Scalar interactions with Matrices - ACT Math

Card 0 of 12

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.

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

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

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

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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,

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Question

Simplify the following

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

Answer

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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Question

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

If , then evaluate $n _{3,1}$ .

Answer

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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Question

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

If , then evaluate $n _{3,1}$ .

Answer

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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Question

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

If , evaluate $n_{2,1}$ .

Answer

If , 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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Question

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

If , evaluate $n_{2,3}$ .

Answer

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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Question

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

If , evaluate $n_{2,1}$ .

Answer

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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Question

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

If , evaluate $n_{2,2}$ .

Answer

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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Question

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

Answer

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

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

$\begin{bmatrix} 03&33 \ -13& -2*3 \end{bmatrix}=$

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

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