Linear Algebra

Welcome to Our Site

Good morning.
I greet you this day, Fri Apr 04 2025 00:52:48 GMT+0000 (Coordinated Universal Time).

For the Classic ACT exam:
The ACT Mathematics test is a timed exam...60 questions in 60 minutes
This implies that you have to solve each question in one minute.
Each of the first 20 questions (less challenging) will typically take less than a minute a solve.
Each of the next 20 questions (medium challenging) may take about a minute to solve.
Each of the last 20 questions (more challenging) may take more than a minute to solve.
The goal is to maximize your time.
You use the time saved on the questions you solve in less than a minute to solve questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.
Also: please note that unless specified otherwise, geometric figures are drawn to scale. So, you can figure out the correct answer by eliminating the incorrect options.
Other suggestions are listed in the solutions/explanations as applicable.

These are the solutions to the ACT past questions on the topics: Linear Algebra.
When applicable, the TI-84 Plus CE calculator (also applicable to TI-84 Plus calculator) solutions are provided for some questions.
The link to the video solutions will be provided for you. Please subscribe to the YouTube channel to be notified of upcoming livestreams. You are welcome to ask questions during the video livestreams.
If you find these resources valuable and helpful in your passing the Mathematics test of the ACT, please consider making a donation:

Cash App: $ExamsSuccess or
cash.app/ExamsSuccess

PayPal: @ExamsSuccess or
PayPal.me/ExamsSuccess

Google charges me for the hosting of this website and my other educational websites. It does not host any of the websites for free.
Besides, I spend a lot of time to type the questions and the solutions well. As you probably know, I provide clear explanations on the solutions.
Your donation is appreciated.

Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome.
Feel free to contact me. Please be positive in your message.
I wish you the best.
Thank you.

(1.) Last semester, each of the 4 students listed below took 6 classes worth 1 credit each.
In the 1st matrix, the row corresponding to each student gives the counts of the letter grades earned by that student.
The second matrix gives the point value of each letter grade.

Number 1

The grade point average of each student is the total number of grade points earned by that student divided by that student's total number of class credits.
To the nearest 0.01, what is Ann's grade point average?

$A.\;\; 1.92 \\[3ex] B.\;\; 2.00 \\[3ex] C.\;\; 3.00 \\[3ex] D.\;\; 3.83 \\[3ex] E.\;\; 4.00 \\[3ex]$

$\underline{Ann} \\[3ex] Grade\;\;Point\;\;Average = \dfrac{\Sigma Number\;\;of\;\;Grade\;\;Points}{\Sigma Number\;\;of\;\;class\;\;credits} \\[5ex] = \dfrac{1(5) + 3(4) + 2(3) + 0(2) + 0(0)}{6(1)} \\[5ex] = \dfrac{5 + 12 + 6}{6} \\[5ex] = \dfrac{23}{6} \\[5ex] = 3.833333333 \\[3ex] \approx 3.83...to\;\;the\;\;nearest\;\;0.01 \\[3ex]$ Student: We were given the grades for 5 classes
But you used the number of credits for 6 classes
May you please explain?
Teacher: Nice observation.
The question specified 6 classes of 1 credit each.
We have to use what the question gave us.
Besides, if you decide to divide by 5(1), the answer is not listed in the option.
Be it as it may, we have to use what we are given.

(3.) The cost of 2 notebooks and a package of pencils is $7.00
The cost of 3 notebooks and 2 packages of pencils is $11.00
What is the cost of 1 notebook and 1 package of pencils?

$F.\;\; \$1.00 \\[3ex] G.\;\; \$3.00 \\[3ex] H.\;\; \$4.00 \\[3ex] J.\;\; \$4.40 \\[3ex] K.\;\; \$4.66 \\[3ex]$

Let:
The cost of 1 notebook = n
The cost of 1 package of pencils = p

$2n + p = 7 ...eqn.(1) \\[3ex] 3n + 2p = 11 ...eqn.(2) \\[5ex] \underline{\text{Elimination by Subtraction Method}} \\[3ex] 2 \cdot eqn.(1) \implies \\[3ex] 2(2n + p) = 2(7) \\[3ex] \implies 4n + 2p = 14 ...eqn.(3) \\[3ex] eqn.(3) - eqn.(2) \implies \\[3ex] (4n - 3n) + (2p - 2p) = 14 - 11 \\[3ex] \implies n = 3 \\[5ex] \underline{\text{Substitution Method}} \\[3ex] \text{Substitute n = 3 into eqn.(1)} \\[3ex] 2(3) + p = 7 \\[3ex] 6 + p = 7 \\[3ex] p = 7 - 6 \\[3ex] p = 1 \\[5ex] \underline{\text{Cost of 1 notebook and 1 package of pencils}} \\[3ex] n + p \\[3ex] 3 + 1 \\[3ex] \$4.00$

(4.) Matrix A has dimensions 3 × 2, and matrix B has dimensions 2 × 3.
One of the following matrices is the matrix product AB.
Which one?

