Matrices - ACT Math

Card 0 of 288

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

Read the following question:

A high school band sold large boxes of cookies for $4.75 each and small boxes of cookies for $3.25 each. The band sold a total of 305 boxes and raised a total of $1,196.75.

Which of the following augmented matrices represents the system of equations that could be set up to solve this problem?

Answer

If we let and represent the number of large and small boxes sold, respectively, since 305 boxes were sold, one linear equation of the 2x2 system will be

The money raised from the sale of large boxes of cookies, each of which cost $4.75, is ; the money raised from the sale of small boxes of cookies, each of which cost $3.25, is . The total money raised is $1,196.75, so the other linear equation of the system is

The augmented matrix of this system will comprise the coefficients of these equations, both of which are in standard form, so the matrix will be

$\begin{bmatrix} 1&1 \ 4.75&3.25 \end{matrix} \begin{vmatrix} 305 \ 1,196.75 \end{bmatrix}$

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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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Question

Simplify:

$\begin{bmatrix} 14&22\ -3 &4 \end{bmatrix}+\begin{bmatrix} -3&12\ 5 &-10 \end{bmatrix}$

Answer

Matrix addition is very easy! All that you need to do is add each correlative member to each other. Think of it like this:

$\begin{bmatrix} 14&22\ -3 &4 \end{bmatrix}+\begin{bmatrix} -3&12\ 5 &-10 \end{bmatrix}=\begin{bmatrix} 14-3&22+12\ -3+5 &4-10 \end{bmatrix}$

Now, just simplify:

$\begin{bmatrix} 11&34\ 2 &-6 \end{bmatrix}$

There is your answer!

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Question

Simplify:

$\begin{bmatrix} -5&7\ 102 & -12 \end{bmatrix}+\begin{bmatrix} 3&107\ -10 & -9 \end{bmatrix}$

Answer

Matrix addition is really easy—don't overthink it! All you need to do is combine the two matrices in a one-to-one manner for each index:

$\begin{bmatrix} -5&7\ 102 & -12 \end{bmatrix}+\begin{bmatrix} 3&107\ -10 & -9 \end{bmatrix}=\begin{bmatrix} -5+3&7+107\ 102-10 & -12-9 \end{bmatrix}$

Then, just simplify all of those simple additions and subtractions:

$\begin{bmatrix} -5+3&7+107\ 102-10 & -12-9 \end{bmatrix}=\begin{bmatrix} -2&114\ 92 & -21 \end{bmatrix}$

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Question

What is the value of

$\begin{bmatrix}2 & 2\ 3& -1 \end{bmatrix}$ $+\begin{bmatrix}0 &1 \ -1& 12 \end{bmatrix}$ ?

Answer

To add matrices you simply add the numbers in the same position as each other.

$\begin{bmatrix}a_1 & b_1\ c_1& d_1 \end{bmatrix}$ $+\begin{bmatrix}a_2 & b_2\ c_2& d_2 \end{bmatrix}$ $=\begin{bmatrix} a_1+a_2&b_1+b_2 \ c_1+c_2&d_1+d_2 \end{bmatrix}$

Plugging the given values into the above formula, we are able to solve the question.

$\begin{bmatrix}2 & 2\ 3& -1 \end{bmatrix}$ $+\begin{bmatrix}0 &1 \ -1& 12 \end{bmatrix}$ $=\begin{bmatrix} 2+0&2+1 \ 3+(-1)&-1+12 \end{bmatrix}$

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Question

With matrix notation, what does M2x3 x N3x4 equal?

Answer

M2x3 x N3x4 = P2x4

In general matrix notation, Mrxc shows that the matrix is named M and r is the number of rows and c is the number of columns. When multiplying two matrices, the number of columns in the first matrix must match the number of rows in the second matrix. In addition, when adding or subtracting matrices, the matrices must be of the same size.

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Question

What is the solution to the following matrix?

$\begin{vmatrix}8 & 3\ -2 & 4 \end{vmatrix}$

Answer

In order to solve the matrix, the determinant rule "ad-bc" must be used. is in the "a" position, is in the "b" position, is in the "c" position, and is in the "d" position. After plugging the numbers into "ad-bc," we get

$(8\times 4)-(-2\times 3)=(32)-(-6)=32+6=38$

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Question

Which of the following augmented matrices can be used to solve this system of equations?

Answer

To set up and augmented matrix for a 3x3 system of equations, all equations must be in standard form . The third equation is already in standard form; the first two are not and must be rewritten as such.

The system is now

Write the augmented matrix with each row comprising the coefficients of one equation in order:

$\begin{bmatrix} 5 & -1 & 0 \ 1 & 1 & -3 \2 & - 3 & 4 \end{matrix} \begin{vmatrix} 0 \0\ 100 \end{bmatrix}$

is the correct choice.

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Question

Which of the following augmented matrices can be used to solve this system of equations?

Answer

To set up and augmented matrix for a 2x2 system of equations, both equations must be in standard form . The second equation is already in standard form.

Rewrite the first equation in standard form as follows:

The system has been rewritten as

Write the augmented matrix with each row comprising the coefficients of one equation in order:

$\begin{bmatrix} 4&-1 \ 2& -7 \end{matrix} \begin{vmatrix} -7\ 8 \end{bmatrix}$

is the correct choice.

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