We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

SAT II Math II : Matrices

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

Create An Account

All SAT II Math II Resources

2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 Next →

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

Report an Error

Example Question #1 : Find The Sum Or Difference Of Two 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

Report an Error

Example Question #1 : 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:

$\frac{103}{2}$

102

$\frac{99}{2}$

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

Report an Error

Example Question #21 : Matrices

$A = \begin{bmatrix} 0&0& 0 & 3\\ 0& 0 & 3& 0\\ 0& 3& 0& 0 \\3& 0&0& 0 \end{bmatrix}$

Evaluate $A^{2}$ .

Possible Answers:

$A ^{2}= \begin{bmatrix} 9& 0&0& 0\\ 0& 9& 0& 0\\ 0& 0 & 9& 0\\ 0&0& 0 & 9 \end{bmatrix}$

$A ^{2}= \begin{bmatrix}9 & 0&0& 0 \\0&0& 0 & 9 \\ 0& 0 & 9& 0 \\ 0& 9& 0& 0 \end{bmatrix}$

$A ^{2}= \begin{bmatrix} 0&0& 0 & 0 \\ 0&0& 0 & 0 \\ 0&0& 0 & 0\\ 0&0& 0 & 0 \end{bmatrix}$

$A ^{2}= \begin{bmatrix}0& 0 & 9& 0 \\ 0& 9& 0& 0 \\ 9 & 0&0& 0 \\0&0& 0 & 9 \end{bmatrix}$

$A ^{2}= \begin{bmatrix}0&0& 0 & 9\\0& 0 & 9& 0 \\ 0& 9& 0& 0 \\ 9 & 0&0& 0 \end{bmatrix}$

Correct answer:

$A ^{2}= \begin{bmatrix} 9& 0&0& 0\\ 0& 9& 0& 0\\ 0& 0 & 9& 0\\ 0&0& 0 & 9 \end{bmatrix}$

Explanation:

$A^{2} = A \cdot A = \begin{bmatrix} 0&0& 0 & 3\\ 0& 0 & 3& 0\\ 0& 3& 0& 0 \\3& 0&0& 0 \end{bmatrix}\begin{bmatrix} 0&0& 0 & 3\\ 0& 0 & 3& 0\\ 0& 3& 0& 0 \\3& 0&0& 0 \end{bmatrix}$

The element in row , column , of $A^{2}$ can be found by multiplying row of by row of - that is, by multiplying elements in corresponding positions and adding the products. Therefore,

$A^{2} = \begin{bmatrix} 0 \cdot 0 + 0 \cdot 0 +0 \cdot 0 +3 \cdot 3 &0 \cdot 0 +0 \cdot 0 + 0 \cdot 3 +3 \cdot 0 &0 \cdot 0 + 0 \cdot 3 +0 \cdot 0 +3 \cdot 0&0 \cdot 3 + 0 \cdot 0 +0 \cdot 0 +3 \cdot 0 \\ 0 \cdot 0 + 0 \cdot 0 +3 \cdot 0 +0 \cdot 3 &0 \cdot 0 +0 \cdot 0 + 3 \cdot 3 +0 \cdot 0 &0 \cdot 0 + 0 \cdot 3 +3 \cdot 0 +0 \cdot 0&0 \cdot 3 + 0 \cdot 0 +3 \cdot 0 +0 \cdot 0\\ 0 \cdot 0 + 3 \cdot 0 +0 \cdot 0 +0 \cdot 3 &0 \cdot 0 +3 \cdot 0 + 0 \cdot 3 +0 \cdot 0 &0 \cdot 0 + 3 \cdot 3 +0 \cdot 0 +0\cdot 0&0 \cdot 3 + 3 \cdot 0 +0 \cdot 0 +0 \cdot 0\\3 \cdot 0 + 0 \cdot 0 +0 \cdot 0 +0 \cdot 3 &3 \cdot 0 +0 \cdot 0 + 0 \cdot 3 +0 \cdot 0 &3 \cdot 0 + 0 \cdot 3 +0 \cdot 0 +0 \cdot 0&3 \cdot 3 + 0 \cdot 0 +0 \cdot 0 +0\cdot 0 \end{bmatrix}$

$= \begin{bmatrix} 0+0+0+9 & 0 +0+0+0 & 0 +0+0+0 & 0 +0+0+0 \\ 0 +0+0+0 & 0+0+9+0 & 0 +0+0+0& 0 +0+0+0 \\ 0 +0+0+0 & 0 +0+0+0& 0+9+0+0 & 0 +0+0+0 \\0 +0+0+0 & 0 +0+0+0& 0 +0+0+0 & 9+0+0+0 \end{bmatrix}$

$= \begin{bmatrix} 9& 0&0& 0\\ 0& 9& 0& 0\\ 0& 0 & 9& 0\\ 0&0& 0 & 9 \end{bmatrix}$

Report an Error

Example Question #22 : Matrices

$A =\begin{bmatrix} - 3 & x\\ 5& 4 \end{bmatrix}$

The determinant of this matrix is equal to 4. Evaluate .

