Non-Homogeneous Cases - Linear Algebra

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$\begin{align}&$\text{Find the values of }$\begin{bmatrix}x\y\z\end{bmatrix}\&$\text{for the function: }$\begin{bmatrix}-4&4&6\-15&0&-8\3&18&-3\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-187\172\170\end{bmatrix}\end{align}$

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$\begin{align}&$\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:}$\&\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-0.058&-0.049&0.013\0.028&0.002&0.049\0.109&-0.034&-0.024\end{bmatrix}\begin{bmatrix}-187\172\170\end{bmatrix}\&\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}4.74\3.59\-30.40\end{bmatrix}\end{align}$

$\begin{align}&$\text{Find }$\begin{bmatrix}x\y\z\end{bmatrix}\&$\text{for the system: }$\begin{bmatrix}-3&-15&6\15&-9&14\-18&-5&-10\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-14\5\-8\end{bmatrix}\end{align}$

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$- ¬\begin{align}&$\text{When faced with a system of equations like the}$\&$\text{one represented by }$\&\begin{bmatrix}-3&-15&6\15&-9&14\-18&-5&-10\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-14\5\-8\end{bmatrix}\&$\text{an approach often taught is to manipulate coefficients to eliminate}$\&$\text{variables. There is another to consider.}$\&$\text{Note that if a matrix has an inverse, }$\&$\text{then the product of that matrix and its inverse is the identity $matrix:}$\&A^{-1}$$A=AA^{-1}$=I\&$\text{Knowing this, it stands to reason that if given an equation like: }$Ax=B\&$\text{We can isolate x as $follows:}$\&A^{-1}$$Ax=A^{-1}$$B\&x=A^{-1}$B\&$\text{With this property we can solve our system:}$\&\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-0.43011&0.48387&0.41935\0.27419&-0.37097&-0.35484\0.6371&-0.68548&-0.67742\end{bmatrix}\begin{bmatrix}-14\5\-8\end{bmatrix}\&\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}5.086\-2.8548\-6.9274\end{bmatrix}\end{align}$

$\begin{align}&$\text{Find the values of }$\begin{bmatrix}x\y\end{bmatrix}\&$\text{for the function: }$\begin{bmatrix}7&15\0&-18\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-8\6\end{bmatrix}\end{align}$

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$\begin{align}&$\text{When faced with a system of equations like the}$\&$\text{one represented by }$\&\begin{bmatrix}7&15\0&-18\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-8\6\end{bmatrix}\&$\text{an approach often taught is to manipulate coefficients to eliminate}$\&$\text{variables. This approach works, though there is another to consider.}$\&$\text{If a matrix has an inverse, i.e. it has a determinant,}$\&$\text{then the product of that matrix and its inverse is the identity $matrix:}$\&A^{-1}$$A=AA^{-1}$=I\&$\text{It then stands to reason that if given an equation like: }$Ax=B\&$\text{We can isolate x as $follows:}$\&A^{-1}$$Ax=A^{-1}$$B\&x=A^{-1}$B\&$\text{Using this, we can solve for our values:}$\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}0.143&0.119\0.000&-0.056\end{bmatrix}\begin{bmatrix}-8\6\end{bmatrix}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-0.43\-0.33\end{bmatrix}\end{align}$

$\begin{align}&$\text{Find }$\begin{bmatrix}x\y\end{bmatrix}\&$\text{for the system: }$\begin{bmatrix}12&-6\-2&-5\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}7\3\end{bmatrix}\end{align}$

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$\begin{align}&$\text{When faced with a system of equations like the}$\&$\text{one represented by }$\&\begin{bmatrix}12&-6\-2&-5\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}7\3\end{bmatrix}\&$\text{an approach often taught is to manipulate coefficients to eliminate}$\&$\text{variables. This approach works, though there is another to consider.}$\&$\text{If a matrix has an inverse, i.e. it has a determinant,}$\&$\text{then the product of that matrix and its inverse is the identity $matrix:}$\&A^{-1}$$A=AA^{-1}$=I\&$\text{It then stands to reason that if given an equation like: }$Ax=B\&$\text{We can isolate x as $follows:}$\&A^{-1}$$Ax=A^{-1}$$B\&x=A^{-1}$B\&$\text{With this property we can solve our system:}$\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}0.069&-0.083\-0.028&-0.167\end{bmatrix}\begin{bmatrix}7\3\end{bmatrix}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}0.24\-0.69\end{bmatrix}\end{align}$

