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Finite Mathematics : Systems of Linear Equations: Matrices

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Example Questions

Example Question #31 : Systems Of Linear Equations: Matrices

$A= \begin{bmatrix}- \frac{4}{7} & \frac{2}{9} \\ \\ -\frac{5}{6}& \frac{3}{11} \end{bmatrix}$ and $B= \begin{bmatrix} \frac{5}{7} & \frac{2}{9} \\ \\ -\frac{5}{6}& -\frac{8}{11} \end{bmatrix}$ .

True or false: A- B = I , where is the two-by-two identity matrix.

Possible Answers:

True

False

Correct answer:

False

Explanation:

First, it must be established that A- B is defined. This is the case if and only and have the same number of rows, which is true, and they have the same number of columns, which is also true. A -B is therefore defined.

Subtraction of two matrices is performed by subtracting corresponding elements, so

$A-B= \begin{bmatrix}- \frac{4}{7} & \frac{2}{9} \\ \\ -\frac{5}{6}& \frac{3}{11} \end{bmatrix}- \begin{bmatrix} \frac{5}{7} & \frac{2}{9} \\ \\ -\frac{5}{6}& -\frac{8}{11} \end{bmatrix}$

$= \begin{bmatrix}- \frac{4}{7}-\frac{5}{7} & \frac{2}{9} - \frac{2}{9} \\ \\ -\frac{5}{6} - \left ( -\frac{5}{6} \right )& \frac{3}{11} -\left ( -\frac{8}{11} \right )\end{bmatrix}$

$= \begin{bmatrix} {\color{Red} -1} & 0 \\ 0 & 1 \end{bmatrix}$

However, $I = \begin{bmatrix} {\color{Red} 1 }& 0 \\ 0 & 1 \end{bmatrix}$ .

Therefore, $A-B \ne I$ .

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Example Question #31 : Systems Of Linear Equations: Matrices

2x+ 5y = 17

4x-y = 13

Use Cramer's rule to evaluate .

Possible Answers:

$x= \frac{21}{11}$

$x= \frac{41}{11}$

None of the other choices gives the correct response.

$x= -\frac{59}{22}$

$x= -\frac{69}{22}$

Correct answer:

$x= \frac{41}{11}$

Explanation:

By Cramer's rule, the value of is equal to $x= \frac{D_{x}}{D}$ , where is the determinant of the matrix of coefficients

$D =\begin{vmatrix} 2 & 5 \\ 4 & -1 \end{vmatrix}$ ,

and $D_{x}$ is the same matrix with the x-coefficients replaced by the constants:

$D_{x}= \begin{vmatrix} 17 & 5 \\ 13 & -1 \end{vmatrix}$

The determinant of a two-by-two matrix is equal to the product of the entries in the main diagonal minus the product of the other two entries. Therefore,

$D_{x}= \begin{vmatrix} 17 & 5 \\ 13 & -1 \end{vmatrix} = 17(-1)-5(13)= -17 -65 = -82$

$D= \begin{vmatrix} 2 & 5 \\ 4 & -1 \end{vmatrix} = 2(-1) - 5(4) = -2 - 20 = -22$

and

$x= \frac{D_{x}}{D} = \frac{-82}{-22} = \frac{41}{11}$ .

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Example Question #33 : Systems Of Linear Equations: Matrices

The initial simplex tableau for a minimization linear programming problem is given below:

$\begin{bmatrix} 2 & 3 & 2 &1 & 0 & 0 & 1,200 \\ 1 & 2 & 4 & 0 & 1 & 0 & 600\\ 0 & 4 & 3 & 0 & 0 & 1 & 900 \\ -1 & -1 & -2 & 0&0&0 & 0 \end{bmatrix}$

After the first pivot of the simplex method is performed, what will be the entry at bottom right?

Possible Answers:

400

1,200

300

600

Correct answer:

300

Explanation:

First, the pivot element must be identified. Select the pivot column by spotting the "most negative" element in the bottom, or objective, row. This is the in Column 3, making that the pivot column.

$\begin{bmatrix} 2 & 3 & 2 &1 & 0 & 0 & 1,200 \\ 1 & 2 & 4 & 0 & 1 & 0 & 600\\ 0 & 4 & 3 & 0 & 0 & 1 & 900 \\ -1 & -1 & {\color{Red} \textbf{-2}} & 0&0&0 & 0 \end{bmatrix}$

Select the pivot row by dividing the rightmost element in each other row by the pivot-column element in that row. The "least positive" quotient determines the pivot row:

$\begin{bmatrix} 2 & 3 & {\color{Blue} \textbf{2}} &1 & 0 & 0 & {\color{Blue} \textbf{1,200}} \\ 1 & 2 & {\color{Blue} \textbf{4}} & 0 & 1 & 0 & {\color{Blue} \textbf{600}}\\ 0 & 4 & {\color{Blue} \textbf{3}} & 0 & 0 & 1 & {\color{Blue} \textbf{900}} \\ -1 & -1 & {\color{Red} \textbf{-2}} & 0&0&0 & 0 \end{bmatrix}\begin{matrix} 1,200 \div 2 = 600 \\ 600 \div 4 = {\color{Red} \textbf{150}} \\ 900 \div 3 = 300\\\; \end{matrix}$

The "least positive" of the three quotients is 150, in Row 2. The pivot element is the 4 in Row 2, Column 3.

Perform the pivot by first multiplying each entry in Row 2 by $\frac{1}{4}$ in order to make the pivot element a 1:

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

$\begin{bmatrix} 2 & 3 & 2 &1 & 0 & 0 & 1,200 \\ \frac{1}{4} & \frac{1}{2} & 1 & 0 & \frac{1}{4} & 0 & 150\\ 0 & 4 & 3 & 0 & 0 & 1 & 900 \\ -1 & -1 & -2 & 0&0&0 & 0 \end{bmatrix}$

Make the other elements in that column zeroes by adding multiples of the pivot row to the other rows, as follows:

$-2R2 + R 1 \rightarrow R1$

$-3R2 + R3 \rightarrow R3$

$2R2 + R4 \rightarrow R4$

$\begin{bmatrix} \frac{3}{2} & 2 & 0 &1 & -\frac{1}{2} & 0 & 900 \\ \frac{1}{4} & \frac{1}{2} & 1 & 0 & \frac{1}{4} & 0 & 150\\ -\frac{3}{4} & \frac{5}{2} & 0 & 0 & -\frac{3}{4} & 1 & 450 \\ -\frac{1}{2} & 0& 0 & 0&\frac{1}{2}&0 & {\color{Red} \textbf{300}} \end{bmatrix}$

The entry at bottom right is 300.

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Finite Mathematics : Systems of Linear Equations: Matrices

Study concepts, example questions & explanations for Finite Mathematics

All Finite Mathematics Resources

Example Questions

Example Question #31 : Systems Of Linear Equations: Matrices

Example Question #31 : Systems Of Linear Equations: Matrices

Example Question #33 : Systems Of Linear Equations: Matrices

All Finite Mathematics Resources

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