Systems of Linear Equations: Matrices - Finite Mathematics

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Question

Find the value of when,

$6\times 7\equiv n\ (\text{mod } 5)$ .

Answer

To find the value of when,

$6\times 7\equiv n\ (\text{mod } 5)$ first multiply six and seven together.

$6\times 7=42$

Now, recall that mod means the remainder after division occurs.

In this case

$6\times 7\equiv n\ (\text{mod } 5)$

$5)\overline{42\ }$

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Therefore, the remainder is two.

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Question

The following tables describe matrix operations.

$\begin{tabular}{ c|cc } $ & * & % \ \hline * & %& * \ % & * & * \ \end{tabular}$ $\begin{tabular}{ c|cc } # & %& * \ \hline % & & % \ & % & * \ \end{tabular}$

Calculate the following using the tables from above.

$(%\ #\ %)\ $\ %$

Answer

This question is testing the matrix operation. Remember to use order of operations and perform the algebraic operation that is inside the parentheses first.

$(%\ #\ %)\ $\ %$

First look at the # table.

$\begin{tabular}{ c|cc } # & %& * \ \hline % & & % \ & % & * \ \end{tabular}$

Multiplying % by % results in .

Now go to the $ table and multiply by %.

$\begin{tabular}{ c|cc } $ & * & % \ \hline * & %& * \ % & * & * \ \end{tabular}$

$\ $\ %=$

Therefore,

$(%\ #\ %)\ $\ %=*$

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Question

Find the value of when,

$3\times 9\equiv n\ (\text{mod } 4)$ .

Answer

To find the value of when,

$3\times 9\equiv n\ (\text{mod } 4)$ first multiply three and nine together.

$3\times 9=27$

Now, recall that mod means the remainder after division occurs.

In this case

$3\times 9\equiv n\ (\text{mod } 4)$

$4)\overline{27\ }$

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Therefore, the remainder is three.

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Question

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

Give the determinant of .

Answer

The determinant of a two-by-two matrix

$A= \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix}$

can be found by evaluating the expression

$\det A = a_{11}a_{22} - a_{12}a_{23}$

Substitute the corresponding elements to get

$\det A = 3(4)- (-1)(-2) = 12 - 2 = 10$ .

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Question

Consider the system of linear equations:

What kind of system is this?

Answer

One way to identify whether the matrix is consistent and independent is to form a matrix of its variable coefficients, and calculate its determinant. The matrix is

$A = \begin{bmatrix} 1 & 8 \ -4 & 32 \end{bmatrix}$

The determinant of a two-by-two matrix

$A= \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix}$

can be found by evaluating the expression

$\det A = a_{11}a_{22} - a_{12}a_{23}$

Substitute the corresponding elements to get

$\det A = 1(32)- 8(-4) = 32 -(-32) = 64$

Since $\det A e 0$ ,

it follows that the system is consistent and independent.

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Question

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

Which of the following is equal to $B^{-1}$ ?

Answer

$B^{-1}$ , the inverse of a two-by-two matrix

$B = \begin{bmatrix} b_{11} & b_{12} \ b_{21} & b_{22} \end{bmatrix}$ ,

can be calculated as follows:

$B^{-1} = \frac{1}{D} \begin{bmatrix} b_{22} & -b_{12} \ -b_{21} & b_{11} \end{bmatrix}$ ,

where $D = b_{11}b_{22} - b_{12} b_{21}$ .

Setting each of the values accordingly,

$B^{-1} = \frac{1}{1} \begin{bmatrix} 3& -2 \ -4 & 3 \end{bmatrix}$ , or

$B^{-1} = \begin{bmatrix} 3& -2 \ -4 & 3 \end{bmatrix}$ .

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Question

Let $A = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix}$ and $B= \begin{bmatrix} 2 & 1 & 3 \ 0 & 1 & 2 \end{bmatrix}$

Find .