$\begin{align}&$\text{Find }$\begin{bmatrix}x\y\end{bmatrix}\&$\text{for the system: }$\begin{bmatrix}12&18\-20&6\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}7\-10\end{bmatrix}\end{align}$

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$\begin{align}&$\text{When faced with a system of equations like the}$\&$\text{one represented by }$\&\begin{bmatrix}12&18\-20&6\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}7\-10\end{bmatrix}\&$\text{an approach often taught is to manipulate coefficients to eliminate}$\&$\text{variables. This approach works, though there is another to consider.}$\&$\text{If a matrix has an inverse, i.e. it has a determinant,}$\&$\text{then the product of that matrix and its inverse is the identity $matrix:}$\&A^{-1}$$A=AA^{-1}$=I\&$\text{It then stands to reason that if given an equation like: }$Ax=B\&$\text{We can isolate x as $follows:}$\&A^{-1}$$Ax=A^{-1}$$B\&x=A^{-1}$B\&$\text{With this property we can solve our system:}$\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}0.014&-0.042\0.046&0.028\end{bmatrix}\begin{bmatrix}7\-10\end{bmatrix}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}0.51\0.05\end{bmatrix}\end{align}$

$\begin{align}&$\text{Find }$\begin{bmatrix}x\y\z\end{bmatrix}\&$\text{for the system: }$\begin{bmatrix}13&-19&-4\-10&-1&16\-9&4&-19\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-138\134\-122\end{bmatrix}\end{align}$

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$\begin{align}&$\text{When faced with a system of equations like the}$\&$\text{one represented by }$\&\begin{bmatrix}13&-19&-4\-10&-1&16\-9&4&-19\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-138\134\-122\end{bmatrix}\&$\text{a common approach is to manipulate coefficients to eliminate}$\&$\text{variables. This works, though there is another 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}$$A=AA^{-1}$=I\&$\text{It then stands to reason that if given an equation like: }$Ax=B\&$\text{We can isolate x as $follows:}$\&A^{-1}$$Ax=A^{-1}$$B\&x=A^{-1}$B\&$\text{With this property we can solve our system:}$\&\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-0.008&-0.063&-0.052\-0.056&-0.048&-0.028\-0.008&0.020&-0.034\end{bmatrix}\begin{bmatrix}-138\134\-122\end{bmatrix}\&\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-1.13\4.81\7.97\end{bmatrix}\end{align}$

$\begin{align}&$\text{Solve the system of equations: }$\begin{bmatrix}-5&-16\19&-3\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-167\87\end{bmatrix}\end{align}$

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$\begin{align}&$\text{When faced with a system of equations like the}$\&$\text{one represented by }$\&\begin{bmatrix}-5&-16\19&-3\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-167\87\end{bmatrix}\&$\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:}$\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-0.009&0.050\-0.060&-0.016\end{bmatrix}\begin{bmatrix}-167\87\end{bmatrix}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}5.93\8.58\end{bmatrix}\end{align}$

$\begin{align}&$\text{Find }$\begin{bmatrix}x\y\end{bmatrix}\&$\text{for the system: }$\begin{bmatrix}4&18\-16&16\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}20\-29\end{bmatrix}\end{align}$

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$\begin{align}&$\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:}$\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}0.045&-0.051\0.045&0.011\end{bmatrix}\begin{bmatrix}20\-29\end{bmatrix}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}2.39\0.58\end{bmatrix}\end{align}$

$\begin{align}&$\text{Solve the system of equations: }$\begin{bmatrix}18&-17&14\17&-20&1\6&-5&6\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}105\30\53\end{bmatrix}\end{align}$

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$\begin{align}&$\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:}$\&\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-2.212&0.615&5.058\-1.846&0.462&4.231\0.673&-0.231&-1.365\end{bmatrix}\begin{bmatrix}105\30\53\end{bmatrix}\&\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}54.31\44.23\-8.62\end{bmatrix}\end{align}$

