Calculus 3 : Matrices

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #111 : Matrices

Find the matrix product of \(\displaystyle A\times b\), where \(\displaystyle A=\begin{vmatrix}-8 &1 &0 \\-8 &-5 &-9 \\2 &-7 &7 \end{vmatrix}\) and \(\displaystyle b=\begin{vmatrix} 4\\-5 \\-8 \end{vmatrix}\) 

Possible Answers:

\(\displaystyle \begin{vmatrix} -37\\65 \\-13 \end{vmatrix}\)

\(\displaystyle \begin{vmatrix} 11\\-5 \\4 \end{vmatrix}\)

\(\displaystyle \begin{vmatrix} -23\\18 \\-42 \end{vmatrix}\)

\(\displaystyle \begin{vmatrix} 86\\25 \\44 \end{vmatrix}\)

\(\displaystyle \begin{vmatrix} -39\\-115 \\82 \end{vmatrix}\)

Correct answer:

\(\displaystyle \begin{vmatrix} -37\\65 \\-13 \end{vmatrix}\)

Explanation:

In order to multiply two matrices, \(\displaystyle A\times b\), the respective dimensions of each must be of the form \(\displaystyle m \times n\) and \(\displaystyle n \times p\) to create an \(\displaystyle m\times p\) (notation is rows x columns) matrix. Unlike the multiplication of individual values, the order of the matrices does matter.

\(\displaystyle (A\times b \neq b \times A)\)

For a multiplication of the form

\(\displaystyle \begin{vmatrix} A_{1,1}&A_{1,2} &... &A_{1,n} \\ A_{2,1}&A_{2,2} &.. &A_{2,n} \\ ...& ...&... &... \\ A_{m,1}&A_{m,2} &... &A_{m,n} \end{vmatrix}\times \begin{vmatrix} b_{1,1}&b_{1,2} &... &b_{1,p} \\ b_{2,1}&b_{2,2} &... &b_{2,p} \\ ...&... &... &... \\ b_{n,1}&b_{n,2} &... &b_{n,p} \end{vmatrix}\)

The resulting matrix is

\(\displaystyle \begin{vmatrix} A_{1,1}b_{1,1}+A_{1,2}b_{2,1}+...+A_{1,n}b_{n,1}&... &A_{1,1}b_{1,p}+A_{1,2}b_{2,p}+...+A_{1,n}b_{n,p} \\ ...&... &... \\ A_{m,1}b_{1,1}+A_{m,2}b_{2,1}+...+A_{m,n}b_{n,1}&... &A_{m,1}b_{1,p}+A_{m,2}b_{2,p}+...+A_{m,n}b_{n,p} \end{vmatrix}\)

The notation may be daunting but numerical examples may elucidate.

We're told that 

\(\displaystyle A=\begin{vmatrix}-8 &1 &0 \\-8 &-5 &-9 \\2 &-7 &7 \end{vmatrix}\) and \(\displaystyle b=\begin{vmatrix} 4\\-5 \\-8 \end{vmatrix}\)

The resulting matrix product is then:

\(\displaystyle A\times b = \begin{vmatrix} -32-5+0\\-32+25+72 \\8+35-56 \end{vmatrix}=\begin{vmatrix} -37\\65 \\-13 \end{vmatrix}\)

Example Question #112 : Matrices

Find the matrix product of \(\displaystyle A\times b\), where \(\displaystyle A=\begin{vmatrix}9 &7 &9 \\7 &-5 &-6 \\-1 &3 &-9 \end{vmatrix}\) and \(\displaystyle b=\begin{vmatrix} -7\\-7 \\8 \end{vmatrix}\) 

Possible Answers:

\(\displaystyle \begin{vmatrix}40 \\62 \\86 \end{vmatrix}\)

\(\displaystyle \begin{vmatrix}-40 \\-62 \\86 \end{vmatrix}\)

\(\displaystyle \begin{vmatrix}40 \\62 \\-86 \end{vmatrix}\)

