
Compute the products AB and BA whichever exists in the following cases:
\[A=\left[ \begin{matrix}
1 & -2 \\
2 & 3 \\
\end{matrix} \right],B=\left[ \begin{align}
& \begin{matrix}
1 & 2 & 3 \\
\end{matrix} \\
& \begin{matrix}
2 & 3 & 1 \\
\end{matrix} \\
\end{align} \right]\]
Answer
533.1k+ views
Hint: The product of the two matrices with $m\times n$ and $a\times b$ number of rows and columns respectively exist when the number of columns of the first matrix is equal to the number of rows of the second matrix. This means $n=a$. If this condition is following then proceed in multiplying the two matrices.
Complete step by step solution:
The two matrices given in the above problem is as follows:
\[A=\left[ \begin{matrix}
1 & -2 \\
2 & 3 \\
\end{matrix} \right],B=\left[ \begin{align}
& \begin{matrix}
1 & 2 & 3 \\
\end{matrix} \\
& \begin{matrix}
2 & 3 & 1 \\
\end{matrix} \\
\end{align} \right]\]
First matrix contains 2 rows and 2 columns so this matrix is $2\times 2$ and the second matrix B contains 2 rows and 3 columns so the second matrix B is of the form $2\times 3$.
Now, if we multiply A and B then we have to multiply the matrix $2\times 2$ by the matrix $2\times 3$. As you can see, the number of columns of the first row is equal to the number of rows of the second matrix so we can compute the product of A and B.
\[\begin{align}
& AB=\left[ \begin{matrix}
1 & -2 \\
2 & 3 \\
\end{matrix} \right]\left[ \begin{matrix}
\begin{matrix}
1 & 2 & 3 \\
\end{matrix} \\
\begin{matrix}
2 & 3 & 1 \\
\end{matrix} \\
\end{matrix} \right] \\
& \Rightarrow AB=\left[ \begin{matrix}
\begin{matrix}
1\left( 1 \right)-2\left( 2 \right) & 1\left( 2 \right)-2\left( 3 \right) & 1\left( 3 \right)-2\left( 1 \right) \\
\end{matrix} \\
\begin{matrix}
2\left( 1 \right)+3\left( 2 \right) & 2\left( 2 \right)+3\left( 3 \right) & 2\left( 3 \right)+3\left( 1 \right) \\
\end{matrix} \\
\end{matrix} \right] \\
& \Rightarrow AB=\left[ \begin{matrix}
\begin{matrix}
1-4 & 2-6 & 3-2 \\
\end{matrix} \\
\begin{matrix}
2+6 & 4+9 & 6+3 \\
\end{matrix} \\
\end{matrix} \right] \\
& \Rightarrow AB=\left[ \begin{matrix}
\begin{matrix}
-3 & -4 & 1 \\
\end{matrix} \\
\begin{matrix}
8 & 13 & 9 \\
\end{matrix} \\
\end{matrix} \right] \\
\end{align}\]
Hence, we have computed the product AB.
Now, can we compute the product BA. The matrix B is of the form $2\times 3$ and matrix A is of the form $2\times 2$ and as you can see that the number of columns of matrix B is 3 and number of rows of the matrix A is 2 and these two are not equal so computation of the product BA is not possible.
Note: So, whenever we asked to compute the product of the two matrices, first of all check whether the multiplication is possible or not by seeing the number of columns of the first matrix and the number of rows of the second matrix if they are equal then multiplication is possible otherwise not.
Also, from this problem, we have learnt that in an algebra product of $ab$ or $ba$ is the same but in the matrix multiplication this is not the case AB might be equal or not equal to BA.
Complete step by step solution:
The two matrices given in the above problem is as follows:
\[A=\left[ \begin{matrix}
1 & -2 \\
2 & 3 \\
\end{matrix} \right],B=\left[ \begin{align}
& \begin{matrix}
1 & 2 & 3 \\
\end{matrix} \\
& \begin{matrix}
2 & 3 & 1 \\
\end{matrix} \\
\end{align} \right]\]
First matrix contains 2 rows and 2 columns so this matrix is $2\times 2$ and the second matrix B contains 2 rows and 3 columns so the second matrix B is of the form $2\times 3$.
Now, if we multiply A and B then we have to multiply the matrix $2\times 2$ by the matrix $2\times 3$. As you can see, the number of columns of the first row is equal to the number of rows of the second matrix so we can compute the product of A and B.
\[\begin{align}
& AB=\left[ \begin{matrix}
1 & -2 \\
2 & 3 \\
\end{matrix} \right]\left[ \begin{matrix}
\begin{matrix}
1 & 2 & 3 \\
\end{matrix} \\
\begin{matrix}
2 & 3 & 1 \\
\end{matrix} \\
\end{matrix} \right] \\
& \Rightarrow AB=\left[ \begin{matrix}
\begin{matrix}
1\left( 1 \right)-2\left( 2 \right) & 1\left( 2 \right)-2\left( 3 \right) & 1\left( 3 \right)-2\left( 1 \right) \\
\end{matrix} \\
\begin{matrix}
2\left( 1 \right)+3\left( 2 \right) & 2\left( 2 \right)+3\left( 3 \right) & 2\left( 3 \right)+3\left( 1 \right) \\
\end{matrix} \\
\end{matrix} \right] \\
& \Rightarrow AB=\left[ \begin{matrix}
\begin{matrix}
1-4 & 2-6 & 3-2 \\
\end{matrix} \\
\begin{matrix}
2+6 & 4+9 & 6+3 \\
\end{matrix} \\
\end{matrix} \right] \\
& \Rightarrow AB=\left[ \begin{matrix}
\begin{matrix}
-3 & -4 & 1 \\
\end{matrix} \\
\begin{matrix}
8 & 13 & 9 \\
\end{matrix} \\
\end{matrix} \right] \\
\end{align}\]
Hence, we have computed the product AB.
Now, can we compute the product BA. The matrix B is of the form $2\times 3$ and matrix A is of the form $2\times 2$ and as you can see that the number of columns of matrix B is 3 and number of rows of the matrix A is 2 and these two are not equal so computation of the product BA is not possible.
Note: So, whenever we asked to compute the product of the two matrices, first of all check whether the multiplication is possible or not by seeing the number of columns of the first matrix and the number of rows of the second matrix if they are equal then multiplication is possible otherwise not.
Also, from this problem, we have learnt that in an algebra product of $ab$ or $ba$ is the same but in the matrix multiplication this is not the case AB might be equal or not equal to BA.
Recently Updated Pages
Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

The pH of the pancreatic juice is A 64 B 86 C 120 D class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