Answer

For the product of two matrices to be defined, the number of columns in the first matrix must be equal to the number of rows in the second. This is the case, since has two columns and has two rows. is defined.

Matrices are multiplied by multiplying each row of the first matrix by each column of the second matrix. This is done by adding the products of the entries in corresponding positions. Thus,

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

$= \begin{bmatrix} 3 (2)+1(0)& 3(1)+1(1) & 3(3)+ 1(2) \ -1(2)+2(0)& -1(1)+2(1) & -1(3)+ 2(2) \end{bmatrix}$

$= \begin{bmatrix} 6&4& 11\ -2& 1& 1\end{bmatrix}$ ,

the correct product.

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Question

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

True or false:

$7K = \begin{bmatrix} 7 & -14 & 21 \ -4 & 2 & 1 \ 2 & 1 & 3 \end{bmatrix}$

Answer

The product of any scalar value and a matrix is the matrix formed when each entry in the matrix is multiplied by that scalar. Thus, if

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

then

$7K = \begin{bmatrix} 7 (1) & 7 ( -2 )& 7(3) \ 7(-4) & 7(2) & 7(1) \ 7(2) & 7(1) & 7(3) \end{bmatrix}$

$= \begin{bmatrix} 7& -14 & 21 \ -28 & 14 & 7 \ 14 & 7 & 21 \end{bmatrix}$

This is not equal to the matrix

$\begin{bmatrix} 7 & -14 & 21 \ -4 & 2 & 1 \ 2 & 1 & 3 \end{bmatrix}$ ,

since the entries in the second and third rows differ. The statement is false.

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Question

$A = \begin{bmatrix} 1 & 0 & 2 & 3 \ 0 & 1 & 1 & 4 \ 0 & 0 & 0 & 0 \end{bmatrix}$

True or false:

is an example of a matrix in reduced row-echelon form.

Answer

A matrix is in reduced row-echelon form if it meets four criteria:

No row comprising only 0's can be above a row with a nonzero entry.

This condition is met, since the only all-zero row is the one at bottom:

$\begin{bmatrix} 1 & 0 & 2 & 3 \ 0 & 1 & 1 & 4 \ \textbf{{\color{Red} 0}} & \textbf{{\color{Red} 0}} & \textbf{{\color{Red} 0}} & \textbf{{\color{Red} 0}} \end{bmatrix}$

The first nonzero entry in each nonzero row is a 1.

Each leading 1 is in a column to the right of the above leading 1.

Both conditions are met:

$\begin{bmatrix} \textbf{{\color{Cyan} 1}} & 0 & 2 & 3 \ 0 & \textbf{{\color{Cyan} 1}} & 1 & 4 \ 0 & 0 & 0 & 0 \end{bmatrix}$

In every column that includes a leading 1, all other entries are 0's.

This condition is met:

$A = \begin{bmatrix} \textbf{{\color{Cyan} 1}} & \textbf{{\color{DarkOrange} 0}} & 2 & 3 \ \textbf{{\color{DarkOrange} 0}} & \textbf{{\color{Cyan} 1}} & 1 & 4 \ \textbf{{\color{DarkOrange} 0}} & \textbf{{\color{DarkOrange} 0}} & 0 & 0 \end{bmatrix}$

meets all four criteria and is therefore in reduced row-echelon form.

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Question

Give the solution set of the system of equations

Answer

Multiply both sides of the first equation by 2 in order to make the x-coefficients each other's opposite:

Add each side of this equation to each side of the other equation:

$\underline{-2x+6y = -40}$

, or

.

This indicates that the two equations are equivalent. Therefore,

The solution set can be written in parametric form as

, arbitrary.

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Question

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

refers to the two-by-two identity matrix.

Which of the following expressions is equal to ?

Answer

For the sum of two matrices to be defined, they must have the same number of rows and columns. is a matrix with three columns; since, in this problem, refers to the two-by-two identity matrix

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

has two columns. Since the number of columns differs, is undefined.

