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Finite Mathematics : Systems of Linear Equations: Matrices

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Example Questions

Example Question #23 : Finite Mathematics

Solve the linear system:

5x + 2y = 80

-10x - 4y = -160

Possible Answers:

$\left ( -\frac{2}{5}y - 16, y \right )$ , arbitrary.

$\left ( -\frac{2}{5}y + 16, y \right )$ , arbitrary.

$\left ( \frac{2}{5}y - 16, y \right )$ , arbitrary.

$\left ( \frac{2}{5}y + 16, y \right )$ , arbitrary.

The system has no solution.

Correct answer:

$\left ( -\frac{2}{5}y + 16, y \right )$ , arbitrary.

Explanation:

First, make the y-coefficients each other's opposite. This can be done by multiplying the first equation by 2 on both sides:

5x + 2y = 80

2(5x + 2y )= 2(80)

10x+4y = 160

The y-coefficients of the two equations are now opposites, so, if the left and right sides of the two equations are added, the y-terms will cancel out, as follows:

$\begin{matrix} \; \; 10x+4y = \; \; \; 160 \\\underline{ -10x - 4y = -160}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; {\color{Red} 0 = \; \; \; \; \; \; 0} \end{matrix}$

The resulting statement is identically true. It follows that the two equations of the system form a dependent system. The solution set can be found by allowing to be arbitrary, and to solve for in either equation. Selecting the first:

5x + 2y = 80

5x = -2y + 80

$x = -\frac{2}{5}y + 16$

The solution set is

$\left ( -\frac{2}{5}y + 16, y \right )$ , arbitrary.

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Example Question #23 : Finite Mathematics

It is suggested (but not necessary) that you use a calculator with matrix capability to work this problem.

A small factory produces Food A and Food B.

For every pound of Food A produced, one ounce of Food A and two ounces of Food B are consumed. For every pound of Food B produced, one half ounce of Food A and one ounce of Food B are consumed. The factory must meet a monthly demand of two thousand pounds of Food A and three thousand pounds of Food B.

Use a Leontief model to determine how many pounds of Food A and Food B must be produced to meet the demand. Then answer the question - to the nearest whole pound, how many pounds of Food A must be produced?

Possible Answers:

2,571

2,857

2, 714

2, 143

2,286

Correct answer:

2,571

Explanation:

The Leontief model for an input-output system requires the following steps.

Let be the technology matrix for the system, the entries of which are the amounts of each product consumed in the manufacture of each product. Letting the first row/column represent Food A and the second row/column represent Food B, this matrix is

$A = \begin{bmatrix} \frac{1}{16} & \frac{1}{8} \\ \\ \frac{1}{32} & \frac{1}{16} \end{bmatrix}$

Note that ounces have been converted to pounds.

Let be the external demand vector, which gives the amount of each product demanded by outside consumers:

$D= \begin{bmatrix} 2,000\\ \\ 3,000 \end{bmatrix}$

The output vector , which gives the amount of each product that must be produced to meet the demand, can be obtained through the matrix equation

$X = (I- A )^{-1}D$

$=\left ( \begin{bmatrix} 1 & 0 \\ \\ 0 & 1 \end{bmatrix}- \begin{bmatrix} \frac{1}{16} & \frac{1}{8} \\ \\ \frac{1}{32} & \frac{1}{16} \end{bmatrix} \right )^{-1} \begin{bmatrix} 2,000\\ \\ 3,000 \end{bmatrix}$

$= \begin{bmatrix} \frac{15}{16} & -\frac{1}{8} \\ \\ -\frac{1}{32} & \frac{15}{16} \end{bmatrix} ^{-1} \begin{bmatrix} 2,000\\ \\ 3,000 \end{bmatrix}$

$= \begin{bmatrix}\frac{15}{14} & \frac{1}{7}\\ \\ \frac{1}{28} & \frac{15}{14}\end{bmatrix} \begin{bmatrix} 2,000\\ \\ 3,000 \end{bmatrix}$

$\approx \begin{bmatrix} 2,571 \\ 3,286 \end{bmatrix}$

The top entry is the number of pounds of Food A that must be produced - 2,571 pounds.

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Example Question #24 : Finite Mathematics

It is suggested (but not necessary) that you use a calculator with matrix capability to work this problem.

A small factory produces Food A and Food B.

