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

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

For the matrix \(\displaystyle A=\begin{vmatrix} 7& 11\\3 &23 \end{vmatrix}\)

The determinant is thus:

\(\displaystyle |A|=161-33=128\)

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

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

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

The determinant is thus:

\(\displaystyle |A|=8-3=5\)

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

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

For the matrix \(\displaystyle A=\begin{vmatrix}12 &4 \\30 &10 \end{vmatrix}\)

The determinant is thus:

\(\displaystyle |A|=120-120=0\)

This result is due to the columns being linearly dependent, i.e. multiples of each other. The first column is three times the second column.

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

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

For the matrix \(\displaystyle A=\begin{vmatrix} 1& 3\\7 &13 \end{vmatrix}\)

The determinant is thus:

\(\displaystyle |A|=13-21=-8\)

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

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

For the matrix \(\displaystyle A=\begin{vmatrix} -15& -2\\ 4& -1\end{vmatrix}\)

The determinant is thus:

\(\displaystyle |A|=15-(-8)=23\)

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

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

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

The determinant is thus:

\(\displaystyle |A|=24-14=10\)

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

The resulting matrix product is then:

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

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

The resulting matrix product is then:

\(\displaystyle A\times b = \begin{vmatrix} 3+0\\0+12 \end{vmatrix}=\begin{vmatrix}3\\12 \end{vmatrix}\)

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

The resulting matrix product is then:

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

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

The resulting matrix product is then:

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