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ACT Math : Algebra

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14 Diagnostic Tests 767 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #5 : Matrices

Define matrix $M = \begin{bmatrix} 3&-5 &4 \\ 5&6 &1 \\ 3&-4 & -7 \end{bmatrix}$ .

If 5N = M , evaluate $n_{2,1}$ .

Possible Answers:

The correct answer is not among the other responses.

Correct answer:

Explanation:

If 5N = M , then $N = \frac{1}{5} M$ .

Scalar multplication of a matrix is done elementwise, so

$n_{2,1} = \frac{1}{5} m_{2,1}$

$m_{2,1}$ is the first element in the second row of , which is 5, so

$n_{2,1} = \frac{1}{5} (5) = 1$

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Example Question #2 : Matrices

Define matrix $M = \begin{bmatrix} 3&-5 &4 \\ 5&6 &1 \\ 3&-4 & -7 \end{bmatrix}$ .

If N = 3M , evaluate $n_{2,3}$ .

Possible Answers:

$-1\frac{1}{3}$

The correct answer is not among the other responses.

$\frac{1}{3}$

Correct answer:

Explanation:

Scalar multplication of a matrix is done elementwise, so

$n_{2,3} = 3 m _{2,3}$

$m _{2,3}$ is the third element in the second row of , which is 1, so

$n_{2,3} = 3 \cdot 1 = 3$

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Example Question #1 : Matrices

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

If M = 4N + I , evaluate $n_{2,1}$ .

Possible Answers:

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

The correct answer is not given among the other responses.

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

$1\frac{1}{4}$

Correct answer:

$1\frac{1}{4}$

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

$m_{2,1} = 4n_{2,1} +i_{2,1}$

$m_{2,1}$ is the first element in the second row, which is 5; similarly, $i_{2,1} = 0$ . The equation becomes

$5= 4n_{2,1} +0$

$5= 4n_{2,1}$

$n_{2,1}= 5 \div 4 = 1\frac{1}{4}$

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Example Question #1 : How To Multiply A Matrix By A Scalar

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

If M = N +3 I , evaluate $n_{2,2}$ .

Possible Answers:

The correct answer is not given among the other responses.

Correct answer:

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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Example Question #1 : How To Multiply A Matrix By A Scalar

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

When multiplying a constant to a matrix, multiply each entry in the matrix by the constant.

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

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

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

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Example Question #1 : Other Matrices

With matrix notation, what does M_2x3 x N_3x4 equal?

Possible Answers:

P_2x4

None of the answers are correct

P_3x4

P_4x2

P_3x3

Correct answer:

P_2x4

Explanation:

M_2x3 x N_3x4 = P_2x4

In general matrix notation, M_rxc 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.

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Example Question #12 : Matrices

What is the solution to the following matrix?

$\begin{vmatrix}8 & 3\\ -2 & 4 \end{vmatrix}$

Possible Answers:

Correct answer:

Explanation:

In order to solve the matrix, the determinant rule "ad-bc" must be used. is in the "a" position, is in the "b" position, is in the "c" position, and is in the "d" position. After plugging the numbers into "ad-bc," we get

$(8\times 4)-(-2\times 3)=(32)-(-6)=32+6=38$

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Example Question #1 : How To Find An Answer With A Matrix

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

y = 5x

x+ y = 3z

2x - 3y + 4z = 100

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

y = 5x

y - y = 5x - y

5x - y = 0

x+ y = 3z

x+ y- 3z = 3z - 3z

x+ y- 3z = 0

The system is now

5x - y = 0

x+ y- 3z = 0

2x - 3y + 4z = 100

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.

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Example Question #2 : How To Find An Answer With A Matrix

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

y = 4x + 7

2x - 7y = 8

Possible Answers:

$\begin{bmatrix} 4&-1 \\ 2& -7 \end{matrix} \begin{vmatrix} -7\\ 8 \end{bmatrix}$

$\begin{bmatrix} 4& 1 \\ 2& -7 \end{matrix} \begin{vmatrix} 7\\ 8 \end{bmatrix}$

$\begin{bmatrix} 4&-1 \\ 2& -7 \end{matrix} \begin{vmatrix} 7\\ 8 \end{bmatrix}$

$\begin{bmatrix} 1&-4 \\ 2& -7 \end{matrix} \begin{vmatrix} 7\\ 8 \end{bmatrix}$

$\begin{bmatrix} 1&4 \\ 2& -7 \end{matrix} \begin{vmatrix} 7\\ 8 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 4&-1 \\ 2& -7 \end{matrix} \begin{vmatrix} -7\\ 8 \end{bmatrix}$

Explanation:

To set up and augmented matrix for a 2x2 system of equations, both equations must be in standard form Ax+By = C . The second equation is already in standard form.