Possible Answers:

3, 143

3,286

3,571

3, 714

3,857

Correct answer:

3,286

Explanation:

The Leontief model for an input-output system requires the following steps.

$A = \begin{bmatrix} \frac{1}{16} & \frac{1}{8} \\ \\ \frac{1}{32} & \frac{1}{16} \end{bmatrix}$

Note that ounces have been converted to pounds.

Let be the external demand vector, which gives the amount of each product demanded by outside consumers:

$D= \begin{bmatrix} 2,000\\ \\ 3,000 \end{bmatrix}$

The output vector , which gives the amount of each product that must be produced to meet the demand, can be obtained through the matrix equation

$X = (I- A )^{-1}D$

$= \begin{bmatrix} \frac{15}{16} & -\frac{1}{8} \\ \\ -\frac{1}{32} & \frac{15}{16} \end{bmatrix} ^{-1} \begin{bmatrix} 2,000\\ \\ 3,000 \end{bmatrix}$

$= \begin{bmatrix}\frac{15}{14} & \frac{1}{7}\\ \\ \frac{1}{28} & \frac{15}{14}\end{bmatrix} \begin{bmatrix} 2,000\\ \\ 3,000 \end{bmatrix}$

$\approx \begin{bmatrix} 2,571 \\ 3,286 \end{bmatrix}$

The bottom entry is the number of pounds of Food B that must be produced - 3,286 pounds.

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Example Question #24 : Finite Mathematics

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

Find $A^{-1}$

Possible Answers:

$\begin{bmatrix} \frac{ 1}{5} & 0 & \frac{ 1}{5} \\ 0 & \frac{1}{3} & 0 \\ \frac{ 1}{5} & 0 & -\frac{4 }{5}\end{bmatrix}$

$\begin{bmatrix} \frac{ 1}{5} & 0 & \frac{ 1}{5} \\ 0 &- \frac{1}{3} & 0 \\ \frac{ 1}{5} & 0 & -\frac{4 }{5}\end{bmatrix}$

$\begin{bmatrix} \frac{ 1}{5} & 0 & -\frac{ 1}{5} \\ 0 & \frac{1}{3} & 0 \\ -\frac{ 1}{5} & 0 & \frac{4 }{5}\end{bmatrix}$

$\begin{bmatrix} \frac{ 1}{5} & 0 & \frac{ 1}{5} \\ 0 & \frac{1}{3} & 0 \\ -\frac{ 1}{5} & 0 & \frac{4 }{5}\end{bmatrix}$

$\begin{bmatrix} \frac{ 1}{5} & 0 & \frac{ 1}{5} \\ 0 & -\frac{1}{3} & 0 \\ -\frac{ 1}{5} & 0 & \frac{4 }{5}\end{bmatrix}$

Correct answer:

$\begin{bmatrix} \frac{ 1}{5} & 0 & \frac{ 1}{5} \\ 0 & \frac{1}{3} & 0 \\ \frac{ 1}{5} & 0 & -\frac{4 }{5}\end{bmatrix}$

Explanation:

To find $A^{-1}$ , the inverse of , set up the augmented matrix $\left [ A\; \; | \; \; I \right ]$ as follows:

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

Apply the Gauss-Jordan elimination process to this matrix by performing row operations on it until the matrix $\left [ I\; \; | \; \; A^{-1} \right ]$ is formed - that is,

$\begin{bmatrix} 1 & 0 & 0 & \times& \times& \times \\ 0 & 1 & 0 & \times& \times& \times \\ 0 & 0 & 1 & \times& \times& \times \end{bmatrix}$ .

Work each of the first three columns in left to right order until the entry in the diagonal position is a one and the other two entries are zeroes.

One possible set of row operations, with the resulting matrices, is as follows:

$R1 \leftrightarrow R3$

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

$-4R1 + R3 \rightarrow R3$

$\begin{bmatrix}1 & 0 & -1 & 0 & 0 & 1 \\ 0 & 3 & 0 & 0 & 1 & 0 \\ 0& 0 & 5 & 1 & 0 & -4 \end{bmatrix}$

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

$\begin{bmatrix}1 & 0 & -1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & \frac{1}{3} & 0 \\ 0& 0 & 5 & 1 & 0 & -4 \end{bmatrix}$

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

$\begin{bmatrix}1 & 0 & -1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & \frac{1}{3} & 0 \\ 0& 0 & 1 &\frac{ 1}{5} & 0 & -\frac{4 }{5}\end{bmatrix}$

