Other Matrices

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Questions 1 - 10

Read the following problem:

The barista at the Teahouse of the December Sun has a problem. He needs to mix twenty pounds of two different kinds of tea together to create a blend called Strawberry Peppermint Delight. The two varieties are Peppermint Nirvana, which costs $12 a pound, and Strawberry Fields, which costs $15 a pound; the new tea will cost $13 a pound, and it will sell for the same price as the two blended teas would separately. How much of each variety will go into the twenty pounds of Strawberry Peppermint Delight?

Which of the following augmented matrices represents the system of equations that could be set up to solve this problem?

$\begin{bmatrix} 1&1 \ 12 & 15 \end{matrix} \begin{vmatrix} 20 \ 260 \end{bmatrix}$

$\begin{bmatrix} 1&1 \ 12 & 15 \end{matrix} \begin{vmatrix} 20 \ 13\end{bmatrix}$

$\begin{bmatrix} 1&1 \ 12 & 15 \end{matrix} \begin{vmatrix} 260 \ 13\end{bmatrix}$

$\begin{bmatrix} 1&1 \ 12 & 15 \end{matrix} \begin{vmatrix} 260 \ 20\end{bmatrix}$

$\begin{bmatrix} 1&1 \ 12 & 15 \end{matrix} \begin{vmatrix} 13 \ 260 \end{bmatrix}$

Explanation

If the barista mixes pounds of Peppermint Nirvana and pounds of Strawberry Fields to make twenty pounds of tea total, then

will be one of the equations in the system.

pounds of Peppermint Nirvana tea for $12 a pound will cost a total of dollars; pounds of Strawberry Fields tea will cost a total of dollars. Tewnty pounds of the Strawberry Peppermint Delight tea for $13 a pound will cost $13 \times 20 = 260$ dollars. Since the tea will sell for the same price blended as separate, the other equation of the system will be

The augmented matrix of this system will comprise the coefficients of these equations, both of which are in standard form, so the matrix will be

$\begin{bmatrix} 1&1 \ 12 & 15 \end{matrix} \begin{vmatrix} 20 \ 260 \end{bmatrix}$ .

Read the following problem:

Which of the following augmented matrices represents the system of equations that could be set up to solve this problem?

$\begin{bmatrix} 1&1 \ 12 & 15 \end{matrix} \begin{vmatrix} 20 \ 260 \end{bmatrix}$

$\begin{bmatrix} 1&1 \ 12 & 15 \end{matrix} \begin{vmatrix} 20 \ 13\end{bmatrix}$

$\begin{bmatrix} 1&1 \ 12 & 15 \end{matrix} \begin{vmatrix} 260 \ 13\end{bmatrix}$

$\begin{bmatrix} 1&1 \ 12 & 15 \end{matrix} \begin{vmatrix} 260 \ 20\end{bmatrix}$

$\begin{bmatrix} 1&1 \ 12 & 15 \end{matrix} \begin{vmatrix} 13 \ 260 \end{bmatrix}$

Explanation

If the barista mixes pounds of Peppermint Nirvana and pounds of Strawberry Fields to make twenty pounds of tea total, then

will be one of the equations in the system.

The augmented matrix of this system will comprise the coefficients of these equations, both of which are in standard form, so the matrix will be

$\begin{bmatrix} 1&1 \ 12 & 15 \end{matrix} \begin{vmatrix} 20 \ 260 \end{bmatrix}$ .

If $\begin{bmatrix} a&b\ c&d \end{bmatrix} + \begin{bmatrix} 3&14\ 12&4 \end{bmatrix} = \begin{bmatrix} 19&2\ 4&5 \end{bmatrix}$ , what is ?

