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SAT II Math I : SAT Subject Test in Math I

Study concepts, example questions & explanations for SAT II Math I

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Example Questions

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Example Question #4 : Ratios And Proportions

If for every $\frac{1}{2}$ inch on a map it represents miles, how many inches will be needed for a distance of 275 miles?

Possible Answers:

5.5

4.5

6.5

Correct answer:

5.5

Explanation:

The ratio of inches to miles is

$\frac{0.5}{25}$

so you can set up a proportion of $\frac{0.5}{25}=\frac{x}{275}$ .

You then cross multiply to get

25x= 137.5 .

Then you solve for the variable by dividing by which gives us

137.5/25=5.5 .

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Example Question #1 : Ratios And Proportions

In a geometric sequence the ratio of the first term to the third term is 9:1. What is the ratio of the second term to the fifth term?

Possible Answers:

27:1

81:1

9:1

3:1

5:2

Correct answer:

27:1

Explanation:

Recall that in a geometric sequence there is a constant ratio between any pair of consecutive terms:

$r \cdot a_{n} =a_{n+1}$

If the ratio of the first term to the third term is 9:1 then the ratio of two consecutive terms is a number that when multiplied by itself is 9:

r^2 = 9

r=3

Therefore the common ratio for this sequence is 3.

There are three terms between term 2 and 5 so the common ratio will be applied to the second term three times to get the fifth term:

$r^3 \cdot a_{2} = a_{5}$

Since for this sequence

$3^{3} = 27$

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Example Question #1 : Matrices

Define $A = \begin{bmatrix} 5& 10\\ -6 & -12 \end{bmatrix}$ .

Give $A^{-1}$ .

Possible Answers:

$A^{-1} = \begin{bmatrix} -\frac{1}{24}& - \frac{1}{12}\\ \frac{1}{20} & -\frac{1}{10 }\end{bmatrix}$

$A^{-1}$ is not defined.

$A^{-1} = \begin{bmatrix} -12& -10\\ 6 & -5\end{bmatrix}$

$A^{-1} = \begin{bmatrix} 5& -10\\ 6 & -12 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} -\frac{1}{10}& - \frac{1}{12}\\ \frac{1}{20} & -\frac{1}{24}\end{bmatrix}$

Correct answer:

$A^{-1}$ is not defined.

Explanation:

The inverse of a 2 x 2 matrix $\begin{bmatrix} A& B\\ C& D \end{bmatrix}$ , if it exists, is the matrix

$\frac{\begin{bmatrix} D& -B\\ -C&A \end{bmatrix}}{\begin{vmatrix} A& B\\ C& D \end{vmatrix}}= \frac{\begin{bmatrix} D& -B\\ -C&A \end{bmatrix}}{AD - BC}$

First, we need to establish that the inverse is defined, which it is if and only if determinant $AD - BC \neq 0$ .

Set A = 5, B = 10, C = -6, D =-12 , and check:

$AD - BC = 5 \cdot \left (-12 \right ) - 10 \cdot \left ( -6\right ) =-60 - (-60 )= -60 + 60 = 0$

The determinant is equal to 0, so does not have an inverse.

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Example Question #41 : Mathematical Relationships

Give the determinant of the matrix $\begin{bmatrix} x&8 \\ 6& x \end{bmatrix}$

Possible Answers:

14x

$x^{2} - 48$

$x^{2} - 14$

-2x

Correct answer:

$x^{2} - 48$

Explanation:

The determinant of the matrix $\begin{bmatrix} A& B\\ C& D \end{bmatrix}$ is

$\begin{vmatrix} A& B\\ C& D \end{vmatrix} = AD - BC$ .

Substitute A = D = x,B = 8, C = 6 :

$\begin{vmatrix} A& B\\ C& D \end{vmatrix} = x \cdot x - 8 \cdot 6 = x^{2} - 48$

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Example Question #3 : Matrices

Simplify:

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

Possible Answers:

$\begin{bmatrix} 17&10\\ -8 &14 \end{bmatrix}$

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

$\begin{bmatrix} 45&-4 \end{bmatrix}$

$\begin{bmatrix} 17 & -5 \\ 3 & 1 \end{bmatrix}$

$\begin{bmatrix} 13\\ 28 \end{bmatrix}$

Correct answer:

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

Explanation:

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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Example Question #4 : Matrices

Simplify:

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

Possible Answers:

$\begin{bmatrix} 90\\ 93 \end{bmatrix}$

$\begin{bmatrix} 112\\ 71 \end{bmatrix}$

$\begin{bmatrix} -8&-100\\ 112 & -3 \end{bmatrix}$

$\begin{bmatrix} 112 & 71\end{bmatrix}$

$\begin{bmatrix} -2&114\\ 92 & -21 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -2&114\\ 92 & -21 \end{bmatrix}$

Explanation:

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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Example Question #8 : Find The Product Of A Matrix And A Scalar

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

Possible Answers:

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

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

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

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

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

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 #9 : Find The Product Of A Matrix And A Scalar

Simplify:

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

Possible Answers:

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

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

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

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

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

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 #10 : Find The Product Of A Matrix And A Scalar

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

What is a+b ?

Possible Answers:

$\frac{191}{4}$

$\frac{148}{5}$

$\frac{41}{5}$

$\frac{448}{5}$

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 #1 : How To Subtract Matrices

Given the following matrices, what is the product of and ?

$\begin{bmatrix} 1&-4 \\ 2&x \end{bmatrix} - \begin{bmatrix} y&4 \\ 8&1 \end{bmatrix} = \begin{bmatrix} -8&-8 \\ -6&4 \end{bmatrix}$

Possible Answers:

Correct answer:

Explanation:

When subtracting matrices, you want to subtract each corresponding cell.

$\begin{bmatrix} 1-y&-4-4 \\ 2-8&x-1 \end{bmatrix}= \begin{bmatrix} -8&-8 \\ -6&4 \end{bmatrix}$

Now solve for and

x-1=4

x=5

1-y=-8

y=9

xy = 45

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SAT II Math I : SAT Subject Test in Math I

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #4 : Ratios And Proportions

Example Question #1 : Ratios And Proportions

Example Question #1 : Matrices

Example Question #41 : Mathematical Relationships

Example Question #3 : Matrices

Example Question #4 : Matrices

Example Question #8 : Find The Product Of A Matrix And A Scalar

Example Question #9 : Find The Product Of A Matrix And A Scalar

Example Question #10 : Find The Product Of A Matrix And A Scalar

Example Question #1 : How To Subtract Matrices

All SAT II Math I Resources

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