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SAT II Math I : Mathematical Relationships

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Example Questions

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Example Question #11 : Matrices

Let $A = \begin{bmatrix} 3&2 \\ 5& 4 \end{bmatrix}$ and $B = A^{-1}$ .

Evaluate $b_{21}$ .

Possible Answers:

$-\frac{5}{2}$

$\frac{3}{2}$

$A^{-1}$ does not exist.

Correct answer:

$-\frac{5}{2}$

Explanation:

The inverse $A ^{-1}$ of any two-by-two matrix can be found according to this pattern:

If $A = \begin{bmatrix} a& b\\ c& d \end{bmatrix}$

then

$A ^{-1}= \frac{1}{D} \begin{bmatrix}d& - b\\ -c& a \end{bmatrix} = \begin{bmatrix}\frac{d}{D}& - \frac{b}{D}\\ -\frac{c}{D}& \frac{a }{D} \end{bmatrix}$ ,

where determinant is equal to ad-cb .

Therefore, if $B = A ^{-1}$ , then $b_{21}$ , the second row/first column entry in the matrix , can be found by setting a = 3, b= 2, c= 5, d=4 , then evaluating:

$b_{21} = -\frac{c}{D} = -\frac{c}{ad-bc} = -\frac{5}{3 \cdot 4 - 2 \cdot 5} = -\frac{5}{12-10} = -\frac{5}{2}$ .

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Example Question #51 : Mathematical Relationships

Solve: $\begin{bmatrix} 1&-1 \\ 3& -9 \end{bmatrix}+\begin{bmatrix} -1&-2 \\ 3& -9 \end{bmatrix}$

Possible Answers:

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

$\begin{bmatrix} 0&-3 \\ 6& -18 \end{bmatrix}$

$\textup{The answer is not given.}$

$\begin{bmatrix} 0&-1 \\ 6& 0 \end{bmatrix}$

$\begin{bmatrix} 2&-3 \\ 6& -18 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 0&-3 \\ 6& -18 \end{bmatrix}$

Explanation:

To compute the matrices, simply add the terms with the correct placement in the matrices. The resulting matrix is two by two.

$\begin{bmatrix} 1&-1 \\ 3& -9 \end{bmatrix}+\begin{bmatrix} -1&-2 \\ 3& -9 \end{bmatrix}= \begin{bmatrix} 1+(-1)&(-1)+(-2) \\ 3+3& (-9)+(-9) \end{bmatrix}$

The answer is: $\begin{bmatrix} 0&-3 \\ 6& -18 \end{bmatrix}$

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Example Question #91 : Sat Subject Test In Math I

$\textup{Let }A = \begin{bmatrix} -2 &x \\ 5 & x- 7 \end{bmatrix}$ .

$\textup{Which of the following values of }x\textup{ makes a matrix without an inverse?}$

Possible Answers:

$A \textup{ has an inverse regardless of the value of }x$

x = -2

x= -5

x = 5

x = 2

Correct answer:

x = 2

Explanation:

A matrix $A = \begin{bmatrix} a& b\\ c& d \end{bmatrix}$ lacks an inverse if and only if its determinant ad-bc is equal to zero. The determinant of $A = \begin{bmatrix} -2 &x \\ 5 & x- 7 \end{bmatrix}$ is

D = -2(x-7) - 5x = -2x+14 - 5x = -7x +14

Set this equal to 0 and solve for :

-7x +14 = 0

-7x +14- 14 = 0 - 14

-7x = - 14

$-7x\div (-7) = - 14 \div (-7)$

x = 2 ,

the correct response.

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Example Question #51 : Mathematical Relationships

Let $A = \begin{bmatrix} 6&4 \\ 8&d \end{bmatrix}$

Which of the following values of makes a matrix without an inverse?

Possible Answers:

$d = 5 \frac{1}{2}$

None of these

d = 3

$d = 5 \frac{1}{3}$

d = 12

Correct answer:

$d = 5 \frac{1}{3}$

Explanation:

A matrix $A = \begin{bmatrix} a& b\\ c& d \end{bmatrix}$ lacks an inverse if and only if its determinant ad-bc is equal to zero. The determinant of $A = \begin{bmatrix} 6&4 \\ 8&d \end{bmatrix}$ is

$D = 6\cdot d - 4 \cdot 8 = 6d - 32$ .

We seek the value of that sets this quantity equal to 0. Setting it as such then solving for :

6d - 32 = 0

6d - 32 + 32 = 0 + 32

6d = 32

$6d \div 6 = 32 \div 6$

$d = 5 \frac{1}{3}$ ,

the correct response.

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Example Question #11 : Matrices

Let equal the following:

$A = \begin{bmatrix} 8 &x \\ x&32 \end{bmatrix}$

Which of the following values of makes a matrix without an inverse?

Possible Answers:

There is one such value: x = 20

There are two such values: x = -2 or x = 2

There is one such value: x = 0

There are two such values: x = -16 or x = 16

There are two such values: x = -4 or x = 4

Correct answer:

There are two such values: x = -16 or x = 16

Explanation:

A matrix $A = \begin{bmatrix} a& b\\ c& d \end{bmatrix}$ lacks an inverse if and only if its determinant ad-bc is equal to zero. The determinant of $A = \begin{bmatrix} 8 &x \\ x&32 \end{bmatrix}$ is

$D = 8 \cdot 32 - x \cdot x = 256 -x^{2}$

Setting this equal to 0 and solving for :

$256 -x^{2} = 0$

$256 -x^{2}+ x^{2} = 0+ x^{2}$

$x^{2} =256$

$x = \pm \sqrt{256}$

$x = \pm 16$

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Example Question #91 : Sat Subject Test In Math I

Let equal the following:

$A = \begin{bmatrix} 4 &x \\ x&-64\end{bmatrix}$ .