Explanation

You can treat matrices just like you treat other members of an equation. Therefore, you can subtract the matrix

$\begin{bmatrix} 3&14\ 12&4 \end{bmatrix}$

from both sides of the equation. This gives you:

$\begin{bmatrix} a&b\ c&d \end{bmatrix}=\begin{bmatrix} 19&2\ 4&5 \end{bmatrix}-\begin{bmatrix} 3&14\ 12&4 \end{bmatrix}$

Now, matrix subtraction is simple. You merely subtract each element, matching the correlative spaces with each other:

$\begin{bmatrix} a&b\ c&d \end{bmatrix} = \begin{bmatrix} 19-3&2-14\ 4-12&5-4 \end{bmatrix}$

Then, you simplify:

$\begin{bmatrix} 19-3&2-14\ 4-12&5-4 \end{bmatrix} = \begin{bmatrix} 16&-12\ -8&1 \end{bmatrix}$

Therefore,

If $\begin{bmatrix} a&b\ c&d \end{bmatrix} + \begin{bmatrix} 3&14\ 12&4 \end{bmatrix} = \begin{bmatrix} 19&2\ 4&5 \end{bmatrix}$ , what is ?

Explanation

You can treat matrices just like you treat other members of an equation. Therefore, you can subtract the matrix

$\begin{bmatrix} 3&14\ 12&4 \end{bmatrix}$

from both sides of the equation. This gives you:

$\begin{bmatrix} a&b\ c&d \end{bmatrix}=\begin{bmatrix} 19&2\ 4&5 \end{bmatrix}-\begin{bmatrix} 3&14\ 12&4 \end{bmatrix}$

Now, matrix subtraction is simple. You merely subtract each element, matching the correlative spaces with each other:

$\begin{bmatrix} a&b\ c&d \end{bmatrix} = \begin{bmatrix} 19-3&2-14\ 4-12&5-4 \end{bmatrix}$

Then, you simplify:

$\begin{bmatrix} 19-3&2-14\ 4-12&5-4 \end{bmatrix} = \begin{bmatrix} 16&-12\ -8&1 \end{bmatrix}$

Therefore,

Read the following question:

A high school band sold large boxes of cookies for $4.75 each and small boxes of cookies for $3.25 each. The band sold a total of 305 boxes and raised a total of $1,196.75.

Which of the following augmented matrices represents the system of equations that could be set up to solve this problem?

$\begin{bmatrix} 1&1 \ 4.75&3.25 \end{matrix} \begin{vmatrix} 305 \ 1,196.75 \end{bmatrix}$

$\begin{bmatrix} 1&4.75 \ 1&3.25 \end{matrix} \begin{vmatrix} 305 \ 1,196.75 \end{bmatrix}$

$\begin{bmatrix} 1&1 \ 4.75&3.25 \end{matrix} \begin{vmatrix}1,196.75 \ 305 \end{bmatrix}$

$\begin{bmatrix} 1&4.75 \ 1&3.25 \end{matrix} \begin{vmatrix}1,196.75 \ 305 \end{bmatrix}$

$\begin{bmatrix}3.25&4.75 \ 305 &1,196.75 \end{matrix} \begin{vmatrix}1 \ 1 \end{bmatrix}$

Explanation

If we let and represent the number of large and small boxes sold, respectively, since 305 boxes were sold, one linear equation of the 2x2 system will be

The money raised from the sale of large boxes of cookies, each of which cost $4.75, is ; the money raised from the sale of small boxes of cookies, each of which cost $3.25, is . The total money raised is $1,196.75, so the other linear equation of the system is

The augmented matrix of this system will comprise the coefficients of these equations, both of which are in standard form, so the matrix will be

$\begin{bmatrix} 1&1 \ 4.75&3.25 \end{matrix} \begin{vmatrix} 305 \ 1,196.75 \end{bmatrix}$

Read the following question:

A high school band sold large boxes of cookies for $4.75 each and small boxes of cookies for $3.25 each. The band sold a total of 305 boxes and raised a total of $1,196.75.

Which of the following augmented matrices represents the system of equations that could be set up to solve this problem?

$\begin{bmatrix} 1&1 \ 4.75&3.25 \end{matrix} \begin{vmatrix} 305 \ 1,196.75 \end{bmatrix}$

$\begin{bmatrix} 1&4.75 \ 1&3.25 \end{matrix} \begin{vmatrix} 305 \ 1,196.75 \end{bmatrix}$

$\begin{bmatrix} 1&1 \ 4.75&3.25 \end{matrix} \begin{vmatrix}1,196.75 \ 305 \end{bmatrix}$

$\begin{bmatrix} 1&4.75 \ 1&3.25 \end{matrix} \begin{vmatrix}1,196.75 \ 305 \end{bmatrix}$

$\begin{bmatrix}3.25&4.75 \ 305 &1,196.75 \end{matrix} \begin{vmatrix}1 \ 1 \end{bmatrix}$

Explanation

If we let and represent the number of large and small boxes sold, respectively, since 305 boxes were sold, one linear equation of the 2x2 system will be

The augmented matrix of this system will comprise the coefficients of these equations, both of which are in standard form, so the matrix will be

$\begin{bmatrix} 1&1 \ 4.75&3.25 \end{matrix} \begin{vmatrix} 305 \ 1,196.75 \end{bmatrix}$

Which of the following augmented matrices can be used to solve this system of equations?