\(\displaystyle \begin{vmatrix}-40 \\-62 \\-86 \end{vmatrix}\)

\(\displaystyle \begin{vmatrix}-40 \\62 \\-86 \end{vmatrix}\)

Correct answer:

\(\displaystyle \begin{vmatrix}-40 \\-62 \\-86 \end{vmatrix}\)

Explanation:

In order to multiply two matrices, \(\displaystyle A\times b\), the respective dimensions of each must be of the form \(\displaystyle m \times n\) and \(\displaystyle n \times p\) to create an \(\displaystyle m\times p\) (notation is rows x columns) matrix. Unlike the multiplication of individual values, the order of the matrices does matter.

\(\displaystyle (A\times b \neq b \times A)\)

For a multiplication of the form

\(\displaystyle \begin{vmatrix} A_{1,1}&A_{1,2} &... &A_{1,n} \\ A_{2,1}&A_{2,2} &.. &A_{2,n} \\ ...& ...&... &... \\ A_{m,1}&A_{m,2} &... &A_{m,n} \end{vmatrix}\times \begin{vmatrix} b_{1,1}&b_{1,2} &... &b_{1,p} \\ b_{2,1}&b_{2,2} &... &b_{2,p} \\ ...&... &... &... \\ b_{n,1}&b_{n,2} &... &b_{n,p} \end{vmatrix}\)

The resulting matrix is

\(\displaystyle \begin{vmatrix} A_{1,1}b_{1,1}+A_{1,2}b_{2,1}+...+A_{1,n}b_{n,1}&... &A_{1,1}b_{1,p}+A_{1,2}b_{2,p}+...+A_{1,n}b_{n,p} \\ ...&... &... \\ A_{m,1}b_{1,1}+A_{m,2}b_{2,1}+...+A_{m,n}b_{n,1}&... &A_{m,1}b_{1,p}+A_{m,2}b_{2,p}+...+A_{m,n}b_{n,p} \end{vmatrix}\)

The notation may be daunting but numerical examples may elucidate.

We're told that 

\(\displaystyle A=\begin{vmatrix}9 &7 &9 \\7 &-5 &-6 \\-1 &3 &-9 \end{vmatrix}\) and \(\displaystyle b=\begin{vmatrix} -7\\-7 \\8 \end{vmatrix}\)

The resulting matrix product is then:

\(\displaystyle A\times b = \begin{vmatrix} -63-49+72\\-49+35-48 \\7-21-72 \end{vmatrix}=\begin{vmatrix}-40 \\-62 \\-86 \end{vmatrix}\)

Example Question #703 : Vectors And Vector Operations

Calculate the determinant of \(\displaystyle A\).

\(\displaystyle A=\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\)

Possible Answers:

\(\displaystyle \det(A)=2\)

\(\displaystyle \det(A)=0\)

\(\displaystyle \det(A)=-1\)

\(\displaystyle \det(A)=-2\)

\(\displaystyle \det(A)=1\)

Correct answer:

\(\displaystyle \det(A)=-2\)

Explanation:

In order to find the determinant, we need to multiply the main diagonal components and then subtract the off main diagonal components.

\(\displaystyle \det(A)=1(4)-(3)(2)=4-6=-2\)

Example Question #704 : Vectors And Vector Operations

Find the product of the two matrices:

\(\displaystyle \mathbf{A}\mathbf{B}\)

Where

\(\displaystyle \mathbf{A}=\begin{pmatrix} 3&1 \\ 2&-2 \end{pmatrix}\)

and

\(\displaystyle \mathbf{B}=\begin{pmatrix} -1&4 \\ 0&5 \end{pmatrix}\)

Possible Answers:

\(\displaystyle \begin{pmatrix} -3&17 \\ -2&-2 \end{pmatrix}\)

\(\displaystyle \begin{pmatrix} 5&10 \\ -9&-10 \end{pmatrix}\)