Possible Answers:

$x = -3 \frac{1}{5}$

x = -2

x = -8

x = 8

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

Correct answer:

$x = -3 \frac{1}{5}$

Explanation:

A matrix $A = \begin{bmatrix} a& b\\ c& d \end{bmatrix}$ has as its determinant D = ad-bc . Setting a= -3, b = x, c = 5, d = 4 , this becomes

$D = ad-bc =-3 \cdot 4 - x \cdot 5 = -12 - 5x$

Set this determinant equal to 4 and solve for :

-12 - 5x = 4

-12 - 5x + 12 = 4 + 12

-5x = 16

$\frac{-5x}{-5 } = \frac{16}{-5}$

$x = -3 \frac{1}{5}$

the correct response.

Report an Error

Example Question #23 : Matrices

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:

has an inverse for all real values of

There is one such value: x = 144

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

There is one such value: x = 13

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 .

Report an Error

Example Question #21 : Matrices

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:

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

A+ I does not have an inverse.

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

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

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

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

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

Report an Error

Example Question #25 : Matrices

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

Evaluate $b_{11}$ .

Possible Answers:

$\frac{1}{10}$

$\frac{1}{16}$

$-\frac{1}{20}$

$-\frac{1}{8}$

$A^{-1}$ 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:

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

Report an Error

Example Question #21 : Matrices

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

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

Possible Answers:

x= 0

x= -4 or x = 4

x = 16

x= -2 or x = 2

None of these

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 .

Report an Error

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 is one such value: x = 32

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

There is one such value: x = 0

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

There is one such value: x = 8

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.

Report an Error

← Previous 1 2 3 Next →

View SAT Subject Test in Mathematics Level 2 Tutors

Luu Chi Hieu
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Computer Science.

View SAT Subject Test in Mathematics Level 2 Tutors

Antonia
Certified Tutor

University of Michigan-Ann Arbor, Bachelor of Engineering, Nuclear Engineering. University of Wisconsin-Madison, Current Grad...

View SAT Subject Test in Mathematics Level 2 Tutors

Reuben
Certified Tutor

University of Alberta, Bachelor of Science, Mathematics.

All SAT II Math II Resources

2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Chemistry Tutors in Atlanta, Spanish Tutors in Dallas Fort Worth, GMAT Tutors in Philadelphia, ACT Tutors in Atlanta, MCAT Tutors in Washington DC, Chemistry Tutors in Houston, LSAT Tutors in Philadelphia, LSAT Tutors in Los Angeles, ISEE Tutors in Houston, Biology Tutors in San Diego

Popular Courses & Classes

GMAT Courses & Classes in Seattle, SAT Courses & Classes in New York City, ACT Courses & Classes in Houston, GMAT Courses & Classes in San Diego, GMAT Courses & Classes in Phoenix, Spanish Courses & Classes in Seattle, GMAT Courses & Classes in Miami, ACT Courses & Classes in Denver, GRE Courses & Classes in Philadelphia, GMAT Courses & Classes in Denver

Popular Test Prep

MCAT Test Prep in Dallas Fort Worth, ACT Test Prep in Houston, SAT Test Prep in San Diego, GMAT Test Prep in Seattle, SAT Test Prep in Phoenix, SSAT Test Prep in Atlanta, SAT Test Prep in Houston, ACT Test Prep in Dallas Fort Worth, SSAT Test Prep in Denver, ACT Test Prep in Philadelphia

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

SAT II Math II : Matrices

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

All SAT II Math II Resources

Example Questions

Example Question #1 : How To Subtract Matrices

Example Question #1 : Find The Sum Or Difference Of Two Matrices

Example Question #1 : Matrices

Example Question #21 : Matrices

Example Question #22 : Matrices

Example Question #23 : Matrices

Example Question #21 : Matrices

Example Question #25 : Matrices

Example Question #21 : Matrices

Example Question #21 : Matrices

All SAT II Math II Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: