Matrices - AP Calculus BC

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Question

Find the matrix product of $A\times b$ , where $A=\begin{vmatrix} 4& -9\6 &9 \end{vmatrix}$ and $b=\begin{vmatrix} 6\5 \end{vmatrix}$

Answer

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

$(A\times b eq b \times A)$

For a multiplication of the form

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

The resulting matrix is

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

The notation may be daunting but numerical examples may elucidate.

We're told that

$A=\begin{vmatrix} 4& -9\6 &9 \end{vmatrix}$ and $b=\begin{vmatrix} 6\5 \end{vmatrix}$

The resulting matrix product is then:

$A\times b = \begin{vmatrix} 24-45\36+45 \end{vmatrix}=\begin{vmatrix} -21\81 \end{vmatrix}$

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Question

Find the matrix product of $A\times b$ , where $A=\begin{vmatrix} -9&-3 \2 &6 \end{vmatrix}$ and $b=\begin{vmatrix} 3\ -8\end{vmatrix}$

Answer

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

$(A\times b eq b \times A)$

For a multiplication of the form

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

The resulting matrix is

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

The notation may be daunting but numerical examples may elucidate.

We're told that

$A=\begin{vmatrix} -9&-3 \2 &6 \end{vmatrix}$ and $b=\begin{vmatrix} 3\ -8\end{vmatrix}$

The resulting matrix product is then:

$A\times b = \begin{vmatrix} -27+24\6-48 \end{vmatrix}=\begin{vmatrix} -3\-42 \end{vmatrix}$

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Question

Find the matrix product of $A\times b$ , where $A=\begin{vmatrix} -8& -9\6 &3 \end{vmatrix}$ and $b=\begin{vmatrix} 7\-4 \end{vmatrix}$

Answer

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

$(A\times b eq b \times A)$

For a multiplication of the form

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

The resulting matrix is

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

The notation may be daunting but numerical examples may elucidate.

We're told that

$A=\begin{vmatrix} -8& -9\6 &3 \end{vmatrix}$ and $b=\begin{vmatrix} 7\-4 \end{vmatrix}$

The resulting matrix product is then:

$A\times b = \begin{vmatrix} -56+36\42-12 \end{vmatrix}=\begin{vmatrix} -20\30 \end{vmatrix}$

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Question

Find the matrix product of $A\times b$ , where $A=\begin{vmatrix} 4&-9 \-2 &-9 \end{vmatrix}$ and $b=\begin{vmatrix} -6\-3 \end{vmatrix}$

Answer

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

$(A\times b eq b \times A)$

For a multiplication of the form

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

The resulting matrix is

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

The notation may be daunting but numerical examples may elucidate.

We're told that

$A=\begin{vmatrix} 4&-9 \-2 &-9 \end{vmatrix}$ and $b=\begin{vmatrix} -6\-3 \end{vmatrix}$

The resulting matrix product is then:

$A\times b = \begin{vmatrix} -24+27\ 12+27\end{vmatrix}=\begin{vmatrix} 3\39 \end{vmatrix}$

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Question

Find the matrix product of $A\times b$ , where $A=\begin{vmatrix} 6&-9 \6 &-1 \end{vmatrix}$ and $b=\begin{vmatrix} 2\ 3\end{vmatrix}$

Answer

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

$(A\times b eq b \times A)$

For a multiplication of the form

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

The resulting matrix is

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

The notation may be daunting but numerical examples may elucidate.

We're told that

$A=\begin{vmatrix} 6&-9 \6 &-1 \end{vmatrix}$ and $b=\begin{vmatrix} 2\ 3\end{vmatrix}$

The resulting matrix product is then:

$A\times b = \begin{vmatrix} 12-27\ 12-3\end{vmatrix}=\begin{vmatrix} -15\9 \end{vmatrix}$

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Question

Find the matrix product of $A\times b$ , where $A=\begin{vmatrix} -9&-1 \4 &-5 \end{vmatrix}$ and $b=\begin{vmatrix}-9\-5 \end{vmatrix}$

Answer

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

$(A\times b eq b \times A)$

For a multiplication of the form

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

The resulting matrix is

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

The notation may be daunting but numerical examples may elucidate.

We're told that

$A=\begin{vmatrix} -9&-1 \4 &-5 \end{vmatrix}$ and $b=\begin{vmatrix}-9\-5 \end{vmatrix}$

The resulting matrix product is then:

$A\times b = \begin{vmatrix}81+5 \-36+25 \end{vmatrix}=\begin{vmatrix} 86\ -11\end{vmatrix}$

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Question

Find the matrix product of $A\times b$ , where $A=\begin{vmatrix} 0&-8 \-7 &8 \end{vmatrix}$ and $b=\begin{vmatrix} 5\5 \end{vmatrix}$

Answer

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

$(A\times b eq b \times A)$

For a multiplication of the form

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

The resulting matrix is

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

The notation may be daunting but numerical examples may elucidate.