$F.\:\: \begin{bmatrix} 290 \end{bmatrix} \\[5ex] G.\:\: \begin{bmatrix} 105 & 90 \\[2ex] 0 & 0 \end{bmatrix} \\[7ex] H.\:\: \begin{bmatrix} 9 & 3 & 0 \\[2ex] 1 & 7 & 5 \end{bmatrix} \\[7ex] J.\:\: \begin{bmatrix} 9 & 0 \\[2ex] 8 & 0 \\[2ex] 5 & 2 \end{bmatrix} \\[10ex] K.\:\: \begin{bmatrix} 81 & 27 & 0 \\[2ex] 72 & 24 & 0 \\[2ex] 47 & 29 & 10 \end{bmatrix} \\[10ex]$

Order of Matrix A	Order of Matrix B	Order of Matrix AB
$\color{purple}{3} \times \color{darkblue}{2}$	$\color{darkblue}{2} \times \color{purple}{3}$	$\color{purple}{3} \times \color{purple}{3}$

The 3 × 3 matrix is the matrix in Option K

(5.) Which of the following matrices is equal to

$5\begin{bmatrix} 1 & 2 \\[2ex] 3 & 4 \end{bmatrix}$

$A.\:\: \begin{bmatrix} 5 & 2 \\[2ex] 3 & 4 \end{bmatrix} \\[5ex] B.\:\: \begin{bmatrix} 5 & 2 \\[2ex] 3 & 20 \end{bmatrix} \\[5ex] C.\:\: \begin{bmatrix} 1 & 10 \\[2ex] 15 & 4 \end{bmatrix} \\[5ex] D.\:\: \begin{bmatrix} 6 & 7 \\[2ex] 8 & 9 \end{bmatrix} \\[5ex] E.\:\: \begin{bmatrix} 5 & 10 \\[2ex] 15 & 20 \end{bmatrix} \\[5ex]$

Because of the timed ACT test, you do not need to complete the question.
One of the options typically stand out as the answer after few computations.

Samuel, what do you mean? This is what I mean.

$5(1) = 5 \\[3ex] 5(2) = 10 \\[3ex] STOP \\[3ex]$ The correct answer is Option E.

$5\begin{bmatrix} 1 & 2 \\[2ex] 3 & 4 \end{bmatrix} = \begin{bmatrix} 5(1) & 5(2) \\[2ex] 5(3) & 5(4) \end{bmatrix} = \begin{bmatrix} 5 & 10 \\[2ex] 15 & 20 \end{bmatrix}$

(7.) Which of the following matrices is equal to

$4\begin{bmatrix} -2 & 6 \\[2ex] 0 & -3 \end{bmatrix}$

$A.\:\: \begin{bmatrix} -8 & 12 \end{bmatrix} \\[5ex] B.\:\: \begin{bmatrix} 16 \\[2ex] -12 \end{bmatrix} \\[5ex] C.\:\: \begin{bmatrix} 2 & 10 \\[2ex] 4 & 1 \end{bmatrix} \\[5ex] D.\:\: \begin{bmatrix} -\dfrac{1}{2} & \dfrac{3}{2} \\[5ex] 0 & -\dfrac{3}{4} \end{bmatrix} \\[8ex] E.\:\: \begin{bmatrix} -8 & 24 \\[2ex] 0 & -12 \end{bmatrix} \\[5ex]$

Because of the timed ACT test, you do not need to complete the question.
One of the options typically stand out as the answer after few computations.

Samuel, what do you mean? This is what I mean.

$4(-2) = -8 \\[3ex] 4(6) = 24 \\[3ex] STOP \\[3ex]$ The correct answer is Option E.

$4\begin{bmatrix} -2 & 6 \\[2ex] 0 & -3 \end{bmatrix} = \begin{bmatrix} 4(-2) & 4(6) \\[2ex] 4(0) & 4(-3) \end{bmatrix} = \begin{bmatrix} -8 & 24 \\[2ex] 0 & -12 \end{bmatrix}$

(9.) Matrices A and B are given below.

$A = \begin{bmatrix} 2 & -6 \\[2ex] -3 & 8 \end{bmatrix} \hspace{5em} B = \begin{bmatrix} -8 & 5 \\[2ex] 3 & 4 \end{bmatrix}$

Which of the following matrices is A − B?

$F.\:\: \begin{bmatrix} -10 & 11 \\[2ex] 6 & -4 \end{bmatrix} \\[5ex] G.\:\: \begin{bmatrix} -6 & 0 \\[2ex] -1 & 12 \end{bmatrix} \\[5ex] H.\:\: \begin{bmatrix} -6 & -1 \\[2ex] 0 & 12 \end{bmatrix} \\[5ex] J.\:\: \begin{bmatrix} 10 & -6 \\[2ex] -11 & 4 \end{bmatrix} \\[5ex] K.\:\: \begin{bmatrix} 10 & -11 \\[2ex] -6 & 4 \end{bmatrix} \\[5ex]$

Because of the timed ACT test, you do not need to complete the question.
One of the options typically stand out as the answer after few computations.

Samuel, what do you mean? This is what I mean.

$2 - (-8) = 2 + 8 = 10 \\[3ex] -6 - 5 = -11 \\[3ex] STOP \\[3ex]$ The correct answer is Option K.

$A - B \\[3ex] = \begin{bmatrix} 2 & -6 \\[2ex] -3 & 8 \end{bmatrix} - \begin{bmatrix} -8 & 5 \\[2ex] 3 & 4 \end{bmatrix} \\[8ex] = \begin{bmatrix} 2 - (-8) & -6 - 5 \\[2ex] -3 - 3 & 8 - 4 \end{bmatrix} \\[8ex] = \begin{bmatrix} 2 + 8 & -11 \\[2ex] -6 & 4 \end{bmatrix} \\[8ex] = \begin{bmatrix} 10 & -11 \\[2ex] -6 & 4 \end{bmatrix}$

Top