$\displaystyle \begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}4&18\\-16&16\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}20\\-29\end{bmatrix}\\&\text{a common approach is to manipulate coefficients to eliminate}\\&\text{variables. This works, though there is another way to consider.}\\&\text{If a matrix has an inverse,}\\&\text{then the product of that matrix and its inverse is the identity matrix:}\\&A^{-1}A=AA^{-1}=I\\&\text{As such, if given an equation like: }Ax=B\\&\text{we can isolate x as follows:}\\&A^{-1}Ax=A^{-1}B\\&x=A^{-1}B\\&\text{With this property we can solve our system:}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0.045&-0.051\\0.045&0.011\end{bmatrix}\begin{bmatrix}20\\-29\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2.39\\0.58\end{bmatrix}\end{align*}$

$\displaystyle \begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-4&4&6\\-15&0&-8\\3&18&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-187\\172\\170\end{bmatrix}\\&\text{a common approach is to manipulate coefficients to eliminate}\\&\text{variables. This works, though there is another way to consider.}\\&\text{If a matrix has an inverse,}\\&\text{then the product of that matrix and its inverse is the identity matrix:}\\&A^{-1}A=AA^{-1}=I\\&\text{As such, if given an equation like: }Ax=B\\&\text{we can isolate x as follows:}\\&A^{-1}Ax=A^{-1}B\\&x=A^{-1}B\\&\text{Using this, we can solve for our values:}\\&\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-0.058&-0.049&0.013\\0.028&0.002&0.049\\0.109&-0.034&-0.024\end{bmatrix}\begin{bmatrix}-187\\172\\170\end{bmatrix}\\&\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4.74\\3.59\\-30.40\end{bmatrix}\end{align*}$

$\displaystyle \begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}18&-17&14\\17&-20&1\\6&-5&6\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}105\\30\\53\end{bmatrix}\\&\text{a common approach is to manipulate coefficients to eliminate}\\&\text{variables. This works, though there is another way to consider.}\\&\text{If a matrix has an inverse,}\\&\text{then the product of that matrix and its inverse is the identity matrix:}\\&A^{-1}A=AA^{-1}=I\\&\text{As such, if given an equation like: }Ax=B\\&\text{we can isolate x as follows:}\\&A^{-1}Ax=A^{-1}B\\&x=A^{-1}B\\&\text{With this, we can solve our system of equations:}\\&\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-2.212&0.615&5.058\\-1.846&0.462&4.231\\0.673&-0.231&-1.365\end{bmatrix}\begin{bmatrix}105\\30\\53\end{bmatrix}\\&\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}54.31\\44.23\\-8.62\end{bmatrix}\end{align*}$

$\displaystyle \begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-2&-17&10\\-19&-3&-19\\20&1&17\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-47\\154\\-98\end{bmatrix}\\&\text{a common approach is to manipulate coefficients to eliminate}\\&\text{variables. This works, though there is another way to consider.}\\&\text{If a matrix has an inverse,}\\&\text{then the product of that matrix and its inverse is the identity matrix:}\\&A^{-1}A=AA^{-1}=I\\&\text{As such, if given an equation like: }Ax=B\\&\text{we can isolate x as follows:}\\&A^{-1}Ax=A^{-1}B\\&x=A^{-1}B\\&\text{Using this, we can solve for our values:}\\&\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-0.022&0.207&0.245\\-0.040&-0.162&-0.158\\0.028&-0.234&-0.220\end{bmatrix}\begin{bmatrix}-47\\154\\-98\end{bmatrix}\\&\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}8.98\\-7.63\\-15.88\end{bmatrix}\end{align*}$

$\displaystyle \begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}16&-11\\-10&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-78\\131\end{bmatrix}\\&\text{a common approach is to manipulate coefficients to eliminate}\\&\text{variables. This works, though there is another way to consider.}\\&\text{If a matrix has an inverse,}\\&\text{then the product of that matrix and its inverse is the identity matrix:}\\&A^{-1}A=AA^{-1}=I\\&\text{As such, if given an equation like: }Ax=B\\&\text{we can isolate x as follows:}\\&A^{-1}Ax=A^{-1}B\\&x=A^{-1}B\\&\text{With this property we can solve our system:}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-0.429&-0.786\\-0.714&-1.143\end{bmatrix}\begin{bmatrix}-78\\131\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-69.50\\-94.00\end{bmatrix}\end{align*}$

$\displaystyle \begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}1&16\\-10&-11\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}136\\-2\end{bmatrix}\\&\text{a common approach is to manipulate coefficients to eliminate}\\&\text{variables. This works, though there is another way to consider.}\\&\text{If a matrix has an inverse,}\\&\text{then the product of that matrix and its inverse is the identity matrix:}\\&A^{-1}A=AA^{-1}=I\\&\text{As such, if given an equation like: }Ax=B\\&\text{we can isolate x as follows:}\\&A^{-1}Ax=A^{-1}B\\&x=A^{-1}B\\&\text{With this property we can solve our system:}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-0.074&-0.107\\0.067&0.007\end{bmatrix}\begin{bmatrix}136\\-2\end{bmatrix}\\&\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-9.83\\9.11\end{bmatrix}\end{align*}$

A quartic polynomial takes the form 

$\displaystyle p(x) = a_{4}x^{4} + a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0}$

where the $\displaystyle a_{n}$ are the coefficients. The value we are trying to identify is $\displaystyle a_{4}$, the quartic term.

