Systems of Linear Equations: Matrices

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Finite Mathematics › Systems of Linear Equations: Matrices

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Use Cramer's rule to evaluate .

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

CORRECT

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

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

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

None of the other choices gives the correct response.

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

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

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

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

False

CORRECT

True

Explanation

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.

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

Find .

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

CORRECT

is undefined.

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

$\begin{bmatrix} 6 & 1 & 0 \ 0 & 2 &0 \end{bmatrix}$

$\begin{bmatrix} 6 & 2 & 7 \ 2 & 3 & 7 \end{bmatrix}$

Explanation

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.

Find the value of when,

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

CORRECT

Explanation

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

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

$4)\overline{27\ }$

--------------

Therefore, the remainder is three.

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

True or false: , where is the two-by-two identity matrix.

False

CORRECT

True

Explanation

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, 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, $A-B \ne I$ .

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

True or false:

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

True

CORRECT

False

Explanation

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.

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

refers to the two-by-two identity matrix.

Which of the following expressions is equal to ?

is undefined.

CORRECT

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

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

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

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

Explanation

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.

is a three-by-four matrix.

Which must be true?

has three rows.

CORRECT

has three rows.

has four rows.

None of the statements in the other choices must be true.

Explanation

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 .

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

Find .

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

CORRECT

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

$\begin{bmatrix} 60 & 30 & 90 \ 0 & -10 &- 20 \end{bmatrix}$

$\begin{bmatrix} 60 & 0\ 30 &-10 \ 90 & -20 \end{bmatrix}$

is not defined.

Explanation

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,

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

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

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

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

From these five choices, select the correct value of .

CORRECT

Explanation

The given linear system can be rewritten as the matrix equation

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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