$R3+ R1 \rightarrow R1$

$\begin{bmatrix}1 & 0 & 0 & \frac{ 1}{5} & 0 & \frac{ 1}{5} \\ 0 & 1 & 0 & 0 & \frac{1}{3} & 0 \\ 0& 0 & 1 &\frac{ 1}{5} & 0 & -\frac{4 }{5}\end{bmatrix}$

This is the desired result. The inverse is seen in the last three columns:

$A^{-1}= \begin{bmatrix} \frac{ 1}{5} & 0 & \frac{ 1}{5} \\ 0 & \frac{1}{3} & 0 \\ \frac{ 1}{5} & 0 & -\frac{4 }{5}\end{bmatrix}$

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Example Question #22 : Finite Mathematics

$A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & -2 & 0 \\ 1 & 2 & - 4 \end{bmatrix}$

Find $A^{-1}$

Possible Answers:

$\begin{bmatrix} 1 & \frac{1}{2} & -\frac{1}{2} \\0 & - \frac{1}{2} & \frac{1}{4} \\ 0 & 0& \frac{1}{4} \end{bmatrix}$

$\begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{2} & - \frac{1}{2} & 0 \\ \frac{1}{2} & -\frac{1}{4} & -\frac{1}{4} \end{bmatrix}$

$\begin{bmatrix} 1 & 1 & 1 \\ 0& - \frac{1}{2} & \frac{1}{2} \\ 0 &0& -\frac{1}{4} \end{bmatrix}$

$\begin{bmatrix} 1 & 0 & 0 \\ 1& - \frac{1}{2} & 0 \\ 1 & \frac{1}{2} & -\frac{1}{4} \end{bmatrix}$

$\begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{2} & - \frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{1}{4} & \frac{1}{4} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{2} & - \frac{1}{2} & 0 \\ \frac{1}{2} & -\frac{1}{4} & -\frac{1}{4} \end{bmatrix}$

Explanation:

To find $A^{-1}$ , the inverse of , set up the augmented matrix $\left [ A\; \; | \; \; I \right ]$ as follows:

$\begin{bmatrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 1 & -2 & 0 & 0 & 1 & 0 \\ 1 & 2 & - 4& 0 & 0 & 1 \end{bmatrix}$

Apply the Gauss-Jordan elimination process to this matrix by performing row operations on it until the matrix $\left [ I\; \; | \; \; A^{-1} \right ]$ is formed - that is,

$\begin{bmatrix} 1 & 0 & 0 & \times& \times& \times \\ 0 & 1 & 0 & \times& \times& \times \\ 0 & 0 & 1 & \times& \times& \times \end{bmatrix}$ .

Work each of the first three columns, right-to-left, until the entry in the diagonal position is a one and the other two entries are zeroes.

One possible sequence of row operations, with the resulting matrices, is as follows:

$-R1 + R2 \rightarrow R2$

$-R1 + R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & -2 & 0 & -1 & 1 & 0 \\ 0& 2 & - 4& -1 & 0 & 1 \end{bmatrix}$

$-\frac{1}{2} R2 \rightarrow R2$

$\begin{bmatrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & \frac{1}{2} & - \frac{1}{2} & 0 \\ 0& 2 & - 4& -1 & 0 & 1 \end{bmatrix}$

$-2R2 + R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & \frac{1}{2} & - \frac{1}{2} & 0 \\ 0& 0 & - 4& -2 & 1 & 1 \end{bmatrix}$

$-\frac{1}{4} R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & \frac{1}{2} & - \frac{1}{2} & 0 \\ 0& 0 & 1& \frac{1}{2} & -\frac{1}{4} &-\frac{1}{4} \end{bmatrix}$

This is the desired result. The inverse is seen in the last three columns:

$A^{-1}= \begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{2} & - \frac{1}{2} & 0 \\ \frac{1}{2} & -\frac{1}{4} & -\frac{1}{4} \end{bmatrix}$

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Example Question #1 : Solve A System Of Equations In Three Variables Using Augmented Matrices

$A = \begin{bmatrix} 0.7 & -0.4 \\ -0.8 & 0.2 \end{bmatrix}$ and $B= \begin{bmatrix} 0. 3 & 0.4 \\ 0.8 & 0.8 \end{bmatrix}$ .