$\begin{align}&$\text{Find the values of }$\begin{bmatrix}x\y\z\end{bmatrix}\&$\text{for the function: }$\begin{bmatrix}-2&-17&10\-19&-3&-19\20&1&17\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-47\154\-98\end{bmatrix}\end{align}$

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$\begin{align}&$\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:}$\&\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-0.022&0.207&0.245\-0.040&-0.162&-0.158\0.028&-0.234&-0.220\end{bmatrix}\begin{bmatrix}-47\154\-98\end{bmatrix}\&\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}8.98\-7.63\-15.88\end{bmatrix}\end{align}$

$\begin{align}&$\text{Find }$\begin{bmatrix}x\y\end{bmatrix}\&$\text{for the system: }$\begin{bmatrix}16&-11\-10&6\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-78\131\end{bmatrix}\end{align}$

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$\begin{align}&$\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:}$\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-0.429&-0.786\-0.714&-1.143\end{bmatrix}\begin{bmatrix}-78\131\end{bmatrix}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-69.50\-94.00\end{bmatrix}\end{align}$

$\begin{align}&$\text{Find }$\begin{bmatrix}x\y\end{bmatrix}\&$\text{for the system: }$\begin{bmatrix}1&16\-10&-11\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}136\-2\end{bmatrix}\end{align}$

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$\begin{align}&$\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:}$\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-0.074&-0.107\0.067&0.007\end{bmatrix}\begin{bmatrix}136\-2\end{bmatrix}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-9.83\9.11\end{bmatrix}\end{align}$

It is recommended that you use a calculator with matrix arithmetic capability for this question.

The graph of a quartic (degree four) polynomial passes through the following five points:

, , , ,

What is the cubic term of ?

Round to four decimal digits.

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A quartic polynomial takes the form

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

where the are the coefficients. The value we are trying to identify is , the quartic term.

An equation can be formed from each given ordered pair by substituting the abscissa for and the ordinate for . The problem can be simplified somewhat by noting that, since the -intercept of the graph of is given as , . Therefore, we are looking for the coefficients of

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

The equations formed from the other four points are

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

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

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

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

Adding two to both sides of each equation:

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

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

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

$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 , where is the matrix of coefficients, , the column matrix of variables, and the matrix of constants; this equation is

$\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 ; since , $x = $A^{-1}$b$ , or

$\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

$\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 . The correct response is $0.0104 $x^{3}$$ .

Write the initial simplex method tableau for the problem of maximizing the expression

subject to the constraints:

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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, , called slack variables, as follows:

becomes

becomes

becomes

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

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

The initial tableau for a system of linear inequalities being solved using the simplex method is

$\begin{bmatrix} 2 & 5 & 1 & 1 & 0 & 0 & 250 \ 1 & 2 & 1 & 0 & 1 & 0 & 300 \ 3 & 1 & 2 & 0 & 0 & 1& 400 \ -3 & -4 & -1 & 0 & 0 & 0 & 0 \end{bmatrix}$

The variables are .

What is represented by the first row?

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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

,

which is a constraint of the system.

The initial tableau for a system of linear inequalities being solved using the simplex method is

$\begin{bmatrix} 2 & 5 & 1 & 1 & 0 & 0 & 250 \ 1 & 2 & 1 & 0 & 1 & 0 & 300 \ 3 & 1 & 2 & 0 & 0 & 1& 400 \ -3 & -4 & -1 & 0 & 0 & 0 & 0 \end{bmatrix}$

The variables are .

Which of the following is the objective function of the system?

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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 $3 $x_{1}$+ 4 $x_{2}$+ $x_{3}$$ .

Given a minimization linear programming problem, the objective function of its dual maximization problem is .

True or false: It follows that in the original minimization problem, the objective function is .

