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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

Example Question #1 : Matrices

If $\begin{bmatrix} a&b\\ c&d \end{bmatrix} + \begin{bmatrix} 3&14\\ 12&4 \end{bmatrix} = \begin{bmatrix} 19&2\\ 4&5 \end{bmatrix}$ , what is a+b+c+d ?

Possible Answers:

Correct answer:

Explanation:

You can treat matrices just like you treat other members of an equation. Therefore, you can subtract the matrix

$\begin{bmatrix} 3&14\\ 12&4 \end{bmatrix}$

from both sides of the equation. This gives you:

$\begin{bmatrix} a&b\\ c&d \end{bmatrix}=\begin{bmatrix} 19&2\\ 4&5 \end{bmatrix}-\begin{bmatrix} 3&14\\ 12&4 \end{bmatrix}$

Now, matrix subtraction is simple. You merely subtract each element, matching the correlative spaces with each other:

$\begin{bmatrix} a&b\\ c&d \end{bmatrix} = \begin{bmatrix} 19-3&2-14\\ 4-12&5-4 \end{bmatrix}$

Then, you simplify:

$\begin{bmatrix} 19-3&2-14\\ 4-12&5-4 \end{bmatrix} = \begin{bmatrix} 16&-12\\ -8&1 \end{bmatrix}$

Therefore, a+b+c+d = 16-12-8+1=-3

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Example Question #5 : Matrices

Simplify:

$\begin{bmatrix} 3&4 \\ 5& 6 \end{bmatrix} \begin{bmatrix} 1&2 \\ 3&4 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 22&15 \\ 34&23 \end{bmatrix}$

$\begin{bmatrix} 3&8 \\ 15&24 \end{bmatrix}$

$\begin{bmatrix} 4&8 \\ 8&24 \end{bmatrix}$

$\begin{bmatrix} 11&21 \\ 14&20 \end{bmatrix}$

$\begin{bmatrix} 15&22 \\ 23&34 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 15&22 \\ 23&34 \end{bmatrix}$

Explanation:

The dimensions of the matrices are 2 by 2.

The end result will also be a 2 by 2.

$\begin{bmatrix} 3&4 \\ 5& 6 \end{bmatrix} \begin{bmatrix} 1&2 \\ 3&4 \end{bmatrix}=\begin{bmatrix} (3)(1)+(4)(3)&(3)(2)+(4)(4) \\ (5)(1)+(6)(3)&(5)(2)+(6)(4) \end{bmatrix}$

Evaluate the matrix.

The correct answer is:

$\begin{bmatrix} 15&22 \\ 23&34 \end{bmatrix}$

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Example Question #2 : Matrices

If $\frac{1}{2}$\bigl(\begin{smallmatrix} 2a\\4b \end{smallmatrix} \bigr)$=$\bigl(\begin{smallmatrix} 53 \\ 98 \end{smallmatrix} \bigr)$$ , what is a+b ?

Possible Answers:

$\frac{103}{2}$

$\frac{99}{2}$

102

Correct answer:

102

Explanation:

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:

a=53

and

2b=98 or b=49

Therefore, a+b=53+49=102

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Example Question #91 : Sat Subject Test In Math I

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

Evaluate $b_{21}$ .

Possible Answers:

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

$\frac{3}{2}$

$-\frac{5}{2}$

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:

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

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

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

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

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

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 #51 : Mathematical Relationships

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

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

Possible Answers:

x = -2

x= -5

x = 5

x = 2

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

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 #14 : Matrices

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 = 12

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

None of these

d = 3

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

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 = -16 or x = 16

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

There is one such value: x = 0

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

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 #51 : Mathematical Relationships

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 = -2 or x = 2

There is one such value: x = -30

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

has an inverse for all real values of

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

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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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 #1 : Matrices

Example Question #5 : Matrices

Example Question #2 : Matrices

Example Question #91 : Sat Subject Test In Math I

Example Question #51 : Mathematical Relationships

Example Question #51 : Mathematical Relationships

Example Question #14 : Matrices

Example Question #11 : Matrices

Example Question #51 : Mathematical Relationships

Example Question #1 : Sequences

All SAT II Math I Resources

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