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Question

Let $A = \begin{bmatrix} 30 & -10 \end{bmatrix}$ and $B= \begin{bmatrix} 2 & 1 & 3 \ 0 & 1 & 2 \end{bmatrix}$ .

Find .

Answer

For the product of two matrices to be defined, the number of columns in the first matrix must be equal to the number of rows in the second. This is the case, since has two columns and has two rows. is defined.

Matrix multiplication is worked by multiplying each row of the first matrix by each column of the second matrix. This is done by adding the products of the entries in corresponding positions. Thus,

$AB= \begin{bmatrix} 30 & -10 \end{bmatrix}\begin{bmatrix} 2 & 1 & 3 \ 0 & 1 & 2 \end{bmatrix}$

$= \begin{bmatrix} 30(2) + (-10)(0) & 30(1) + (-10)(1) & 30 (3)+ (-10)(2) \end{bmatrix}$

$= \begin{bmatrix} 60 & 20& 70 \end{bmatrix}$

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Question

is a three-by-four matrix.

Which must be true?

Answer

The product of two matrices and , where has rows and columns and has rows and columns, is a matrix with rows and columns. It follows that must have the same number of rows as . Since has three rows, so does . Nothing can be inferred about the number of rows of .

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Question

$K = \begin{bmatrix} 3 & 0 & 2 \ 7 & 1 & 2 \ -4 & 1 & 3 \end{bmatrix}$

True or false:

$-3K = \begin{bmatrix} -9 & 0 & -6 \ -21 & -3&-6 \ -12 & -3 & -9 \end{bmatrix}$

Answer

The product of any scalar value and a matrix is the matrix formed when each entry in the matrix is multiplied by that scalar. Thus, if

$K = \begin{bmatrix} 3 & 0 & 2 \ 7 & 1 & 2 \ -4 & 1 & 3 \end{bmatrix}$ ,

then

$-3 K = \begin{bmatrix} -3 \cdot 3 & -3 \cdot 0 & -3 \cdot 2 \-3 \cdot 7 & -3 \cdot 1 & -3 \cdot 2 \ -3 \cdot( -4 )& -3 \cdot 1 & -3 \cdot 3 \end{bmatrix} = \begin{bmatrix} -9 & 0 & -6 \ -21 & -3&-6 \ \textbf{{\color{Red} 12}} & -3 & -9 \end{bmatrix} e \begin{bmatrix} -9 & 0 & -6 \ -21 & -3&-6 \ \textbf{{\color{Red} -12}} & -3 & -9 \end{bmatrix}$

The statement is false, since the entry in Row 3, Column 1 is incorrect.

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Question

$\begin{bmatrix} 14 \ 25 \end{bmatrix} = \begin{bmatrix} x + y \ x - y \end{bmatrix}$

True or false: there is no solution that makes this matrix equation true.

Answer

For two matrices to be equal, two conditions must hold:

The matrices must have the same dimension. This can be seen to be the case, since both matrices have two rows and one column.

All corresponding entries must be equal. For this to happen, it must hold that

This is a system of two equations in two variables, which can be solved as follows:

Add both sides of the equations:

$\begin{matrix} x+y=14 \ \underline{x-y=25} \ 2x; ; ; ; ; = 39 \end{matrix}$

It follows that

Substitute back:

Thus, there exists a solution to the system, and, consequently, to the original matrix equation. The statement is therefore false.

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Question

$C = \begin{bmatrix} 1 & 0 & 0 & 0 \ 0 & 2 & 0 & 0 \ 0 &0 &1 & 0 \ 0 & 0 & 0 & 1\end{bmatrix}$

True or false: is an example of a matrix in reduced row-echelon form.

Answer

A matrix is in reduced row-echelon form if it meets four criteria:

No row comprising only 0's can be above a row with a nonzero entry.

The first nonzero entry in each nonzero row is a 1.

Each leading 1 is in a column to the right of the above leading 1.

In every column that includes a leading 1, all other entries are 0's.