An equation can be formed from each given ordered pair by substituting the abscissa for $\displaystyle x$ and the ordinate for $\displaystyle p(x)$. The problem can be simplified somewhat by noting that, since the $\displaystyle y$-intercept of the graph of $\displaystyle p$ is given as $\displaystyle (0 , -2 )$, $\displaystyle a_{0} = -2$. Therefore, we are looking for the coefficients of 

$\displaystyle p(x) = a_{4}x^{4} + a_{3}x^{3} + a_{2}x^{2} + a_{1}x -2$.

The equations formed from the other four points are

$\displaystyle 256 a_{4} -64a_{3} + 16a_{2} - 4a_{1} -2 = 0$

$\displaystyle 16 a_{4} -8a_{3} + 4a_{2} - 2a_{1} -2 = 6$

$\displaystyle 16 a_{4} +8a_{3} + 4a_{2} +2a_{1} -2 = 5$

$\displaystyle 256 a_{4} +64a_{3} + 16a_{2} + 4a_{1} -2 = -1$

Adding two to both sides of each equation:

$\displaystyle 256 a_{4} -64a_{3} + 16a_{2} - 4a_{1} =2$

$\displaystyle 16 a_{4} -8a_{3} + 4a_{2} - 2a_{1} =8$

$\displaystyle 16 a_{4} +8a_{3} + 4a_{2} +2a_{1} =7$

$\displaystyle 256 a_{4} +64a_{3} + 16a_{2} + 4a_{1} = 1$

This is a system of four linear equations in four variables, which can be rewritten as the matrix multiplication equation $\displaystyle A x = b$, where $\displaystyle A$ is the matrix of coefficients, $\displaystyle x$, the column matrix of variables, and $\displaystyle b$ the matrix of constants; this equation is 

$\displaystyle \begin{bmatrix} 256 & -64 & 16 & -4 \\ 16 & -8 & 4 & -2 \\ 16 & 8 & 4 & 2 \\ 256 & 64 & 16 & 4 \end{bmatrix}\begin{bmatrix} a_{4} \\a_{3} \\ a _{2} \\ a_{1} \end{bmatrix}=\begin{bmatrix}2 \\8\\7\\1 \end{bmatrix}$

We are trying to calculate $\displaystyle x$; since $\displaystyle A x = b$,  $\displaystyle x = A^{-1}b$, or

$\displaystyle \begin{bmatrix} a_{4} \\a_{3} \\ a _{2} \\ a_{1} \end{bmatrix}=\begin{bmatrix} 256 & -64 & 16 & -4 \\ 16 & -8 & 4 & -2 \\ 16 & 8 & 4 & 2 \\ 256 & 64 & 16 & 4 \end{bmatrix} ^{-1}\begin{bmatrix}2 \\8\\7\\1 \end{bmatrix}$

This calculates to

$\displaystyle \begin{bmatrix} a_{4} \\a_{3} \\ a _{2} \\ a_{1} \end{bmatrix}= \begin{bmatrix}-0.1484 \\ 0.0104\\2.4469\\-0.2917 \end{bmatrix}$.

Since the cubic term is requested, select $\displaystyle a_{3} = 0.0104$. The correct response is $\displaystyle 0.0104 x^{3}$.

The bottom row of an initial simplex tableau is the one that represents the objective function to be minimized; the other rows represent constraints on the variables in the system. In a system of three variables, the first three entries of one of the rows represent the coefficients of those variables; the next three represent those of slack variables introduced into the system, which were not part of the original inequality. The last entry gives the number the linear expression is less than or equal to. Therefore, the first row represents the inequality

$\displaystyle 2x_{1}+ 5x_{2}+ x_{3} \le 250$,

which is a constraint of the system.

The objective function in a simplex method - the function to be minimized - is represented by the bottom row of the initial tableau, with the coefficients the additive inverses of the entries in that row. The objective function of the system is therefore $\displaystyle 3 x_{1}+ 4 x_{2}+ x_{3}$.

When writing the initial simplex method tableau for a system of linear inequalities, first, recast each inequality as an equation by introducing three other variables, $\displaystyle s_{1}, s_{2}, s_{3}$, called slack variables, as follows:

$\displaystyle x_{1} - x_{3} \le 2,000$ 

becomes

$\displaystyle x_{1} - x_{3}+ s_{1}= 2,000$

$\displaystyle x_{2}+ x_{3} \le 3,000$

becomes 

$\displaystyle x_{2}+ x_{3}+ s_{2} = 3,000$

$\displaystyle x_{1}- 3x_{2} \le 4,000$

becomes 

$\displaystyle x_{1}- 3x_{2} +s_{3}= 4,000$

The system of three inequalities in three equations becomes a system of three equations in six variables. The coefficients form the first three rows of the augmented matrix that sets up the initial simplex tableau.

The bottom row comprises the additive inverses of the coefficients of the expression to be maximized, or the objective function. 

The initial tableau is therefore

$\displaystyle \begin{bmatrix} 1 & 0 & -1 & 1 & 0 & 0 & 2,000 \\ 0 & 1 & 1 & 0 & 1 & 0 & 3,000 \\ 1 & 0 & -3 & 0 & 0 & 1 & 4,000 \\ -4 & -2 & -1 & 0 & 0 & 0 & 0 \end{bmatrix}$.