AP Calculus BC - Matrices

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

Explanation

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:

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

$\begin{vmatrix}8\10 \end{vmatrix}$

$\begin{vmatrix}0\10 \end{vmatrix}$

$\begin{vmatrix}8\0 \end{vmatrix}$

$\begin{vmatrix}10\10 \end{vmatrix}$

$\begin{vmatrix}10\8 \end{vmatrix}$

Explanation

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:

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

Find the determinant of the matrix $A=\begin{vmatrix} 1&4 &2 \-3 &0 &5 \2 &7 & 0\end{vmatrix}$

Explanation

The determinant of a 3x3 matrix can be found via a means of reduction into three 2x2 matrices as follows:

$det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a\begin{Vmatrix} e&f \ h&i \end{Vmatrix}-b\begin{Vmatrix} d&f \ g&i \end{Vmatrix}+c\begin{Vmatrix} d&e \ g&h \end{Vmatrix}$

These can then be further reduced via the method of finding the determinant of a 2x2 matrix:

$det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)$

For the matrix $A=\begin{vmatrix} 1&4 &2 \-3 &0 &5 \2 &7 & 0\end{vmatrix}$

The determinant is thus:

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

$\begin{vmatrix} -12\-11 \end{vmatrix}$

$\begin{vmatrix} -11\-12 \end{vmatrix}$

$\begin{vmatrix} -8\-6 \end{vmatrix}$

$\begin{vmatrix} -6\-8 \end{vmatrix}$

$\begin{vmatrix} 12\24 \end{vmatrix}$

Explanation

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

For a multiplication of the form

The resulting matrix is

The notation may be daunting but numerical examples may elucidate.

We're told that

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

The resulting matrix product is then:

$A\times b = \begin{vmatrix} -16+4\-12+1 \end{vmatrix}=\begin{vmatrix} -12\-11 \end{vmatrix}$

Find the determinant of the matrix $A=\begin{vmatrix} 4& 2&0 \3 &5 &9 \1 &1 & 6\end{vmatrix}$

Explanation

The determinant of a 3x3 matrix can be found via a means of reduction into three 2x2 matrices as follows:

$det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a\begin{Vmatrix} e&f \ h&i \end{Vmatrix}-b\begin{Vmatrix} d&f \ g&i \end{Vmatrix}+c\begin{Vmatrix} d&e \ g&h \end{Vmatrix}$

These can then be further reduced via the method of finding the determinant of a 2x2 matrix:

$det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)$

For the matrix $A=\begin{vmatrix} 4& 2&0 \3 &5 &9 \1 &1 & 6\end{vmatrix}$

The determinant is thus:

Find the determinant of the matrix $A=\begin{vmatrix}6 &4 &2 \11 & -8& 2\3 &-9 &4 \end{vmatrix}$

Explanation

The determinant of a 3x3 matrix can be found via a means of reduction into three 2x2 matrices as follows:

$det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a\begin{Vmatrix} e&f \ h&i \end{Vmatrix}-b\begin{Vmatrix} d&f \ g&i \end{Vmatrix}+c\begin{Vmatrix} d&e \ g&h \end{Vmatrix}$

These can then be further reduced via the method of finding the determinant of a 2x2 matrix:

$det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)$

For the matrix $A=\begin{vmatrix}6 &4 &2 \11 & -8& 2\3 &-9 &4 \end{vmatrix}$

The determinant is thus:

Find the determinant of the matrix $A=\begin{vmatrix} 10& -13\-2 &-4 \end{vmatrix}$

Explanation

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 $A=\begin{vmatrix} 10& -13\-2 &-4 \end{vmatrix}$

The determinant is thus:

Find the determinant of the matrix $A=\begin{vmatrix} -1&-12 &12 \15 &11 &3 \14 &2 &-8 \end{vmatrix}$

Explanation

The determinant of a 3x3 matrix can be found via a means of reduction into three 2x2 matrices as follows:

$det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a\begin{Vmatrix} e&f \ h&i \end{Vmatrix}-b\begin{Vmatrix} d&f \ g&i \end{Vmatrix}+c\begin{Vmatrix} d&e \ g&h \end{Vmatrix}$

These can then be further reduced via the method of finding the determinant of a 2x2 matrix:

$det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)$

For the matrix $A=\begin{vmatrix} -1&-12 &12 \15 &11 &3 \14 &2 &-8 \end{vmatrix}$