Matrices - Calculus 3

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Question

Calculate the determinant of Matrix .

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

Answer

In order to find the determinant of , we first need to copy down the first two columns into columns 4 and 5.

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

The next step is to multiply the down diagonals.

$A_d=1\cdot2\cdot5+2\cdot1\cdot 3+3\cdot0\cdot4=10+6+0=16$

The next step is to multiply the up diagonals.

$A_u=3\cdot2\cdot3+4\cdot1\cdot1+5\cdot0\cdot2=18+4=22$

The last step is to substract from .

$\det(A)=A_d-A_u=16-22=-6$

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Question

Calculate the determinant of Matrix .

$A=\begin{bmatrix} 0 & 1&5 \ 10& -3& 2\ 1& 2& -5 \end{bmatrix}$

Answer

In order to find the determinant of , we first need to copy down the first two columns into columns 4 and 5.

$A=\begin{bmatrix} 0 & 1&5 & 0 &1\ 10& -3& 2&10 &-3\ 1& 2& -5 &1 &2\end{bmatrix}$

The next step is to multiply the down diagonals.

$A_d=0\cdot-3\cdot-5+1\cdot2\cdot 1+5\cdot10\cdot2=0+2+100=102$

The next step is to multiply the up diagonals.

$A_u=1\cdot-3\cdot5+2\cdot2\cdot0+(-5)\cdot10\cdot1=-15+0-50=-65$

The last step is to substract from .

$\det(A)=A_d-A_u=102-(-65)=167$

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Question

Calculate the determinant of .

$A=\begin{bmatrix} 1 & 3 \ 2 & 4 \end{bmatrix}$

Answer

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

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

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Question

Calculate the determinant of .

$A=\begin{bmatrix} 3 & 4\ -2 & 10 \end{bmatrix}$

Answer

All we need to do is multiply the main diagonal and substract it from the off diagonal.

$\det(A)=3\cdot10-4(-2)=30+8=38$

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Question

Which of the following is a way to represent the computation $<2,6,1> \cdot <5,2,4>$ using matrices?

Answer

This choice is the only pair with a well-defined operation; the others are meaningless in terms of matrix multiplication.

Computing this we get

$\begin{bmatrix} 2 & 6 & 1 \end{bmatrix} \begin{bmatrix} 5\ 2\ 4\ \end{bmatrix} = (2)(5)+(6)(2)+(1)(4) =26$ .

Which is the same as

$<2,6,1> \cdot <5,2,4> = (2)(5)+(6)(2)+(1)(4) =26$ .

This operation is true for (real) -dimensional vectors in general;

$\begin{bmatrix} a & b & c \end{bmatrix} \begin{bmatrix} d\ e\ f\ \end{bmatrix} = ad+be+cf$

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Question

Find the determinant of the matrix $A=\begin{vmatrix} 1&2 \ 5& 9\end{vmatrix}$

Answer

The determinant of a 2x2 matrix can be found by cross multiplying terms as follows:

$det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc$

For the matrix $A=\begin{vmatrix} 1&2 \ 5& 9\end{vmatrix}$

The determinant is thus:

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Question

Find the determinant of the matrix $A=\begin{vmatrix} 4&2 \-1 &3 \end{vmatrix}$

Answer

The determinant of a 2x2 matrix can be found by cross multiplying terms as follows:

$det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc$

For the matrix $A=\begin{vmatrix} 4&2 \-1 &3 \end{vmatrix}$

The determinant is thus:

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Question

Find the determinant of the matrix $A=\begin{vmatrix} 5& 2\ 3&1 \end{vmatrix}$

Answer

The determinant of a 2x2 matrix can be found by cross multiplying terms as follows:

$det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc$

For the matrix $A=\begin{vmatrix} 5& 2\ 3&1 \end{vmatrix}$

The determinant is thus:

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Question

Find the determinant of the matrix $A=\begin{vmatrix} 4&4 \4 &2 \end{vmatrix}$

Answer

The determinant of a 2x2 matrix can be found by cross multiplying terms as follows:

$det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc$

For the matrix $A=\begin{vmatrix} 4&4 \4 &2 \end{vmatrix}$

The determinant is thus:

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Question

Find the determinant of the matrix $A=\begin{vmatrix} 9&3 \21 &11 \end{vmatrix}$

Answer

The determinant of a 2x2 matrix can be found by cross multiplying terms as follows:

