How to find scalar interactions with a matrix - ACT Math

Card 0 of 36

Question

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

Answer

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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Question

Simplify:

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

Answer

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.

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

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

Compare your answer with the correct one above

Question

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

What is ?

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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:

and

or

Therefore,

Compare your answer with the correct one above

Question

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

Answer

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.

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

Compare your answer with the correct one above

Question

Simplify:

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

Answer

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.

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

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

Compare your answer with the correct one above

Question

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

What is ?

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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:

and

or

Therefore,

Compare your answer with the correct one above

Question

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

Answer

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.

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

Compare your answer with the correct one above

Question

Simplify:

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

Answer

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.

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

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

Compare your answer with the correct one above

Question

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

What is ?

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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:

and

or

Therefore,

Compare your answer with the correct one above

Question

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

Answer

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.

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

Compare your answer with the correct one above

Question

Simplify:

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

Answer

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.

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

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

Compare your answer with the correct one above

Question

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

What is ?

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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:

and

or

Therefore,

Compare your answer with the correct one above

Question

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

Answer

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.

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

Compare your answer with the correct one above

Question

Simplify:

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

Answer

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.

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

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

Compare your answer with the correct one above

Question

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

What is ?

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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:

and

or

Therefore,

Compare your answer with the correct one above

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How to find scalar interactions with a matrix - ACT Math

Evaluate:

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.

Simplify:

Scalar multiplication and addition of matrices are both very easy. Just like regular scalar values, you do multiplication first: The addition of matrices is very easy. You merely need to add them directly together, correlating the spaces directly.

What is ?

You can begin by treating this equation just like it was: That is, you can divide both sides by : Now, for scalar multiplication of matrices, you merely need to multiply the scalar by each component: Then, simplify: Therefore,

If , what is ?

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 : Now, this means that your equation looks like: This simply means: and or Therefore,

Evaluate:

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.

Simplify:

Scalar multiplication and addition of matrices are both very easy. Just like regular scalar values, you do multiplication first: The addition of matrices is very easy. You merely need to add them directly together, correlating the spaces directly.

What is ?

You can begin by treating this equation just like it was: That is, you can divide both sides by : Now, for scalar multiplication of matrices, you merely need to multiply the scalar by each component: Then, simplify: Therefore,

If , what is ?

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 : Now, this means that your equation looks like: This simply means: and or Therefore,

Evaluate:

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.

Simplify:

Scalar multiplication and addition of matrices are both very easy. Just like regular scalar values, you do multiplication first: The addition of matrices is very easy. You merely need to add them directly together, correlating the spaces directly.

What is ?

You can begin by treating this equation just like it was: That is, you can divide both sides by : Now, for scalar multiplication of matrices, you merely need to multiply the scalar by each component: Then, simplify: Therefore,

If , what is ?

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 : Now, this means that your equation looks like: This simply means: and or Therefore,

Evaluate:

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.

Simplify:

Scalar multiplication and addition of matrices are both very easy. Just like regular scalar values, you do multiplication first: The addition of matrices is very easy. You merely need to add them directly together, correlating the spaces directly.

What is ?

You can begin by treating this equation just like it was: That is, you can divide both sides by : Now, for scalar multiplication of matrices, you merely need to multiply the scalar by each component: Then, simplify: Therefore,

If , what is ?

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 : Now, this means that your equation looks like: This simply means: and or Therefore,

Evaluate:

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.

Simplify:

Scalar multiplication and addition of matrices are both very easy. Just like regular scalar values, you do multiplication first: The addition of matrices is very easy. You merely need to add them directly together, correlating the spaces directly.

What is ?

You can begin by treating this equation just like it was: That is, you can divide both sides by : Now, for scalar multiplication of matrices, you merely need to multiply the scalar by each component: Then, simplify: Therefore,

If , what is ?

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 : Now, this means that your equation looks like: This simply means: and or Therefore,

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

Simplify:

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

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

What is ?

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

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

Simplify:

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

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

What is ?

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

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

Simplify:

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

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

What is ?

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

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

Simplify:

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

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

What is ?

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

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

Simplify:

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

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

What is ?

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