$\begin{bmatrix} 5 & -1 & 0 \ 1 & 1 & -3 \2 & - 3 & 4 \end{matrix} \begin{vmatrix} 0 \0\ 100 \end{bmatrix}$

$\begin{bmatrix} 5 & -1 & 0 \ 1 & 1 & 3 \2 & - 3 & 4 \end{matrix} \begin{vmatrix} 0 \0\ 100 \end{bmatrix}$

$\begin{bmatrix} 5 & -1 & 0 \ 1 & 1 & 0 \2 & - 3 & 4 \end{matrix} \begin{vmatrix} 0 \3\ 100 \end{bmatrix}$

$\begin{bmatrix} 0 & 1 & 0 \ 1 & 1 & -3 \2 & - 3 & 4 \end{matrix} \begin{vmatrix} 5 \0\ 100 \end{bmatrix}$

$\begin{bmatrix} 0 & 1 & 0 \ 1 & 1 & 0 \2 & - 3 & 4 \end{matrix} \begin{vmatrix} 5 \3\ 100 \end{bmatrix}$

Explanation

To set up and augmented matrix for a 3x3 system of equations, all equations must be in standard form . The third equation is already in standard form; the first two are not and must be rewritten as such.

The system is now

Write the augmented matrix with each row comprising the coefficients of one equation in order:

$\begin{bmatrix} 5 & -1 & 0 \ 1 & 1 & -3 \2 & - 3 & 4 \end{matrix} \begin{vmatrix} 0 \0\ 100 \end{bmatrix}$

is the correct choice.

Which of the following augmented matrices can be used to solve this system of equations?

$\begin{bmatrix} 5 & -1 & 0 \ 1 & 1 & -3 \2 & - 3 & 4 \end{matrix} \begin{vmatrix} 0 \0\ 100 \end{bmatrix}$

$\begin{bmatrix} 5 & -1 & 0 \ 1 & 1 & 3 \2 & - 3 & 4 \end{matrix} \begin{vmatrix} 0 \0\ 100 \end{bmatrix}$

$\begin{bmatrix} 5 & -1 & 0 \ 1 & 1 & 0 \2 & - 3 & 4 \end{matrix} \begin{vmatrix} 0 \3\ 100 \end{bmatrix}$

$\begin{bmatrix} 0 & 1 & 0 \ 1 & 1 & -3 \2 & - 3 & 4 \end{matrix} \begin{vmatrix} 5 \0\ 100 \end{bmatrix}$

$\begin{bmatrix} 0 & 1 & 0 \ 1 & 1 & 0 \2 & - 3 & 4 \end{matrix} \begin{vmatrix} 5 \3\ 100 \end{bmatrix}$

Explanation

The system is now

Write the augmented matrix with each row comprising the coefficients of one equation in order:

$\begin{bmatrix} 5 & -1 & 0 \ 1 & 1 & -3 \2 & - 3 & 4 \end{matrix} \begin{vmatrix} 0 \0\ 100 \end{bmatrix}$

is the correct choice.

With matrix notation, what does M2x3 x N3x4 equal?

P3x4

P3x3

P4x2

P2x4

None of the answers are correct

Explanation

M2x3 x N3x4 = P2x4

In general matrix notation, Mrxc shows that the matrix is named M and r is the number of rows and c is the number of columns. When multiplying two matrices, the number of columns in the first matrix must match the number of rows in the second matrix. In addition, when adding or subtracting matrices, the matrices must be of the same size.

With matrix notation, what does M2x3 x N3x4 equal?

P3x4

P3x3

P4x2

P2x4

None of the answers are correct

Explanation

M2x3 x N3x4 = P2x4

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