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

Possible Answers:

True

False

Correct answer:

True

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.

Addition of two matrices is performed by adding corresponding elements, so

$A +B= \begin{bmatrix} 0.7 & -0.4 \\ -0.8 & 0.2 \end{bmatrix}+ \begin{bmatrix} 0. 3 & 0.4 \\ 0.8 & 0.8 \end{bmatrix}$

$= \begin{bmatrix} 0.7+0.3 & -0.4+0.4 \\ -0.8+0.8 & 0.2+0.8 \end{bmatrix}$

$= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$

This is the two-by-two identity . Therefore, A+ B = I .

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Example Question #31 : Finite Mathematics

You are given that the inverse of the matrix

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

$A^{-1} = \begin{bmatrix} 8 &9 & -11 \\ 5 & 6 & -7 \\ 1 & 1 & -1 \end{bmatrix}$

Using this information, solve the linear system

x-2y + 3z = 24

-2x+ 3y + z = 40

-x+ y+ 3z = -10

From these five choices, select the correct value of .

Possible Answers:

640

160

430

662

Correct answer:

430

Explanation:

The given linear system can be rewritten as the matrix equation

AX = B ,

where

$A = \begin{bmatrix} 1 & -2 & 3 \\ -2 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}$ , $X= \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ , and $B = \begin{bmatrix} 24 \\ 40 \\ -10 \end{bmatrix}$ .

This equation can be restated as

$A^{-1}(AX) =A^{-1} B$

$(A^{-1}A)X =A^{-1} B$

$IX =A^{-1} B$

$X =A^{-1} B$

We are already given $A^{-1}$ , so:

$X =A^{-1} B$

$X = \begin{bmatrix} 8 &9 & -11 \\ 5 & 6 & -7 \\ 1 & 1 & -1 \end{bmatrix} \begin{bmatrix} 24 \\ 40 \\ -10 \end{bmatrix}$

Multiply two matrices by multiplying the rows in the first by the column in the second; this is done by adding the products of entries in corresponding positions, as follows:

$X = \begin{bmatrix} 8 (24)+9(40) +( -11)(-10) \\ 5(24) + 6(40) +( -7)(-10) \\ 1 (24)+ 1(40 )+( -1)(-10) \end{bmatrix}$

$X = \begin{bmatrix} 662 \\ 430 \\ 74 \end{bmatrix}$

We are only interested in the value of , which is 430.

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Example Question #32 : Finite Mathematics

It is suggested (but not necessary) that you use a calculator with matrix capability to work this problem.

A village is famous for being the most prolific producer in the world of ambrosia, butterbeer, and silphium. Every year, they try to meet the world's demand for 20,000 kg of ambrosia, 30,000 kg of butterbeer, and 15,000 kg of silphium. However, the production of these items requires the consumption of these items as well:

The production of 1 kg of ambrosia requires the use of 0.05 kg of butterbeer and 0.08 kg of silphium;

The production of 1 kg of butterbeer requires the use of 0.04 kg of ambrosia and 0.09 kg of silphium; and,

The production of 1 kg of silphium requires the use of 0.06 kg of ambrosia and 0.07 kg of butterbeer.

Use a Leontief model to determine how much silphium (nearest hundred kilograms) must be produced to meet the demand.

Possible Answers:

$18,500 \textup{ kg }$

$18,800 \textup{ kg }$

$18,700 \textup{ kg }$

$18,600 \textup{ kg }$

$18,400 \textup{ kg }$

Correct answer:

$18,700 \textup{ kg }$

Explanation:

The Leontief model for an input-output system requires the following steps.

Let be the technology matrix for the system, the entries of which are the amounts of each product consumed in the manufacture of each product. Letting the rows/columns represent, in order, ambrosia, butterbeer, and silphium,

$A = \begin{bmatrix} 0 & 0.05 & 0.08 \\ 0.04 & 0 & 0.09 \\ 0.06 & 0.07 & 0\end{bmatrix}$

Let be the external demand vector, which gives the amount of each product demanded by outside consumers:

$D= \begin{bmatrix} 20,000 \\ 30,000 \\ 15,000 \end{bmatrix}$

The output vector , which gives the amount of each product that must be produced to meet the demand, can be obtained through the matrix equation