We're told that

$A=\begin{vmatrix} 0&-8 \-7 &8 \end{vmatrix}$ and $b=\begin{vmatrix} 5\5 \end{vmatrix}$

The resulting matrix product is then:

$A\times b = \begin{vmatrix} 0-40\-35+40 \end{vmatrix}=\begin{vmatrix} -40\5 \end{vmatrix}$

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Question

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

Answer

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

$(A\times b eq b \times A)$

For a multiplication of the form

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

The resulting matrix is

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

The notation may be daunting but numerical examples may elucidate.

We're told that

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

The resulting matrix product is then:

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

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Question

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

Answer

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

$(A\times b eq b \times A)$

For a multiplication of the form

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

The resulting matrix is

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

The notation may be daunting but numerical examples may elucidate.

We're told that

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

The resulting matrix product is then:

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

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Question

Find the product of the two matrices:

$\mathbf{A}\mathbf{B}$

Where

$\mathbf{A}=\begin{pmatrix} 3&1 \ 2&-2 \end{pmatrix}$

and

$\mathbf{B}=\begin{pmatrix} -1&4 \ 0&5 \end{pmatrix}$

Answer

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

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

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

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Question

Evaluate the following matrix operation:

$2\mathbf{A}+4\mathbf{B}$

where

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

Answer

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

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

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Question

Find the determinant of the matrix A:

$\mathbf{A}= \begin{pmatrix} 7&4 \ 5&2 \end{pmatrix}$

Answer

The determinant of a matrix

$\begin{pmatrix} a&b \ c&d \end{pmatrix}$

is defined as:

Here, that becomes:

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Question

$A=\begin{pmatrix} 1\ 2\ 3 \end{pmatrix}$

What is ?

Answer

To find

, we write the rows of

as columns.

Hence,

$A^t=\begin{pmatrix} 1&2 &3 \end{pmatrix}$

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Question

$A=\begin{pmatrix} 1&4 \ 5&7 \end{pmatrix}$

What is ?

Answer

To find , we write the rows of as columns.

Hence,

$A^t=\begin{pmatrix} 1&5 \ 4& 7 \end{pmatrix}$

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Question

$A=\begin{pmatrix} 4&3 \ 2&5 \end{pmatrix}$

What is ?

Answer

To find , we write the rows of as columns.

Hence,

$A^t=\begin{pmatrix} 4&2 \ 3& 5 \end{pmatrix}$

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Question

Perform the following matrix operation:

$\begin{bmatrix}1 & 5 \ 2 & 4 \end{bmatrix}-\begin{bmatrix}2 & 4\ 9 & 1 \end{bmatrix}$

Answer

To perform the matrix operation, you must follow the formula

$\begin{bmatrix}a &b \c & d \end{bmatrix}-\begin{bmatrix} e&f \g & h \end{bmatrix}=\begin{bmatrix}{a-e} & b-f\c-g & d-h \end{bmatrix}$ .

Using the values we have, this becomes

$\begin{bmatrix}1-2 & 5-4\ 2-9&4-1 \end{bmatrix}$

which simplified is

$\begin{bmatrix}-1 &1 \ -7 & 3 \end{bmatrix}$

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Question

Find the determinant of the 3x3 matrix:

$\begin{bmatrix}1 &2 &4 \5 & 0&3 \ 4&1 &0 \end{bmatrix}$

Answer

The formula for the determinant is as follows:

$\left | A\right |=a(ei-fh)-b(di-fg)+c(dh-eg)$ ,

where

$A=\begin{bmatrix}a &b &c \ d& e& f\ g &h &i \end{bmatrix}$ .

Using the matrix in the problem, we get

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Question

Find the determinant of the matrix.

$\begin{bmatrix} 1& 4&5 \ 2& 3&4 \ 0&1 &1 \end{bmatrix}$

Answer

The formula for the determinant of a 3x3 matrix

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

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

Using the matrix we were given, we get

.

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Question

Perform the matrix operation.

$\begin{bmatrix} 2& 3\6 & 7 \end{bmatrix}+\begin{bmatrix} 1& 5\ 0& 1 \end{bmatrix}$

Answer

The formula for adding a pair of 2x2 matrices is

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

Using the matrices in the problem statement, we get

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

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Question

Find the determinant of the 3x3 matrix $\begin{bmatrix} 3&1 &4 \ 4&9 &-4 \ 5&8 &0 \end{bmatrix}$

Answer

The formula for the determinant of a 3x3 matrix $A=\begin{bmatrix} a&b &c \ d& e&f \ g&h &i \end{bmatrix}$ is $\left | A\right |=a(ei-hf)-b(di-gf)+c(dh-ge)$ . Using the matrix in the problem statement, we get

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