
What is \[\left[ {\begin{array}{*{20}{c}}
1 \\
{ - 1} \\
2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
2&1&{ - 1}
\end{array}} \right] = \]
A. \[\left[ { - 1} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}
2 \\
{ - 1} \\
{ - 2}
\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}
2&1&{ - 1} \\
{ - 2}&{ - 1}&1 \\
4&2&{ - 2}
\end{array}} \right]\]
D. Not defined
Answer
164.1k+ views
Hint: In the given problem we first check out the columns and rows of the given matrices. The number of columns of the first matrix must be equal to the number of rows of the second matrix in order to multiply the two given matrices.
Formula Used:
If \[A = {[{a_{ij}}]_{m \times n}}\] and \[B = {[{b_{ij}}]_{n \times p}}\] then we can say that \[A \times B = C\] where the value of C is
\[C = {[{c_{ij}}]_{m \times p}}\]
Here \[{c_{ij}} = \mathop \sum \limits_{j = 1}^m {a_{ij}}{b_{jk}} = {a_{i1}}{b_{1k}} + {a_{i2}}{b_{2k}} + ........ + {a_{im}}{b_{mk}}\]
Complete step by step Solution:
We are given two matrices \[{\left[ {\begin{array}{*{20}{c}}
1 \\
{ - 1} \\
2
\end{array}} \right]_{3 \times 1}}\]and \[{\left[ {\begin{array}{*{20}{c}}
2&1&{ - 1}
\end{array}} \right]_{1 \times 3}}\]
There are three rows and one column in the first matrix and in the second matrix there is one column row and three columns.
So, after multiplication we will get a matrix in which we have three rows and three columns.
\[
\left[ {\begin{array}{*{20}{c}}
1 \\
{ - 1} \\
2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
2&1&{ - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{2 \times 1}&{1 \times 1}&{1 \times - 1} \\
{ - 1 \times 2}&{ - 1 \times 1}&{ - 1 \times - 1} \\
{2 \times 2}&{2 \times 1}&{2 \times - 1}
\end{array}} \right] \\
\\
\]
\[ = \left[ {\begin{array}{*{20}{c}}
2&1&{ - 1} \\
{ - 2}&{ - 1}&1 \\
4&2&{ - 2}
\end{array}} \right]\]
Therefore, the correct option is (C).
Note: To solve the given problem, one must know to multiply two matrices. Before multiplying two matrices it is important to check whether the number of columns of the first matrix and number of rows of the second matrix are equal.
Formula Used:
If \[A = {[{a_{ij}}]_{m \times n}}\] and \[B = {[{b_{ij}}]_{n \times p}}\] then we can say that \[A \times B = C\] where the value of C is
\[C = {[{c_{ij}}]_{m \times p}}\]
Here \[{c_{ij}} = \mathop \sum \limits_{j = 1}^m {a_{ij}}{b_{jk}} = {a_{i1}}{b_{1k}} + {a_{i2}}{b_{2k}} + ........ + {a_{im}}{b_{mk}}\]
Complete step by step Solution:
We are given two matrices \[{\left[ {\begin{array}{*{20}{c}}
1 \\
{ - 1} \\
2
\end{array}} \right]_{3 \times 1}}\]and \[{\left[ {\begin{array}{*{20}{c}}
2&1&{ - 1}
\end{array}} \right]_{1 \times 3}}\]
There are three rows and one column in the first matrix and in the second matrix there is one column row and three columns.
So, after multiplication we will get a matrix in which we have three rows and three columns.
\[
\left[ {\begin{array}{*{20}{c}}
1 \\
{ - 1} \\
2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
2&1&{ - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{2 \times 1}&{1 \times 1}&{1 \times - 1} \\
{ - 1 \times 2}&{ - 1 \times 1}&{ - 1 \times - 1} \\
{2 \times 2}&{2 \times 1}&{2 \times - 1}
\end{array}} \right] \\
\\
\]
\[ = \left[ {\begin{array}{*{20}{c}}
2&1&{ - 1} \\
{ - 2}&{ - 1}&1 \\
4&2&{ - 2}
\end{array}} \right]\]
Therefore, the correct option is (C).
Note: To solve the given problem, one must know to multiply two matrices. Before multiplying two matrices it is important to check whether the number of columns of the first matrix and number of rows of the second matrix are equal.
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