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The statement is false. To show the reason for this, let the original minimization problem be as follows - and note that the objective function has three variables, so there will be three linear constraints other than the trivial ones:

Minimize subject to:

To find the dual maximization problem:

First, form the matrix of coefficients of the constraints (other than the positivity constraints) and the objective function, with the objective function on the bottom:

$\begin{bmatrix} $a_{11}$ & $a_{12}$ & $a_{13}$ & $b_{1}$ \ $a_{21}$ & $a_{22}$ & $a_{23}$& $b_{2}$ \ $a_{31}$ & $a_{32}$ & $a_{33}$ & $b_{3}$ \ $n_{1}$ & $n_{2}$ & $n_{3}$&0\end{bmatrix}$

Next, transpose the matrix:

$\begin{bmatrix} $a_{11}$ & $a_{21}$ & $a_{31}$ & $n_{1}$ \ $a_{12}$ & $a_{22}$ & $a_{32}$& $n_{2}$ \ $a_{13}$ & $a_{23}$ & $a_{33}$ & $n_{3}$ \ $b_{1}$ __MATH_EXPR_28__b_{2}$ & $b_{3}$&0\end{bmatrix}$

This is the coefficient matrix of the dual maximization problem. The bottom row represents the coefficients of the objective function. This function is given to be , so

.

The coefficient matrix of the original problem is

$\begin{bmatrix} $a_{11}$ & $a_{12}$ & $a_{13}$ & 4\ $a_{21}$ & $a_{22}$ & $a_{23}$& 3 \ $a_{31}$ & $a_{32}$ & $a_{33}$ & 6 \ $n_{1}$ & $n_{2}$ & $n_{3}$&0\end{bmatrix}$

Since the entries in the bottom row are still unknown, no information about the objective function of the original minimization problem has been revealed.

Give the partial fractions decomposition of .

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The denominator of the fraction, , can be factored as

.

The partial fractions decomposition of a rational expression is the sum of fractions with these factors, with the numerator above each denominator being degree one less. Thus, the partial fractions decomposition of the given expression takes the form

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

$= $$\frac{Ax+B}{x^{2}$$-3}+ $\frac{C}{x+ 4}$$

for some $A,B, C \in \mathbb{R}$ .

Express the fractions with a common denominator:

$$\frac{4}{ $(x^{2}$$-3) (x+ 4)} = $\frac{(Ax+B)(x+ $4)}{(x^{2}$$-3) (x+ 4)}+ $$\frac{C(x^{2}$$$-3)}{(x^{2}$-3) (x+ 4)}$

Eliminate the denominators and perform some algebra:

$4 = (Ax+B)(x+ 4) + $C(x^{2}$-3)$

$4 = $Ax^{2}$ +4Ax+ Bx + 4B + $Cx^{2}$ -3C$

$4 = $(A+C)x^{2}$ +(4A + B)x +( 4B -3C)$

Set the coefficients equal to form the linear system

This can be solved using Gauss-Jordan elimination on the augmented matrix

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

Perform the following row operations on the matrix to get it into reduced row-echelon form:

$-4R1+R2 \rightarrow R2$

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

$-4R2+R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & 1 & 0 \ 0 & 1 & -4 & 0 \ 0 & 0 & 13 &4 \end{bmatrix}$

$$\frac{1}{13}$R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & 1 & 0 \ \ 0 & 1 & -4 & 0 \ \ 0 & 0 & 1 & $\frac{4}{13}$ \end{bmatrix}$

$-R3 + R1 \rightarrow R1$

$4R3 + R2 \rightarrow R2$

$\begin{bmatrix} 1 & 0 & 0 & - $\frac{4}{13}$ \ \ 0 & 1 & 0 & $\frac{16}{13}$ \ \ 0 & 0 & 1 & $\frac{4}{13}$ \end{bmatrix}$

$A = - $\frac{4}{13}$ ,B = $\frac{16}{13}$ , C= $\frac{4}{13}$$ , so the partial fractions decomposition is

,

or, simplified,

.

Give the partial fractions decomposition of

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The denominator of the fraction, , can be factored as

The partial fractions decomposition of a rational expression is the sum of fractions with these factors, with the numerator above each denominator being degree one less. Thus, the partial fractions decomposition of the given expression takes the form

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

$= $\frac{A}{x-2}$+ $\frac{B}{x+2}$+ $\frac{C}{x+3}$$

for some $A,B, C \in \mathbb{R}$ .

