Matrices

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ACT Math › Matrices

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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

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

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,

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,

Define matrix $M = \begin{bmatrix} 3&-5 &4 \ 5&6 &1 \ 3&-4 & -7 \end{bmatrix}$ , and let be the 3x3 identity matrix.

If , evaluate $n_{2,2}$ .

The correct answer is not given among the other responses.

Explanation

The 3x3 identity matrix is

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

Both scalar multplication of a matrix and matrix addition are performed elementwise, so

$m_{2,2} = n_{2,2} +3i_{2,2}$

$m_{2,2}$ is the second element in the second row, which is 6; similarly, $i_{2,2} = 1$ . The equation becomes

$6= n_{2,2} +3 (1)$

$6= n_{2,2} +3$

$n_{2,2} =3$

Simplify the following

$3\cdot \begin{bmatrix} -3& 5&0 \ 4& -1& 2 \end{bmatrix}$

$\begin{bmatrix} -9& 15&0 \ 12& -3& 6 \end{bmatrix}$

$\begin{bmatrix} -9& 5&0 \ 12& -1& 2 \end{bmatrix}$

$\begin{bmatrix} 6 \ 15 \end{bmatrix}$

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

Explanation

When multplying any matrix by a scalar quantity (3 in our case), we simply multiply each term in the matrix by the scalar.

Therefore, every number simply gets multiplied by 3, giving us our answer.

$\begin{bmatrix} -9& 15&0 \ 12& -3& 6 \end{bmatrix}$

Simplify the following

$3\cdot \begin{bmatrix} -3& 5&0 \ 4& -1& 2 \end{bmatrix}$

$\begin{bmatrix} -9& 15&0 \ 12& -3& 6 \end{bmatrix}$

$\begin{bmatrix} -9& 5&0 \ 12& -1& 2 \end{bmatrix}$

$\begin{bmatrix} 6 \ 15 \end{bmatrix}$

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

Explanation

When multplying any matrix by a scalar quantity (3 in our case), we simply multiply each term in the matrix by the scalar.

Therefore, every number simply gets multiplied by 3, giving us our answer.

$\begin{bmatrix} -9& 15&0 \ 12& -3& 6 \end{bmatrix}$

Define matrix $M = \begin{bmatrix} 3&-5 &4 \ 5&6 &1 \ 3&-4 & -7 \end{bmatrix}$ , and let be the 3x3 identity matrix.

If , evaluate $n_{2,2}$ .

The correct answer is not given among the other responses.

Explanation

The 3x3 identity matrix is

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

Both scalar multplication of a matrix and matrix addition are performed elementwise, so

$m_{2,2} = n_{2,2} +3i_{2,2}$

$m_{2,2}$ is the second element in the second row, which is 6; similarly, $i_{2,2} = 1$ . The equation becomes

$6= n_{2,2} +3 (1)$

$6= n_{2,2} +3$

$n_{2,2} =3$

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