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 \(\displaystyle A=\begin{vmatrix} 5& 0\\3 &9 \end{vmatrix}\) and \(\displaystyle b=\begin{vmatrix} 1\\2 \end{vmatrix}\)

The resulting matrix product is then:

\(\displaystyle A\times b = \begin{vmatrix}5+0 \\3+19 \end{vmatrix}=\begin{vmatrix} 5\\22 \end{vmatrix}\)

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 \(\displaystyle A=\begin{vmatrix} 5& 7\\-3 &4 \end{vmatrix}\) and \(\displaystyle b=\begin{vmatrix}0 \\ 6\end{vmatrix}\) 

The resulting matrix product is then:

\(\displaystyle A\times b = \begin{vmatrix} 0+42\\0+24 \end{vmatrix}=\begin{vmatrix} 42\\24 \end{vmatrix}\)

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 \(\displaystyle A=\begin{vmatrix} 4&3 &-3 \\1 &6 &2 \\-2 &0 &5 \end{vmatrix}\) and \(\displaystyle b=\begin{vmatrix} 0\\ 3\\2 \end{vmatrix}\)

The resulting matrix product is then:

\(\displaystyle A\times b = \begin{vmatrix} 0+9-6\\0+18+4 \\0+0+10 \end{vmatrix}=\begin{vmatrix} 3\\22 \\10 \end{vmatrix}\)

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 \(\displaystyle A=\begin{vmatrix} -4& -1& 8\\2 &3 &5 \\1 &4 &-1 \end{vmatrix}\) and \(\displaystyle b=\begin{vmatrix} 3\\2 \\1 \end{vmatrix}\)

The resulting matrix product is then:

\(\displaystyle A\times b = \begin{vmatrix} -12-2+8\\6+6+5 \\3+8-1 \end{vmatrix}=\begin{vmatrix} -6\\17 \\ 10\end{vmatrix}\)

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 \(\displaystyle A=\begin{vmatrix} 1& 6& 14\\1 &1 &3 \\2 &1 &5 \end{vmatrix}\) and \(\displaystyle b=\begin{vmatrix} 2\\0 \\6 \end{vmatrix}\)

The resulting matrix product is then:

\(\displaystyle A\times b = \begin{vmatrix}2+0+84 \\2+0+18 \\4+0+30 \end{vmatrix}=\begin{vmatrix} 86\\ 20\\34 \end{vmatrix}\)

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

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

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

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

For the matrix \(\displaystyle A=\begin{vmatrix} 1&1 &1 \\1 &2 &3 \\3 &2 &1 \end{vmatrix}\)

The determinant is thus:

\(\displaystyle |A|=1[2(1)-2(3)]-1[1(1)-3(3)]+1[1(2)-3(2)]=0\)

(It is of note that if a matrix has a zero determinant, then its columns are linearly dependent. In other words, one column could be created by some via some combination of the other two.

Note how if you multiply the second column by two and then subtract the first column, the third column results.)

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

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

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

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

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

The determinant is thus:

\(\displaystyle |A|=6[-8(4)-(-9)(2)]-4[11(4)-3(2)]+2[11(-9)-3(-8)]=-386\)

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

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

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

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

For the matrix \(\displaystyle A=\begin{vmatrix} 9&-1 &10 \\2 &3 &1 \\4 &8 &7 \end{vmatrix}\)

The determinant is thus:

\(\displaystyle |A|=9[3(7)-8(1)]-(-1)[2(7)-4(1)]+10[2(8)-4(3)]=167\)

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

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

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

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

For the matrix \(\displaystyle A=\begin{vmatrix} 2&3 &5 \\7 &11 &13 \\17 &19 &23 \end{vmatrix}\)

The determinant is thus:

\(\displaystyle |A|=2[11(23)-19(13)]-3[7(23)-17(13)]+5[7(19)-17(11)]=-78\)

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

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

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

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

For the matrix \(\displaystyle A=\begin{vmatrix} 83& 79& 73\\71 &67 &61 \\ 53&47 &43 \end{vmatrix}\)

The determinant is thus:

\(\displaystyle |A|=83[67(43)-47(61)]-79[71(43)-53(61)]+73[71(47)-53(67)]=-240\)