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ACT Math : Matrices

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Example Questions

Example Question #1 : Matrices

Evaluate: $-3\begin{bmatrix} 2& 3 & 4\\ 4&-5 & 10 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} -12& -1\\ 15& 0\\ -30& 1 \end{bmatrix}$

$\begin{bmatrix} -6& -9& -12\\ -12& 15& -30 \end{bmatrix}$

$\begin{bmatrix} -1& 0& 1\\ -12& 15&-30 \end{bmatrix}$

$\begin{bmatrix} -12&-6 \\ 15& -9\\ -30&-12 \end{bmatrix}$

$\begin{bmatrix} -18& 6& -42 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -6& -9& -12\\ -12& 15& -30 \end{bmatrix}$

Explanation:

This problem involves a scalar multiplication with a matrix. Simply distribute the negative three and multiply this value with every number in the 2 by 3 matrix. The rows and columns will not change.

$-3\begin{bmatrix} 2& 3 & 4\\ 4&-5 & 10 \end{bmatrix}=\begin{bmatrix} -6&-9 &-12 \\ -12& 15&-30 \end{bmatrix}$

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Example Question #2 : Matrices

$5\begin{bmatrix} a \\ b \end{bmatrix} = 14\begin{bmatrix} 20 \\ 12 \end{bmatrix}$

What is a+b ?

Possible Answers:

$\frac{41}{5}$

$\frac{191}{4}$

$\frac{448}{5}$

$\frac{148}{5}$

Correct answer:

$\frac{448}{5}$

Explanation:

You can begin by treating this equation just like it was:

5x=14y

That is, you can divide both sides by :

$\begin{bmatrix} a \\ b \end{bmatrix} = \frac{14}{5}\begin{bmatrix} 20 \\ 12 \end{bmatrix}$

Now, for scalar multiplication of matrices, you merely need to multiply the scalar by each component:

$\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} \frac{14}{5}*20 \\ \frac{14}{5}*12 \end{bmatrix}$

Then, simplify:

$\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 56 \\ \frac{168}{5} \end{bmatrix}$

Therefore, $a+b=56+\frac{168}{5}=\frac{280}{5}+\frac{168}{5}=\frac{448}{5}$

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Example Question #2 : Matrices

If $\frac{1}{2}$\bigl(\begin{smallmatrix} 2a\\4b \end{smallmatrix} \bigr)$=$\bigl(\begin{smallmatrix} 53 \\ 98 \end{smallmatrix} \bigr)$$ , what is a+b ?

Possible Answers:

$\frac{103}{2}$

$\frac{99}{2}$

102

Correct answer:

102

Explanation:

Begin by distributing the fraction through the matrix on the left side of the equation. This will simplify the contents, given that they are factors of :

$\frac{1}{2}\begin{bmatrix} 2a\\4b \end{bmatrix}=\begin{bmatrix} a\\2b \end{bmatrix}$

Now, this means that your equation looks like:
$\begin{bmatrix} a\\2b \end{bmatrix}=\begin{bmatrix} 53 \\ 98 \end{bmatrix}$

This simply means:

a=53

and

2b=98 or b=49

Therefore, a+b=53+49=102

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Example Question #2 : Matrices

Simplify:

$\begin{bmatrix} 13 & 4\\ 71 &3 \end{bmatrix} + 4\begin{bmatrix} 5 & 3\\ 11 &22 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 49 & 206\end{bmatrix}$

$\begin{bmatrix} 49\\ 206\end{bmatrix}$

$\begin{bmatrix} 31 & 93\\ 3 &21 \end{bmatrix}$

$\begin{bmatrix} 33 & 16\\ 115 & 91\end{bmatrix}$

$\begin{bmatrix} 20 & 12\\ 44 &88 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 33 & 16\\ 115 & 91\end{bmatrix}$

Explanation:

Scalar multiplication and addition of matrices are both very easy. Just like regular scalar values, you do multiplication first:

$\begin{bmatrix} 13 & 4\\ 71 &3 \end{bmatrix} + 4\begin{bmatrix} 5 & 3\\ 11 &22 \end{bmatrix}= \begin{bmatrix} 13 & 4\\ 71 &3 \end{bmatrix}+\begin{bmatrix} 20 & 12\\ 44 &88 \end{bmatrix}$

The addition of matrices is very easy. You merely need to add them directly together, correlating the spaces directly.

$\begin{bmatrix} 13 & 4\\ 71 &3 \end{bmatrix} + \begin{bmatrix} 20 & 12\\ 44 &88 \end{bmatrix} = \begin{bmatrix} 13 +20 & 4+12\\ 71+44 &3 +88\end{bmatrix}$

$\begin{bmatrix} 13 +20 & 4+12\\ 71+44 &3 +88\end{bmatrix}= \begin{bmatrix} 33 & 16\\ 115 & 91\end{bmatrix}$

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Example Question #1 : Matrices

Simplify the following

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

Possible Answers:

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

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

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

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

Correct answer:

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

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.

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

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Example Question #5 : Scalar Interactions With 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 N = M -6 I , then evaluate $n _{3,1}$ .

Possible Answers:

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

$n _{3,1} = m _{3,1} -6 i _{3,1}$

$m _{3,1}$ is the first element in the third row of , which is 3; similarly, $i _{3,1} = 0$ . Therefore,

$n _{3,1} = 3 -6 (0)= 3$

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Example Question #2 : 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 N = 7M + I , then evaluate $n _{3,1}$ .

Possible Answers:

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

$n _{3,1} = 7 m _{3,1} + i _{3,1}$

$m _{3,1}$ is the first element in the third row of , which is 3; similarly, $i _{3,1} = 0$ . Therefore,

$n _{3,1} = 7 (3) + 0 = 21$

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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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ACT Math : Matrices

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : Matrices

Example Question #2 : Matrices

Example Question #2 : Matrices

Example Question #2 : Matrices

Example Question #1 : Matrices

Example Question #5 : Scalar Interactions With Matrices

Example Question #2 : Matrices

Example Question #5 : Matrices

Example Question #2 : Matrices

Example Question #1 : Matrices

All ACT Math Resources

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