Rewrite the first equation in standard form as follows:

y = 4x + 7

4x + 7 = y

4x + 7- y - 7 = y - y - 7

4x - y = - 7

The system has been rewritten as

4x - y = - 7

2x - 7y = 8

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

$\begin{bmatrix} 4&-1 \\ 2& -7 \end{matrix} \begin{vmatrix} -7\\ 8 \end{bmatrix}$

is the correct choice.

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Example Question #3 : Other Matrices

Read the following question:

A high school choir sold large boxes of cookies for $5.75 each, medium boxes for $4.75 each, and small boxes for $3.25 each. The band sold a total of 445 boxes and raised a total of $1,924.25. There were twenty more medium boxes sold than large boxes.

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

Possible Answers:

$\begin{bmatrix} 1 & -1 & 5.75 \\ 1 &-1 & 4.75 \\0 & 1 & 3.25 \end{matrix} \begin{vmatrix} -20 \\445\\ 1,924.25\end{bmatrix}$

$\begin{bmatrix} 1 & -1 & 5.75 \\ 1 &-1 & 4.75 \\0 & 1 & 3.25 \end{matrix} \begin{vmatrix} 20 \\445\\ 1,924.25\end{bmatrix}$

$\begin{bmatrix} 1 & -1 & 0 \\ 1 & 1 & 1 \\5.75 & 4.75 & 3.25 \end{matrix} \begin{vmatrix} -20 \\445\\ 1,924.25\end{bmatrix}$

$\begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 & 1 \\5.75 & 4.75 & 3.25 \end{matrix} \begin{vmatrix} 20 \\445\\ 1,924.25\end{bmatrix}$

$\begin{bmatrix} 1 & 1 & 5.75 \\ 1 &-1 & 4.75 \\0 & 1 & 3.25 \end{matrix} \begin{vmatrix} 20 \\445\\ 1,924.25\end{bmatrix}$

Correct answer:

$\begin{bmatrix} 1 & -1 & 0 \\ 1 & 1 & 1 \\5.75 & 4.75 & 3.25 \end{matrix} \begin{vmatrix} -20 \\445\\ 1,924.25\end{bmatrix}$

Explanation:

If we let , , and represent the number of large, medium, and small boxes sold, respectively, since twenty more medium boxes than large were sold, one equation of the 3x3 system will be

y = x+20

or, in standard form,

x-y = -20

Since 445 boxes were sold, another linear equation will be

x+y+z = 445

The money raised from the sale of large boxes of cookies, each of which cost $5.75, is 5.75x ; the money raised from the sale of small boxes of cookies, each of which cost $4.75, is 4.75y ; and the money raised from the sale of small boxes of cookies, each of which cost $3.25, is 3.25z .

The total money raised is $1,924.25, so the other linear equation of the system is

5.75x+ 4.75y+ 3.25z= 1,924.25

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

$\begin{bmatrix} 1 & -1 & 0 \\ 1 & 1 & 1 \\5.75 & 4.75 & 3.25 \end{matrix} \begin{vmatrix} -20 \\445\\ 1,924.25\end{bmatrix}$ ,

which is the correct choice.

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ACT Math : Algebra

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #5 : Matrices

Example Question #2 : Matrices

Example Question #1 : Matrices

Example Question #1 : How To Multiply A Matrix By A Scalar

Example Question #1 : How To Multiply A Matrix By A Scalar

Example Question #1 : Other Matrices

Example Question #12 : Matrices

Example Question #1 : How To Find An Answer With A Matrix

Example Question #2 : How To Find An Answer With A Matrix

Example Question #3 : Other Matrices

All ACT Math Resources

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