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SAT II Math II : SAT Subject Test in Math II

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

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Example Questions

Example Question #103 : Mathematical Relationships

Let $A = \begin{bmatrix}x & 8 \\ 18 & x \end{bmatrix}$ .

Which of the following real value(s) of makes a matrix without an inverse?

Possible Answers:

There are two such values: x = 8 and x = 18

There is one such value: x = 13

There is one such value: x = 144

has an inverse for all real values of

There are two such values: x = -12 and x = 12

Correct answer:

There are two such values: x = -12 and x = 12

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 $A = \begin{bmatrix}x & 8 \\ 18 & x \end{bmatrix}$ is

$D = x \cdot x - 8 \cdot 18 = x^{2}- 144$ .

Setting this equal to 0:

$x^{2}- 144 = 0$

$x^{2}- 144+144 = 0 +144$

$x^{2} = 144$

Taking the square root of both sides:

$x = \pm \sqrt{144}$

$x = \pm 12$

The matrix therefore has no inverse if either x = -12 or x = 12 .

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Example Question #104 : Mathematical Relationships

Let be the two-by-two identity matrix and $A = \begin{bmatrix} 4& -7\\ 8& 3 \end{bmatrix}$ .

Which matrix is equal to the inverse of A+I ?

Possible Answers:

$\begin{bmatrix} \frac{71}{68}&\frac{7}{68} \\ -\frac{2}{17}& \frac{18}{17} \end{bmatrix}$

$\begin{bmatrix} \frac{41}{44}& -\frac{7}{44}\\ \frac{2}{11}& \frac{12}{11} \end{bmatrix}$

A+ I does not have an inverse.

$\begin{bmatrix} -\frac{1}{9}&- \frac{7}{36}\\\frac{ 2}{9}& -\frac{5 }{36}\end{bmatrix}$

$\begin{bmatrix} \frac{ 1}{19}& \frac{7}{76}\\-\frac{ 2}{19}& \frac{5 }{76}\end{bmatrix}$

Correct answer:

$\begin{bmatrix} \frac{ 1}{19}& \frac{7}{76}\\-\frac{ 2}{19}& \frac{5 }{76}\end{bmatrix}$

Explanation:

$A = \begin{bmatrix} 4& -7\\ 8& 3 \end{bmatrix}$ ; the two-by-two identity matrix is $I = \begin{bmatrix} 1& 0\\ 0 & 1 \end{bmatrix}$ . Add the two by adding elements in corresponding positions:

$A +I = \begin{bmatrix} 4+1& -7+0\\ 8+0& 3+1 \end{bmatrix} = \begin{bmatrix} 5& -7 \\ 8 & 4 \end{bmatrix}$ .

The inverse of a two-by-two matrix $B = \begin{bmatrix} a& b\\ c& d \end{bmatrix}$ is $B ^{-1}= \begin{bmatrix} \frac{ d}{\det B}& -\frac{b}{\det B}\\-\frac{ c}{\det B}& \frac{a }{\det B}\end{bmatrix}$ , where

$\det B = ad -bc$ .

We can find $(A+I)^{-1}$ by setting a= 5, b = -7 , c = 8, d = 4 . The determinant of $(A+I)^{-1}$ is

$ad -bc = 5\cdot 4 - (-7) \cdot 8 = 20 - (-56 ) = 76$

Replacing:

$(A+I)^{-1}= \begin{bmatrix} \frac{ 4}{76}& \frac{7}{76}\\-\frac{ 8}{76}& \frac{5 }{76}\end{bmatrix}$ ;

simplifying the fractions, this is

$(A+I)^{-1}= \begin{bmatrix} \frac{ 1}{19}& \frac{7}{76}\\-\frac{ 2}{19}& \frac{5 }{76}\end{bmatrix}$

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Example Question #131 : Sat Subject Test In Math Ii

Let $A = \begin{bmatrix} 5&4 \\ 10& -8 \end{bmatrix}$ and $B = A^{-1}$ .

Evaluate $b_{11}$ .

Possible Answers:

$-\frac{1}{20}$

$\frac{1}{16}$

$\frac{1}{10}$

$-\frac{1}{8}$

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

Correct answer:

$\frac{1}{10}$

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 $b_{11}$ , the first row/first column entry in the matrix , can be found by setting a = 5, b= 4, c= 10, d=-8 , then evaluating:

$b_{11} = \frac{d}{D} = \frac{d}{ad-bc} = \frac{-8}{5(-8)-4(10)} = \frac{-8}{-40 -40} = \frac{-8}{-80} = \frac{1}{10}$

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Example Question #105 : Mathematical Relationships

$A = \begin{bmatrix} 8& x\\ x& 4 \end{bmatrix}$

For which of the following real values of does have determinant of sixteen?

Possible Answers:

x = 16

x= -2 or x = 2

None of these

x= 0

x= -4 or x = 4

Correct answer:

x= -4 or x = 4

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 $A = \begin{bmatrix} 8& x\\ x& 4 \end{bmatrix}$ is

$D = 8 \cdot 4 - x \cdot x = 32 - x^{2}$

We seek the value of that sets this quantity equal to 16. Setting it as such then solving for :

$32 - x^{2} = 16$

$32 - x^{2} - 32 = 16 - 32$

$- x^{2} = -16$

$x^{2} = 16$

$x = \pm 4$

Therefore, either x= -4 or x = 4 .

