ACT Math - How to find scalar interactions with a matrix

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

What is ?

$\frac{448}{5}$

$\frac{41}{5}$

$\frac{191}{4}$

$\frac{148}{5}$

Explanation

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

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

$\frac{103}{2}$

$\frac{99}{2}$

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:

and

Therefore,

Simplify:

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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.

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

How to find scalar interactions with a matrix

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