\(\displaystyle \begin{pmatrix} 5&-9 \\ 10&-10 \end{pmatrix}\)

\(\displaystyle \begin{pmatrix} 3&-2 \\ 17&-2 \end{pmatrix}\)

Correct answer:

\(\displaystyle \begin{pmatrix} -3&17 \\ -2&-2 \end{pmatrix}\)

Explanation:

\(\displaystyle \mathbf{AB}=\begin{pmatrix} 3&1 \\ 2&-2 \end{pmatrix}\textup{}\)\(\displaystyle \begin{pmatrix} -1&4 \\ 0&5 \end{pmatrix}\)

\(\displaystyle =\begin{pmatrix} (3)(-1)+(1)(0)&(3)(4)+(1)(5) \\ (2)(-1)+(-2)(0)&(2)(4)+(-2)(5) \end{pmatrix}\)

\(\displaystyle =\begin{pmatrix} (-3+0)&(12+5) \\ (-2+0)&(8-10) \end{pmatrix}=\begin{pmatrix} -3&17 \\ -2&-2 \end{pmatrix}\)

Example Question #705 : Vectors And Vector Operations

Evaluate the following matrix operation:

\(\displaystyle 2\mathbf{A}+4\mathbf{B}\)

where

\(\displaystyle \mathbf{A}=\begin{pmatrix} 1&4 \\ -3&0 \end{pmatrix}, \, \;\mathbf{B}=\begin{pmatrix} 2&-2 \\ 1&-1 \end{pmatrix}\)

Possible Answers:

\(\displaystyle \begin{pmatrix} -6&16 \\ -10&4 \end{pmatrix}\)

\(\displaystyle \begin{pmatrix} 10&0 \\ -2&-4 \end{pmatrix}\)

\(\displaystyle \begin{pmatrix} 8&12 \\ -10&-2 \end{pmatrix}\)

\(\displaystyle \begin{pmatrix} 2&-8 \\ -2&-1 \end{pmatrix}\)

\(\displaystyle \begin{pmatrix} -1&6 \\ -4&1 \end{pmatrix}\)

Correct answer:

\(\displaystyle \begin{pmatrix} 10&0 \\ -2&-4 \end{pmatrix}\)

Explanation:

\(\displaystyle 2\mathbf{A}+4\mathbf{B}=2\begin{pmatrix} 1&4 \\ -3&0 \end{pmatrix}+4 \begin{pmatrix} 2&-2 \\ 1&-1 \end{pmatrix}= \begin{pmatrix} 2&8 \\ -6&0 \end{pmatrix}+ \begin{pmatrix} 8&-8 \\ 4&-4 \end{pmatrix}\)

\(\displaystyle =\begin{pmatrix} 2+8&8-8 \\ -6+4&0-4 \end{pmatrix}= \begin{pmatrix} 10&0 \\ -2&-4 \end{pmatrix}\)

Example Question #111 : Matrices

Find the determinant of the matrix A:

\(\displaystyle \mathbf{A}= \begin{pmatrix} 7&4 \\ 5&2 \end{pmatrix}\)

Possible Answers:

-6

-24

28

14

18

Correct answer:

-6

Explanation:

The determinant of a matrix

\(\displaystyle \begin{pmatrix} a&b \\ c&d \end{pmatrix}\)

is defined as:

\(\displaystyle ad-bc\)

Here, that becomes:

\(\displaystyle (7)(2)-4(5)=14-20=-6\)

Example Question #112 : Matrices

\(\displaystyle A=\begin{pmatrix} 1&4 \\ 5&7 \end{pmatrix}\)

What is \(\displaystyle A^t\) ?