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Example Question #21 : Matrices

Let equal the following:

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

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

Possible Answers:

There are two such values: x = -20 or x = 20

There is one such value: x = 0

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

There is one such value: x = 8

There is one such value: x = 32

Correct answer:

There is one such value: x = 0

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

$D = x \cdot 32 - 8 \cdot x = 32x - 8x = 24x$

Set this equal to 0 and solve for :

24x = 0

$24x\div 24 = 0 \div 24$

x = 0 ,

the only such value.

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Example Question #1 : Sequences

Evaluate:

$\sum_{i = 1}^{\infty} \left (\frac{4}{3} \right )^{i}$

Possible Answers:

$\frac{7}{4}$

The series diverges

$\frac{5}{3}$

$\frac{5}{4}$

Correct answer:

The series diverges

Explanation:

An infinite series $\sum_{i = 1}^{\infty}a^{i}$ converges to a sum if and only if |a| < 1 . However, in the series $\sum_{i = 1}^{\infty} \left (\frac{4}{3} \right )^{i}$ , this is not the case, as $\left | \frac{4}{3} \right |= \frac{4}{3} > 1$ . This series diverges.

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Example Question #2 : Sequences

Give the next term in this sequence:

$0,0, \frac{1}{2}, \frac{2}{3}, \frac{4}{5}, \frac{7}{8}, \frac{12}{13}, \frac{20}{21},$ _______________

Possible Answers:

$\frac{31}{32}$

$\frac{25}{26}$

$\frac{33}{34}$

$\frac{34}{35}$

Correct answer:

$\frac{33}{34}$

Explanation:

The key to finding the next term lies in the denominators of the third term onwards. They are terms of the Fibonacci sequence, which begin with the terms 1 and 1 and whose subsequent terms are each formed by adding the previous two.

The th term of the sequence is the number $\frac{F_{n}-1}{F_{n}}$ , where $F_{n}$ is the th number in the Fibonacci sequence (since the first two Fibonacci numbers are both 1, the first two terms being 0 fits this pattern). The Fibonacci number following 13 and 21 is their sum, 34, so the next number in the sequence is

$\frac{34-1}{34}= \frac{33}{34}$ .

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Example Question #1 : Sequences

Give the next term in this sequence:

12, 13, 17, 26, 42, 67, __________

Possible Answers:

116

103

100

Correct answer:

103

Explanation:

$12\textbf{ + 1} = 13$

$13\textbf{ + 4}= 17$

$17\textbf{ + 9}= 26$

$26\textbf{ + 16} = 42$

$42\textbf{ + 25}=67$

Each term is derived from the next by adding a perfect square integer; the increment increases from one square to the next higher one each time. To maintain the pattern, add the next perfect square, 36:

$67\textbf{ + 36 } =103$

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Example Question #1 : Sequences

Give the next term in this sequence:

1, 1,3, 5, 11, 21, 43, 85, 171, _____________

Possible Answers:

341

342

340

339

343

Correct answer:

341

Explanation:

Each term is derived from the previous term by doubling the latter and alternately adding and subtracting 1, as follows:

$1 \textbf{ x 2 - 1 } = 1$

$1 \textbf{ x 2 + 1 } = 3$

$3 \textbf{ x 2 - 1 } = 5$

$5 \textbf{ x 2 + 1 } = 11$

$11 \textbf{ x 2 - 1 } =21$

$21 \textbf{ x 2 + 1 } = 43$

$43\textbf{ x 2 - 1 } = 85$

$85 \textbf{ x 2 + 1 } = 171$

The next term is derived as follows:

$171 \textbf{ x 2 - 1 } = 341$

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Example Question #1 : Sequences

Give the next term in this sequence:

$1, \sqrt{2}, \sqrt{6}, 2\sqrt{6}, 2 \sqrt{30}, 12 \sqrt{5},$ _____________

Possible Answers:

$12\sqrt{30}$

The correct answer is not among the other responses.

$72\sqrt{5}$

$84\sqrt{5}$

$12 \sqrt{35}$

Correct answer:

$12 \sqrt{35}$

Explanation:

The pattern becomes more clear if each term is rewritten as a single radical expression:

$1 = \sqrt {1} = \sqrt{1!}$

$\sqrt{2} = \sqrt{2!}$

$\sqrt{6} = \sqrt{3!}$

$2\sqrt{6} =\sqrt{4} \cdot \sqrt{6} = \sqrt {24}=\sqrt{4!}$

$2 \sqrt{30}= \sqrt{4} \cdot \sqrt{30}= \sqrt{120} = \sqrt {5!}$

$12\sqrt{5} = \sqrt{144} \cdot \sqrt{5} = \sqrt{720} = \sqrt{6!}$

The th term is $\sqrt{n!}$ . The next (seventh) term is therefore

$\sqrt{7!}= \sqrt{5,040}= \sqrt{144} \times \sqrt{35} = 12 \sqrt{35}$

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SAT II Math II : SAT Subject Test in Math II

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

All SAT II Math II Resources

Example Questions

Example Question #103 : Mathematical Relationships

Example Question #104 : Mathematical Relationships

Example Question #131 : Sat Subject Test In Math Ii

Example Question #105 : Mathematical Relationships

Example Question #21 : Matrices

Example Question #1 : Sequences

Example Question #2 : Sequences

Example Question #1 : Sequences

Example Question #1 : Sequences

Example Question #1 : Sequences

All SAT II Math II Resources

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