The first nonzero entry in the second row is a 2, violating the second criterion:

$\begin{bmatrix} 1 & 0 & 0 & 0 \ 0 & \textbf{{\color{Red} 2}} & 0 & 0 \ 0 &0 &1 & 0 \ 0 & 0 & 0 & 1\end{bmatrix}$

is not in reduced row-echelon form.

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Question

$E= \begin{bmatrix} 1 & 4 & 0 & 0 & 6 \ 0 & 0 &1 &0 & -7 \end{bmatrix}$

True or false: is an example of a matrix in reduced row-echelon form.

Answer

A matrix is in reduced row-echelon form if it meets four criteria:

No row comprising only 0's can be above a row with a nonzero entry.

This is vacuously true, since there are no zero rows.

The first nonzero entry in each nonzero row is a 1.

Each leading 1 is in a column to the right of the above leading 1.

Both conditions are met:

$\begin{bmatrix} \textbf{{\color{Cyan} 1}} & 4 & 0 & 0 & 6 \ 0 & 0 & \textbf{{\color{Cyan} 1}} &0 & -7 \end{bmatrix}$

In every column that includes a leading 1, all other entries are 0's.

Both conditions are met:

$\begin{bmatrix} \textbf{{\color{Cyan} 1}} & 4 & \textbf{{\color{Red} 0}} & 0 & 6 \ \textbf{{\color{Red} 0}} & 0 & \textbf{{\color{Cyan} 1}} &0 & -7 \end{bmatrix}$

meets all four conditions and is therefore a matrix in reduced row-echelon form.

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Question

$A = \begin{bmatrix} 3 & -2 \ -5 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 1 \ 4 & -3 \end{bmatrix}$ .

True or false:

$A+ B = \begin{bmatrix} 8 & -1 \ -1 & 1 \end{bmatrix}$ .

Answer

First, it must be established that 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. is therefore defined.

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

$A + B = \begin{bmatrix} 3 & -2 \ -5 & 4 \end{bmatrix}+ \begin{bmatrix} 5 & 1 \ 4 & -3 \end{bmatrix}$

$= \begin{bmatrix} 3+5 & -2+1 \ -5+4 & 4+(-3) \end{bmatrix}$

$= \begin{bmatrix} 8 & -1 \ -1 & 1 \end{bmatrix}$

The statement is true.

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Question

$A= \begin{bmatrix} 2 & -6 & -7 \ 3 & 5 & -3 \ 3 & 1 & 9 \end{bmatrix}$ and $B =\begin{bmatrix} -4 & 1 & -9 \ 3 & 6 & -4 \ -2 & 9 & 3 \end{bmatrix}$ ..

True or false:

$A-B = \begin{bmatrix} 6 & -7 & 2 \ 0 & -1 & 1 \ 5 & -8 & 6 \end{bmatrix}$ .

Answer

First, it must be established that 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. is therefore defined.

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

$A - B= \begin{bmatrix} 2 & -6 & -7 \ 3 & 5 & -3 \ 3 & 1 & 9 \end{bmatrix} - \begin{bmatrix} -4 & 1 & -9 \ 3 & 6 & -4 \ -2 & 9 & 3 \end{bmatrix}$

$= \begin{bmatrix} 2-(-4) & -6-1 & -7-(-9) \ 3-3 & 5 -6& -3-(-4) \ 3-(-2) & 1-9 & 9-3 \end{bmatrix}$

$= \begin{bmatrix} 6 & -7 & 2 \ 0 & -1 & 1 \ 5 & -8 & 6 \end{bmatrix}$

The statement is true.

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Question

Solve the linear system:

Answer

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

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:

$\begin{matrix} ; ; ; 10x+4y = 120 \\underline{ -10x - 4y = 120}\ ; ; ; ; ; ; ; ; ; ; ; ; ; ; {\color{Red} 0 = ; 24 0} \end{matrix}$

The resulting statement is identically false. It follows that the two equations of the system are inconsistent with each other. The system has no solution.

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