Possible Answers:

\(\displaystyle A^t=\begin{pmatrix} 4&7 \\ 1& 5 \end{pmatrix}\)

\(\displaystyle A^t=\begin{pmatrix} 1&-4 \\ -5& 7 \end{pmatrix}\)

\(\displaystyle A^t=\begin{pmatrix} 1&5 \\ 4& 7 \end{pmatrix}\)

\(\displaystyle A^t=\begin{pmatrix} 5&1 \\ 7& 4 \end{pmatrix}\)

Correct answer:

\(\displaystyle A^t=\begin{pmatrix} 1&5 \\ 4& 7 \end{pmatrix}\)

Explanation:

To find \(\displaystyle A^t\), we write the rows of \(\displaystyle A\) as columns.

Hence,

\(\displaystyle A^t=\begin{pmatrix} 1&5 \\ 4& 7 \end{pmatrix}\)

Example Question #113 : Matrices

\(\displaystyle A=\begin{pmatrix} 4&3 \\ 2&5 \end{pmatrix}\)

What is \(\displaystyle A^t\)?

Possible Answers:

\(\displaystyle A^t=\begin{pmatrix} 4&2 \\ 3& 5 \end{pmatrix}\)

\(\displaystyle A^t=\begin{pmatrix} 3&5 \\ 4&2 \end{pmatrix}\)

\(\displaystyle A^t=\begin{pmatrix} 4&-3 \\ -2&5 \end{pmatrix}\)

\(\displaystyle A^t=\begin{pmatrix} 2&4 \\ 5&3 \end{pmatrix}\)

Correct answer:

\(\displaystyle A^t=\begin{pmatrix} 4&2 \\ 3& 5 \end{pmatrix}\)

Explanation:

To find \(\displaystyle A^t\), we write the rows of \(\displaystyle A\) as columns.

Hence,

\(\displaystyle A^t=\begin{pmatrix} 4&2 \\ 3& 5 \end{pmatrix}\)

Example Question #711 : Vectors And Vector Operations

Find the determinant of the matrix.  

\(\displaystyle \begin{bmatrix} 1& 4&5 \\ 2& 3&4 \\ 0&1 &1 \end{bmatrix}\)

Possible Answers:

\(\displaystyle -1\)

\(\displaystyle 1\)

\(\displaystyle 2\)

\(\displaystyle 0\)

Correct answer:

\(\displaystyle 1\)

Explanation:

The formula for the determinant of a 3x3 matrix

\(\displaystyle A=\begin{bmatrix} a&b &c \\ d &e & f\\ g&h &i \end{bmatrix}\) is  

\(\displaystyle \left | A\right |=a(ei-hf)-b(di-gf)+c(dh-ge)\).

Using the matrix we were given, we get 

\(\displaystyle 1(3-4)-4(2-0)+5(2-0)=1\).

Example Question #115 : Matrices

Perform the matrix operation. 

\(\displaystyle \begin{bmatrix} 2& 3\\6 & 7 \end{bmatrix}+\begin{bmatrix} 1& 5\\ 0& 1 \end{bmatrix}\)

Possible Answers:

\(\displaystyle \begin{bmatrix} 3& 8\\ 6&8 \end{bmatrix}\)

\(\displaystyle \begin{bmatrix} 1&3 \\ 2&4 \end{bmatrix}\)

\(\displaystyle \begin{bmatrix} 9&1 \\ 2&3 \end{bmatrix}\)

\(\displaystyle \begin{bmatrix} 5&6 \\ 1&2 \end{bmatrix}\)

Correct answer:

\(\displaystyle \begin{bmatrix} 3& 8\\ 6&8 \end{bmatrix}\)

Explanation:

The formula for adding a pair of 2x2 matrices is 

\(\displaystyle \begin{bmatrix} a& b\\ c&d \end{bmatrix}+\begin{bmatrix} e&f \\ g &h \end{bmatrix}=\begin{bmatrix} (a+e)&(b+f) \\ (c+g)& (d+h) \end{bmatrix}\).

Using the matrices in the problem statement, we get 

\(\displaystyle \begin{bmatrix} (2+1)&(3+5) \\ (6+0)&(7+1) \end{bmatrix}=\begin{bmatrix} 3&8 \\ 6& 8 \end{bmatrix}\)

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