Express the fractions with a common denominator: $\frac{4}{ (x-2) (x+2)(x+3) }$

$= $\frac{A(x+2)(x+3)}{ (x-2) (x+2)(x+3)}$+ $\frac{B(x-2)(x+3)}{ (x-2) (x+2) (x+3)}$+ $\frac{C(x-2) (x+2)}{ (x-2) (x+2)(x+3)}$$

Eliminate the denominators and perform some algebra:

$4 = $Ax^{2}$+ 5Ax+ 6A + B $x^{2}$ +Bx - 6B + $Cx^{2}$ -4C$

$4 = (A+B+C) $x^{2}$+( 5A+B) x+ (6A - 6B -4C)$

Set the coefficients equal to form the linear system

This can be solved using Gauss-Jordan elimination on the augmented matrix

$\begin{bmatrix} 1 & 1 & 1 & 0 \ 5 & 1 & 0 & 0 \ 6 & -6 & -4 & 4 \end{bmatrix}$

Perform the following row operations on the matrix to get it into reduced row-echelon form:

$-5R1+R2 \rightarrow R2$

$-6R1+R3 \rightarrow R3$

$\begin{bmatrix} 1 & 1 & 1 & 0 \ 0 & -4 & -5 & 0 \ 0 & -12 & -10 & 4 \end{bmatrix}$

$-$\frac{1}{4}$ R2 \rightarrow R2$

$\begin{bmatrix} 1 & 1 & 1 & 0 \ \0 &1 & $\frac{5}{4}$ & 0 \ \ 0 & -12 & -10 & 4 \end{bmatrix}$

$-R2+R1 \rightarrow R1$

$12R2 + R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & -$\frac{1}{4}$ & 0 \ \0 &1 & $\frac{5}{4}$ & 0 \ \ 0 & 0 & 5 & 4 \end{bmatrix}$

$$\frac{1}{5}$ R32 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & -$\frac{1}{4}$ & 0 \ \0 &1 & $\frac{5}{4}$ & 0 \ \ 0 & 0 & 1 & $\frac{4}{5}$ \end{bmatrix}$

$$\frac{1}{4}$R3 + R1 \rightarrow R1$

$-$\frac{5}{4}$R3 + R2 \rightarrow R2$

$\begin{bmatrix} 1 & 0 &0& $\frac{1}{5}$\ \0 &1 &0 & -1 \ \ 0 & 0 & 1 & $\frac{4}{5}$ \end{bmatrix}$

$A= $\frac{1}{5}$, B= -1, C= $\frac{4}{5}$$ , so the partial fractions decomposition is

,

or, simplified,

.

Solve the linear system:

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Write the augmented matrix of the system using the coefficients of the equations:

$\begin{matrix} x+ 2y + z = 9 \ 2x + y - z = 9 \ 0x+ 4y + 7z= 6 \end{matrix} \Rightarrow \begin{bmatrix} 1 & 2 & 1 & 9 \ 2 & 1 & -1 & 9 \ 0 & 4 & 7 & 6 \end{bmatrix}$

Perform the Gauss-Jordan elimination method on this matrix to get it in reduced row-echelon form. Use the following row operations:

$-2R1 + R2 \rightarrow R2$

$\begin{bmatrix} 1 & 2 & 1 & 9 \ 0 & -3 & -3 &- 9 \ 0 & 4 & 7 & 6 \end{bmatrix}$

$-$\frac{1}{3}$R2 \rightarrow R2$

$\begin{bmatrix} 1 & 2 & 1 & 9 \ 0 & 1 &1&3 \ 0 & 4 & 7 & 6 \end{bmatrix}$

$-2R2 +R1 \rightarrow R1$

$-4R2 +R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & -1 & 3 \ 0 & 1 &1&3 \ 0 & 0 & 3 &- 6 \end{bmatrix}$

$$\frac{1}{3}$ R3 \rightarrow R3$

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

$R3 + R1 \rightarrow R1$

$-R3 + R2 \rightarrow R2$

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

The matrix is now in reduced row-echelon form. This matrix can be interpreted to mean:

.

The only solution is .