Which of the following real values of makes a matrix without an inverse?

Possible Answers:

There are two such values: x = -4 or x = 4

There is one such value: x = -30

There are two such values: x = -2 or x = 2

There are two such values: x = -16 or x = 16

has an inverse for all real values of

Correct answer:

has an inverse for all real values of

Explanation:

A matrix $A = \begin{bmatrix} a& b\\ c& d \end{bmatrix}$ lacks an inverse if and only if its determinant ad-bc is equal to zero. The determinant of $A = \begin{bmatrix} 4 &x \\ x&-64\end{bmatrix}$ is

$D = 4 \cdot 64 - x (-x) = 256 +x^{2}$ , so

$256 +x^{2} = 0$

$256 +x^{2} - 256 = 0 - 256$

$x^{2} = - 256$

Since the square of all real numbers is nonnegative, this equation has no real solution. It follows that the determinant cannot be 0 for any real value of , and that must have an inverse for all real .

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Example Question #1 : Sequences

The first two numbers of a sequence are, in order, 1 and 4. Each successive element is formed by adding the previous two. What is the sum of the first six elements of the sequence?

Possible Answers:

Correct answer:

Explanation:

The first six elements are as follows:

$a_{1} = 1 \;$

$a_{2}= 4$

$a_{3} = 1 + 4 = 5$

$a_{4} = 4 + 5 = 9$

$a_{5} = 5+ 9 = 14$

$a_{6} = 9 + 14 = 23$

Add them:

1 + 4 + 5 + 9 + 14 + 23 = 56

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Example Question #1 : Sequences

The first and third terms of a geometric sequence are 3 and 108, respectively. All What is the sixth term?

Possible Answers:

23,328

Insufficient information is given to answer the question.

$-265\frac{1}{2}$

-23,328

$265\frac{1}{2}$

Correct answer:

Insufficient information is given to answer the question.

Explanation:

Let the common ratio of the sequence be . Then The first three terms of the sequence are $3,3r,3r^{2}$ . The third term is 108, so

$3r^{2} =108$

$r^{2} = 36$

r = 6 or r = -6 .

The common ratio can be either - not enough information exists for us to determine which.

The sixth term is

$a_{1}r^{6-1} = 3r^{5}$

If r = 6 , the seventh term is $3 \cdot 6^{5}= 3 \cdot 7,776 = 23,328$ .

If r =- 6 , the seventh term is $3 \cdot \left (-6 \right )^{5}= 3 \cdot \left (-7,776 \right )= -23,328$ .

Therefore, not enough information exists to determine the sixth term of the sequence.

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Example Question #1 : Sequences

The first and third terms of a geometric sequence are 2 and 50, respectively. What is the seventh term?

Possible Answers:

-31,250

Insufficient information is given to answer the question.

31,250

146

Correct answer:

31,250

Explanation:

Let the common ratio of the sequence be . Then The first three terms of the sequence are $2,2r,2r^{2}$ . The third term is 50, so

$2r^{2} = 50$

$r^{2} = 25$

r = 5 or r = -5 .

Not enough information is given to choose which one is the common ratio. But the seventh term is

$a_{1}r^{7-1} = 2r^{6}$

If r = 5 , the seventh term is $2 \cdot 5^{6}= 2 \cdot 15,625 = 31,250$ .

If r = - 5 , the seventh term is $2 \cdot\left ( -5 \right )^{6}= 2 \cdot 15,625 = 31,250$ .

Either way, the seventh term is 31,250.

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Example Question #4 : Sequences

The sum of 3 odd consecutive numbers is 345. What is the largest number in the sequence?

Possible Answers:

117

103

119

121

Correct answer:

117

Explanation:

When you are dealing with arithmetic means, it is best to define one number in the sequence as x and every other number relative to x.

Because we are trying to find the largest of three numbers, let's define x as the largest number in the equation. Because each number is a consecutive odd number, we must subtract 2 to get to the next number in the sequence.

x: largest number in sequence

x-2: middle number in sequence

x-4: smallest number in sequence

Now, let's make an equation finds the sum of all the numbers in the sequence and set it equal to 354.

$\small x+(x-2)+(x-4)=345$

$\small 3x-6+6=345+6$

$\small \frac{3x}{3}=\frac{351}{3}$

$\small x=117$

117 is the largest number in the sequence.

To check yourself, you can add up the numbers in the sequence {113, 115, 117}.

$\small 113+115+117=345$

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SAT II Math I : Mathematical Relationships

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

All SAT II Math I Resources

Example Questions

Example Question #11 : Matrices

Example Question #51 : Mathematical Relationships

Example Question #91 : Sat Subject Test In Math I

Example Question #51 : Mathematical Relationships

Example Question #11 : Matrices

Example Question #91 : Sat Subject Test In Math I

Example Question #1 : Sequences

Example Question #1 : Sequences

Example Question #1 : Sequences

Example Question #4 : Sequences

The sum of 3 odd consecutive numbers is 345. What is the largest number in the sequence?

117 is the largest number in the sequence.

All SAT II Math I Resources

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