
If $A = \left[ {\begin{array}{*{20}{c}}
1 \\
2 \\
3
\end{array}} \right]$ , then what is the value of $AA'$ ?
A. $14$
B. $\left[ {\begin{array}{*{20}{c}}
1 \\
4 \\
3
\end{array}} \right]$
C. $\left[ {\begin{array}{*{20}{c}}
1&2&3 \\
2&4&6 \\
3&6&9
\end{array}} \right]$
D. None of these
Answer
162.9k+ views
Hint: In the above question, we are provided with a matrix $A$ respectively. Evaluate the transpose matrix, $A'$ by interchanging either the rows with the columns or the columns with the rows. Then, perform Matrix Multiplication and evaluate the value of $A'$.
Complete step by step Solution:
Given Matrix:
$A = \left[ {\begin{array}{*{20}{c}}
1 \\
2 \\
3
\end{array}} \right]$
Calculating the transpose of matrix $A$ , by interchanging the rows and columns,
$A' = \left[ {\begin{array}{*{20}{c}}
1&2&3
\end{array}} \right]$
Number of columns of matrix $A$ = Number of rows of matrix $B = 1$
Hence, Matrix Multiplication of $A$ and $A'$ is possible.
Evaluating the value of $AA'$ and substituting the values,
$AA' = \left[ {\begin{array}{*{20}{c}}
1 \\
2 \\
3
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
1&2&3
\end{array}} \right]$
Performing Matrix Multiplication,
$AA' = \left[ {\begin{array}{*{20}{c}}
{1 \times 1}&{1 \times 2}&{1 \times 3} \\
{2 \times 1}&{2 \times 2}&{2 \times 3} \\
{3 \times 1}&{3 \times 2}&{3 \times 3}
\end{array}} \right]$
On further simplifying, we get:
$AA' = \left[ {\begin{array}{*{20}{c}}
1&2&3 \\
2&4&6 \\
3&6&9
\end{array}} \right]$
Hence, the correct option is (C).
Note:Transpose of a matrix is calculated by interchanging either the rows with the columns or the columns with the rows. It is denoted by $X'$. Thus, if the order of a matrix $X$ is $\left( {m \times n} \right)$ , then the order of matrix $X'$ is \[\left( {n \times m} \right)\].
Complete step by step Solution:
Given Matrix:
$A = \left[ {\begin{array}{*{20}{c}}
1 \\
2 \\
3
\end{array}} \right]$
Calculating the transpose of matrix $A$ , by interchanging the rows and columns,
$A' = \left[ {\begin{array}{*{20}{c}}
1&2&3
\end{array}} \right]$
Number of columns of matrix $A$ = Number of rows of matrix $B = 1$
Hence, Matrix Multiplication of $A$ and $A'$ is possible.
Evaluating the value of $AA'$ and substituting the values,
$AA' = \left[ {\begin{array}{*{20}{c}}
1 \\
2 \\
3
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
1&2&3
\end{array}} \right]$
Performing Matrix Multiplication,
$AA' = \left[ {\begin{array}{*{20}{c}}
{1 \times 1}&{1 \times 2}&{1 \times 3} \\
{2 \times 1}&{2 \times 2}&{2 \times 3} \\
{3 \times 1}&{3 \times 2}&{3 \times 3}
\end{array}} \right]$
On further simplifying, we get:
$AA' = \left[ {\begin{array}{*{20}{c}}
1&2&3 \\
2&4&6 \\
3&6&9
\end{array}} \right]$
Hence, the correct option is (C).
Note:Transpose of a matrix is calculated by interchanging either the rows with the columns or the columns with the rows. It is denoted by $X'$. Thus, if the order of a matrix $X$ is $\left( {m \times n} \right)$ , then the order of matrix $X'$ is \[\left( {n \times m} \right)\].
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