-3$
2. $�=�-1$

Let's solve this system by setting the two expressions for $�$ equal to each other:

${�}^2+2�-3=�-1$

Now, simplify and solve for $�$:

${�}^2+2�-3-�+1=0$

Combine like terms:

${�}^2+�-2=0$

Now, factor the quadratic equation:

$(�-1)(�+2)=0$

Set each factor equal to zero:

1. $�-1=0$ --> $�=1$
2. $�+2=0$ --> $�=-2$

Now that we have values for $�$, we can find the corresponding values of $�$:

For $�=1$: $�=1^2+2(1)-3=0$

For $�=-2$: $�=(-2{)}^2+2(-2)-3=1$

So, the system has two solutions: $(1,0)$ and $(-2,1)$.

Problem 9: The system of equations is:

1. ${�}^2+{�}^2=25$
2. ${�}^2-2{�}^2=1$

Let's solve this system by isolating ${�}^2$ in both equations:

1. ${�}^2=25-{�}^2$
2. ${�}^2=1+2{�}^2$

Now, set these two expressions for ${�}^2$ equal to each other:

$25-{�}^2=1+2{�}^2$

Now, simplify and solve for $�$:

$25-{�}^2-1-2{�}^2=0$

Combine like terms:

$-3{�}^2+24=0$

Now, divide by -3:

${�}^2-8=0$

Factor the quadratic equation:

$(�-2\sqrt{2})(�+2\sqrt{2})=0$

Set each factor equal to zero:

1. $�-2\sqrt{2}=0$ --> $�=2\sqrt{2}$
2. $�+2\sqrt{2}=0$ --> $�=-2\sqrt{2}$

Now that we have values for $�$, we can find the corresponding values of $�$ using the expressions for ${�}^2$:

For $�=2\sqrt{2}$: ${�}^2=25-(2\sqrt{2}{)}^2=25-8=17$ $�=\sqrt{17}$ or $�=-\sqrt{17}$

For $�=-2\sqrt{2}$: ${�}^2=25-(-2\sqrt{2}{)}^2=25-8=17$ $�=\sqrt{17}$ or $�=-\sqrt{17}$

So, the system has four solutions: $(2\sqrt{2},\sqrt{17})$, $(-2\sqrt{2},\sqrt{17})$, $(2\sqrt{2},-\sqrt{17})$, and $(-2\sqrt{2},-\sqrt{17})$

Problem 10: Graph the inequality $�<{�}^2+9$.

This inequality represents the area below the parabola $�={�}^2+9$. The graph is the region under the curve.

Problem 11: Graph the inequality ${�}^2+{�}^2>4�$ and ${�}^2+1$.

This inequality represents the area outside the circle with center (0, 2) and radius 2 (given by ${�}^2+(�-2{)}^2=4$) and the area above the line $�={�}^2+1$. The solution is the region outside the circle and above the parabola.

Problem 12: Perform a partial fraction decomposition for $-8�-30{�}^2+10�+25$.

$-8�-30{�}^2+10�+25=-2(�+5)+4(�-5)-5(�+5)$

So, the partial fraction decomposition is $-2(�+5)+4(�-5)-5(�+5)$.

Problem 13: Perform a partial fraction decomposition for $13�+2(3�+1{)}^2$.

$13�+2(3�+1{)}^2=13�+2(9{�}^2+6�+1)=13�+18{�}^2+12�+2=18{�}^2+25�+2$

Problem 14: Perform a partial fraction decomposition for $\frac{{�}^4-{�}^3+2�-1}{�({�}^2+1{)}^2}$.

The partial fraction decomposition is:

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

Problem 15: Perform the matrix operation $5[4-29]+12[-6412-8]$.

$5[4-29]+12[-6412-8]=[20-1045]+[-7248144-96]=[-5238189-104]$

Problem 16: Perform the matrix multiplication:

$\left[\begin{array}{ccc}1& -2& 12\\ 4& 9& 0\\ -7& -5& -4\end{array}\right]$

multiplied by

$\left[\begin{array}{ccc}3& 15& -4\\ -3& 10& -4\\ 31& 5& -8\end{array}\right]$

Problem 17: Find the inverse of the matrix $\left[\begin{array}{cc}12& 14\\ 13& 15\end{array}\right]$.

Problem 18: Calculate the determinant of the matrix $\mid \begin{array}{cc}0& 4\\ 0& 0\end{array}\mid$.

Problem 19: Calculate the determinant of the matrix $\mid \begin{array}{ccc}1& 2& 1\\ 2& 1& 0\\ 1& 0& 1\end{array}\mid$.

Problem 20: If $\text{det}(�)=-6$, switching rows 1 and 3, multiplying the second row by 12, and taking the inverse would result in a determinant of $1\mathrm{/}6$.

Problem 21: Rewrite the system of linear equations as an augmented matrix.

The system of equations can be written as an augmented matrix as follows:

$$<span\; style="background-color:\; white;">[</span>\begin{array}{ccccc}\mathrm{<span\; style="background-color:\; white;">14</span>}& <span\; style="background-color:\; white;">-</span>\mathrm{<span\; style="background-color:\; white;">2</span>}& \mathrm{<span\; style="background-color:\; white;">13</span>}& \mathrm{<span\; style="background-color:\; white;">\mid </span>}& \mathrm{<span\; style="background-color:\; white;">140</span>}\\ <span\; style="background-color:\; white;">-</span>\mathrm{<span\; style="background-color:\; white;">2</span>}& \mathrm{<span\; style="background-color:\; white;">3</span>}& <span\; style="background-color:\; white;">-</span>\mathrm{<span\; style="background-color:\; white;">6</span>}& \mathrm{<span\; style="background-color:\; white;">\mid </span>}& <span\; style="background-color:\; white;">-</span>\mathrm{<span\; style="background-color:\; white;">1</span>}\\ \mathrm{<span\; style="background-color:\; white;">1</span>}& <span\; style="background-color:\; white;">-</span>\mathrm{<span\; style="background-color:\; white;">5</span>}& \mathrm{<span\; style="background-color:\; white;">12</span>}& \mathrm{<span\; style="background-color:\; white;">\mid </span>}& \mathrm{<span\; style="background-color:\; white;">11</span>}\end{array}<span\; style="background-color:\; white;">]</span>$$

Problem 22: Rewrite the augmented matrix as a system of linear equations.

The augmented matrix can be rewritten as the following system of linear equations:

1. $14�-2�+13�=140$
2. $-2�+3�-6�=-1$
3. $�-5�+12�=11$