Compute matrix multiplication of (1×2) matrix and a (2×3) matrix.
Answer
592.5k+ views
Hint:In the given problem we have to compute matrix multiplication of two different matrices with different dimensions so we assume two random matrices of given dimensions and then solve the multiplication by dot product of rows and columns in the standard way.
Complete solution step by step:
Firstly we have to suppose two different matrices with given dimensions of rows and columns
The first matrix is of (1×2) size which translates into – one row and two columns. This means we assume a matrix with random elements inside like this
\[{\left[ {\begin{array}{*{20}{c}}
4&5
\end{array}} \right]_{1 \times 2}}\]
This is read as a ‘1 by 2’ matrix.
The second matrix is of (2×3) size which translates into – two rows and 3 columns. This means we
assume a matrix with random elements inside like this
\[{\left[ {\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
1&6&9
\end{array}} \\
{\begin{array}{*{20}{c}}
5&2&4
\end{array}}
\end{array}} \right]_{2 \times 3}}\]
This is read as a ‘2 by 3’ matrix.
We can only multiply two matrices if only – “The number of columns in the first matrix is equal to the number of rows in the second matrix”. This condition must be satisfied for matrix multiplication. Here in our question we are given a (1×2) and (2×3) matrix so the number of columns in the first i.e. 2 is equal to the number of rows in the second matrix. So applying dot product multiplication, we multiply element(s) of first row with elements of first column one by one and then add the values so that the resulting matrix is of (1×3) as written below
\[
\left[ {\begin{array}{*{20}{c}}
4&5
\end{array}} \right] \bullet \left[ {\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
1&6&9
\end{array}} \\
{\begin{array}{*{20}{c}}
5&2&4
\end{array}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{4 \times 1 + 5 \times 5}&{4 \times 6 + 5 \times 2}&{4 \times 9 + 5 \times 4}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{4 + 25}&{24 + 10}&{36 + 20}
\end{array}} \right] \\
\Rightarrow \left[ {\begin{array}{*{20}{c}}
4&5
\end{array}} \right] \bullet \left[ {\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
1&6&9
\end{array}} \\
{\begin{array}{*{20}{c}}
5&2&4
\end{array}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{29}&{34}&{56}
\end{array}} \right] \\
\]
This is how matrix multiplication of (1×2) and (2×3) matrices is done.
Note: In general, Multiplying Matrix (a × b) by matrix (b × c) the result comes out to be an (a × c) matrix.Also matrix multiplication is not commutative always e.g. when matrices $A\,{\text{and}}\,B$ are multiplied, then $A \cdot B\; \ne \;B \cdot A$ can be true i.e. changing the order of multiplication gives a different answer.
Complete solution step by step:
Firstly we have to suppose two different matrices with given dimensions of rows and columns
The first matrix is of (1×2) size which translates into – one row and two columns. This means we assume a matrix with random elements inside like this
\[{\left[ {\begin{array}{*{20}{c}}
4&5
\end{array}} \right]_{1 \times 2}}\]
This is read as a ‘1 by 2’ matrix.
The second matrix is of (2×3) size which translates into – two rows and 3 columns. This means we
assume a matrix with random elements inside like this
\[{\left[ {\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
1&6&9
\end{array}} \\
{\begin{array}{*{20}{c}}
5&2&4
\end{array}}
\end{array}} \right]_{2 \times 3}}\]
This is read as a ‘2 by 3’ matrix.
We can only multiply two matrices if only – “The number of columns in the first matrix is equal to the number of rows in the second matrix”. This condition must be satisfied for matrix multiplication. Here in our question we are given a (1×2) and (2×3) matrix so the number of columns in the first i.e. 2 is equal to the number of rows in the second matrix. So applying dot product multiplication, we multiply element(s) of first row with elements of first column one by one and then add the values so that the resulting matrix is of (1×3) as written below
\[
\left[ {\begin{array}{*{20}{c}}
4&5
\end{array}} \right] \bullet \left[ {\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
1&6&9
\end{array}} \\
{\begin{array}{*{20}{c}}
5&2&4
\end{array}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{4 \times 1 + 5 \times 5}&{4 \times 6 + 5 \times 2}&{4 \times 9 + 5 \times 4}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{4 + 25}&{24 + 10}&{36 + 20}
\end{array}} \right] \\
\Rightarrow \left[ {\begin{array}{*{20}{c}}
4&5
\end{array}} \right] \bullet \left[ {\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
1&6&9
\end{array}} \\
{\begin{array}{*{20}{c}}
5&2&4
\end{array}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{29}&{34}&{56}
\end{array}} \right] \\
\]
This is how matrix multiplication of (1×2) and (2×3) matrices is done.
Note: In general, Multiplying Matrix (a × b) by matrix (b × c) the result comes out to be an (a × c) matrix.Also matrix multiplication is not commutative always e.g. when matrices $A\,{\text{and}}\,B$ are multiplied, then $A \cdot B\; \ne \;B \cdot A$ can be true i.e. changing the order of multiplication gives a different answer.
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