
If $A = \left[ {\begin{array}{*{20}{c}}
1&2&3
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
{ - 5}&4&0 \\
0&2&{ - 1} \\
1&{ - 3}&2
\end{array}} \right]$ , then what is the value of $AB$ ?
A. $\left[ {\begin{array}{*{20}{c}}
{ - 5}&4&0 \\
0&4&{ - 2} \\
3&{ - 9}&6
\end{array}} \right]$
B. $\left[ {\begin{array}{*{20}{c}}
3 \\
1 \\
1
\end{array}} \right]$
C. $\left[ {\begin{array}{*{20}{c}}
{ - 2}&{ - 1}&4
\end{array}} \right]$
D. $\left[ {\begin{array}{*{20}{c}}
{ - 5}&8&0 \\
0&4&{ - 3} \\
1&{ - 6}&6
\end{array}} \right]$
Answer
161.7k+ views
Hint: In the above question, we are provided with two matrices $A$ and $B$ respectively. Perform Matrix Multiplication and evaluate the value of $AB$ . Matrix Multiplication of two matrices is possible only when the number of columns of the first matrix is equal to the number of rows of the second matrix.
Complete step by step solution: Given matrices:
$A = \left[ {\begin{array}{*{20}{c}}
1&2&3
\end{array}} \right]$
And
$B = \left[ {\begin{array}{*{20}{c}}
{ - 5}&4&0 \\
0&2&{ - 1} \\
1&{ - 3}&2
\end{array}} \right]$
Number of columns of matrix $A$ = Number of rows of matrix $B = 3$
Thus, Matrix Multiplication is defined for the given matrices.
Evaluating the value of $AB$ and substituting the values of the matrices,
$AB = \left[ {\begin{array}{*{20}{c}}
1&2&3
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
{ - 5}&4&0 \\
0&2&{ - 1} \\
1&{ - 3}&2
\end{array}} \right]$
Performing Matrix Multiplication,
$AB = \left[ {\begin{array}{*{20}{c}}
{\left( { - 5} \right) + 0 + 3}&{4 + 4 + \left( { - 9} \right)}&{0 + \left( { - 2} \right) + 6}
\end{array}} \right]$
Simplifying further, we get:
$AB = \left[ {\begin{array}{*{20}{c}}
{ - 2}&{ - 1}&4
\end{array}} \right]$
Thus, Option (C) is correct.
Note: The number of columns in the first matrix must match the number of rows in the second matrix in order for the two matrices to be multiplied. This means that a matrix $A$ of order $\left( {j \times k} \right)$ and another matrix $B$ of order $\left( {l \times m} \right)$ can only be multiplied when $k = l$ . In the above question, the order of the product matrix, hence, will be $\left( {1 \times 3} \right)$ . In the given options, there is only one option with the same order, which is, therefore, the correct option. Hence, you did not need to evaluate the entire value of the product. You could also have narrowed it down using the order of the product matrix.
Complete step by step solution: Given matrices:
$A = \left[ {\begin{array}{*{20}{c}}
1&2&3
\end{array}} \right]$
And
$B = \left[ {\begin{array}{*{20}{c}}
{ - 5}&4&0 \\
0&2&{ - 1} \\
1&{ - 3}&2
\end{array}} \right]$
Number of columns of matrix $A$ = Number of rows of matrix $B = 3$
Thus, Matrix Multiplication is defined for the given matrices.
Evaluating the value of $AB$ and substituting the values of the matrices,
$AB = \left[ {\begin{array}{*{20}{c}}
1&2&3
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
{ - 5}&4&0 \\
0&2&{ - 1} \\
1&{ - 3}&2
\end{array}} \right]$
Performing Matrix Multiplication,
$AB = \left[ {\begin{array}{*{20}{c}}
{\left( { - 5} \right) + 0 + 3}&{4 + 4 + \left( { - 9} \right)}&{0 + \left( { - 2} \right) + 6}
\end{array}} \right]$
Simplifying further, we get:
$AB = \left[ {\begin{array}{*{20}{c}}
{ - 2}&{ - 1}&4
\end{array}} \right]$
Thus, Option (C) is correct.
Note: The number of columns in the first matrix must match the number of rows in the second matrix in order for the two matrices to be multiplied. This means that a matrix $A$ of order $\left( {j \times k} \right)$ and another matrix $B$ of order $\left( {l \times m} \right)$ can only be multiplied when $k = l$ . In the above question, the order of the product matrix, hence, will be $\left( {1 \times 3} \right)$ . In the given options, there is only one option with the same order, which is, therefore, the correct option. Hence, you did not need to evaluate the entire value of the product. You could also have narrowed it down using the order of the product matrix.
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