$X = (I- A )^{-1}D$

$= \left ( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & 0.05 & 0.08 \\ 0.04 & 0 & 0.09 \\ 0.06 & 0.07 & 0\end{bmatrix} \right )^{-1} \begin{bmatrix} 20,000 \\ 30,000 \\ 15,000 \end{bmatrix}$

$= \begin{bmatrix} 1 &-0.05 & -0.08 \\ -0.04 & 1 & -0.09 \\ -0.06 & -0.07 & 1 \end{bmatrix} ^{-1} \begin{bmatrix} 20,000 \\ 30,000 \\ 15,000 \end{bmatrix}$

$\approx \begin{bmatrix} 1.007394521 &0.056366243 & 0.085664523 \\0.046025673 & 1.008915193 & 0.094484421 \\ 0.063665468 & 0.074006038 &1.011753781 \end{bmatrix} \begin{bmatrix} 20,000 \\ 30,000 \\ 15,000 \end{bmatrix}$

$\approx \begin{bmatrix} 23,123.85 \\ 32,605.24\\ 18,669.80 \end{bmatrix}$

Silphium is represented by the bottom entry; the amount the village must produce rounds to 18,700 kg.

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Example Question #33 : Finite Mathematics

It is suggested (but not necessary) that you use a calculator with matrix capability to work this problem.

The production of 1 kg of ambrosia requires the use of 0.05 kg of butterbeer and 0.08 kg of silphium;

The production of 1 kg of butterbeer requires the use of 0.04 kg of ambrosia and 0.09 kg of silphium; and,

The production of 1 kg of silphium requires the use of 0.06 kg of ambrosia and 0.07 kg of butterbeer.

Use a Leontief model to determine how much ambrosia (nearest hundred kilograms) must be produced to meet the demand.

Possible Answers:

$23,500 \textup{ kg}$

$23,300 \textup{ kg}$

$23,100 \textup{ kg}$

$23,400 \textup{ kg}$

$23,200 \textup{ kg}$

Correct answer:

$23,100 \textup{ kg}$

Explanation:

The Leontief model for an input-output system requires the following steps.

$A = \begin{bmatrix} 0 & 0.05 & 0.08 \\ 0.04 & 0 & 0.09 \\ 0.06 & 0.07 & 0\end{bmatrix}$

Let be the external demand vector, which gives the amount of each product demanded by outside consumers:

$D= \begin{bmatrix} 20,000 \\ 30,000 \\ 15,000 \end{bmatrix}$

The output vector , which gives the amount of each product that must be produced to meet the demand, can be obtained through the matrix equation

$X = (I- A )^{-1}D$

$= \begin{bmatrix} 1 &-0.05 & -0.08 \\ -0.04 & 1 & -0.09 \\ -0.06 & -0.07 & 1 \end{bmatrix} ^{-1} \begin{bmatrix} 20,000 \\ 30,000 \\ 15,000 \end{bmatrix}$

$\approx \begin{bmatrix} 23,123.85 \\ 32,605.24\\ 18,669.80 \end{bmatrix}$

Ambrosia is represented by the top entry; the amount the village must produce rounds to 23,100 kg.

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Example Question #21 : Systems Of Linear Equations: Matrices

$A= \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

True or false: $A^{2} = \begin{bmatrix} 1 & 4 \\ 9 & 16 \end{bmatrix}$

Possible Answers:

True

False

Correct answer:

False

Explanation:

$A^{2} = AA=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

Two matrices are multiplied by multiplying the rows of the first by the columns of the second - that is, adding the products of the entries in corresponding positions:

$A^{2} = \begin{bmatrix} 1(1)+2(3) & 1(2)+2(4) \\3(1)+4(3) & 3(2)+4(4) \end{bmatrix}$

$= \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix}$

$\ne \begin{bmatrix} 1 & 4 \\ 9 & 16 \end{bmatrix}$

The statement is false.

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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 #23 : Finite Mathematics

Example Question #23 : Finite Mathematics

Example Question #24 : Finite Mathematics

Example Question #24 : Finite Mathematics

Example Question #22 : Finite Mathematics

Example Question #1 : Solve A System Of Equations In Three Variables Using Augmented Matrices

Example Question #31 : Finite Mathematics

Example Question #32 : Finite Mathematics

Example Question #33 : Finite Mathematics

Example Question #21 : Systems Of Linear Equations: Matrices

All Finite Mathematics Resources

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