$det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc$

For the matrix $A=\begin{vmatrix} 9&3 \21 &11 \end{vmatrix}$

The determinant is thus:

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Question

Find the determinant of the matrix $A=\begin{vmatrix} 11&3 \2 &4 \end{vmatrix}$

Answer

The determinant of a 2x2 matrix can be found by cross multiplying terms as follows:

$det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc$

For the matrix $A=\begin{vmatrix} 11&3 \2 &4 \end{vmatrix}$

The determinant is thus:

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Question

Find the determinant of the matrix $A=\begin{vmatrix} 10&13 \3 &23 \end{vmatrix}$

Answer

The determinant of a 2x2 matrix can be found by cross multiplying terms as follows:

$det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc$

For the matrix $A=\begin{vmatrix} 10&13 \3 &23 \end{vmatrix}$

The determinant is thus:

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Question

Find the determinant of the matrix $A=\begin{vmatrix}4 &5 \6 &7 \end{vmatrix}$

Answer

The determinant of a 2x2 matrix can be found by cross multiplying terms as follows:

$det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc$

For the matrix $A=\begin{vmatrix}4 &5 \6 &7 \end{vmatrix}$

The determinant is thus:

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Question

Find the determinant of the matrix $A=\begin{vmatrix} 5&0 \ 7& 2\end{vmatrix}$

Answer

The determinant of a 2x2 matrix can be found by cross multiplying terms as follows:

$det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc$

For the matrix $A=\begin{vmatrix} 5&0 \ 7& 2\end{vmatrix}$

The determinant is thus:

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Question

Find the determinant of the matrix $A=\begin{vmatrix} 4&2 \ 40& 20\end{vmatrix}$

Answer

The determinant of a 2x2 matrix can be found by cross multiplying terms as follows:

$det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc$

For the matrix $A=\begin{vmatrix} 4&2 \ 40& 20\end{vmatrix}$

The determinant is thus:

Note that the zero-value determinant means that the two columns are not lineraly independent; one can be found by multiplying the other by some value: in this case, the second column is one half of the first.

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Question

Find the determinant of the matrix $A=\begin{vmatrix} -5& 3\2 &2 \end{vmatrix}$

Answer

The determinant of a 2x2 matrix can be found by cross multiplying terms as follows:

$det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc$

For the matrix $A=\begin{vmatrix} -5& 3\2 &2 \end{vmatrix}$

The determinant is thus:

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Question

Find the determinant of the matrix $A=\begin{vmatrix} -8&3 \-3 &5 \end{vmatrix}$

Answer

The determinant of a 2x2 matrix can be found by cross multiplying terms as follows:

$det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc$

For the matrix $A=\begin{vmatrix} -8&3 \-3 &5 \end{vmatrix}$

The determinant is thus:

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Question

Find the matrix product of $A\times b$ , where $A=\begin{vmatrix} 2&0 \3 &1 \end{vmatrix}$ and $b=\begin{vmatrix} 4\-2 \end{vmatrix}$ .

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 $A=\begin{vmatrix} 2&0 \3 &1 \end{vmatrix}$ and $b=\begin{vmatrix} 4\-2 \end{vmatrix}$

The resulting matrix product is then:

$A\times b = \begin{vmatrix}8+0\ 12-2\end{vmatrix}=\begin{vmatrix}8\10 \end{vmatrix}$

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Question

Find the matrix product of $A\times b$ , where $A=\begin{vmatrix} 2&3 \ 4&5 \end{vmatrix}$ and $b=\begin{vmatrix} 6\7 \end{vmatrix}$ .

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 $A=\begin{vmatrix} 2&3 \ 4&5 \end{vmatrix}$ and $b=\begin{vmatrix} 6\7 \end{vmatrix}$

The resulting matrix product is then:

$A\times b = \begin{vmatrix}12+21\24+35 \end{vmatrix}=\begin{vmatrix}33\59 \end{vmatrix}$

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Question

Find the matrix product of $A\times b$ , where $A=\begin{vmatrix}2 & 1\ 0&9 \end{vmatrix}$ and $b=\begin{vmatrix} 3\ -2\end{vmatrix}$ .

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 $A=\begin{vmatrix}2 & 1\ 0&9 \end{vmatrix}$ and $b=\begin{vmatrix} 3\ -2\end{vmatrix}$

The resulting matrix product is then:

$A\times b = \begin{vmatrix}6-2\0-18 \end{vmatrix}=\begin{vmatrix}4\-18 \end{vmatrix}$

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