1. Scalar multiplication
To multiply a matrix by a single number (i.e. a scalar), each individual element of the matrix is multiplied by the factor.
Example:
\( \scriptsize 4 \; \times \; \begin{pmatrix} 3 & 2 & 5 \\ 6 & 1 & 7 \end{pmatrix}\)= \( \scriptsize \begin{pmatrix} 12 & 8 & 20 \\ 24 & 4 & 28 \end{pmatrix}\)
This also means that in reverse, we can take a common factor out of every element.
Therefore,
\( \scriptsize \begin{pmatrix} 10 & 25 & 45 \\ 35 & 15 & 50 \end{pmatrix}\)can be written as \( \scriptsize 5 \; \times \; \begin{pmatrix} 2 & 5 & 9 \\ 7 & 3 & 10 \end{pmatrix}\)
2. Multiplication of two matrices.
Two matrices can be multiplied together only when the number of columns in the first is equal to the number of rows in the second.
If A = \(\scriptsize \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} \)
and b = \(\scriptsize \begin{pmatrix} b_{1} \\ b_{2} \\ b_{3} \end{pmatrix}\)
Then A.b = \(\scriptsize \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix}. \scriptsize \begin{pmatrix} b_{1} \\ b_{2} \\ b_{3} \end{pmatrix}\)
= \(\scriptsize \begin{pmatrix} a_{11} b_{1} & a_{12}b_{2} & a_{13}b_{3} \\ a_{21}b_{1} & a_{22}b_{2} & a_{23}b_{3} \end{pmatrix} \)
i.e each element in top row of A is multiplied by the corresponding element in the first column of b and the products added. Similarly, the second row of the product is found by multiplying each element in the second row of A by the corresponding element in the first column of b.
Example:
\(\scriptsize \begin{pmatrix}4 & 7 & 6 \\ 2 & 3 & 1 \end{pmatrix}. \scriptsize \begin{pmatrix} 8 \\ 5 \\ 9 \end{pmatrix}\)=\(\scriptsize \begin{pmatrix}4 \times 8 & 7 \times 5 & 6 \times 9 \\ 2 \times 8 & 3 \times 5 & 1 \times 9 \end{pmatrix}\)
= \(\scriptsize \begin{pmatrix} 32 + 35 + 54 \\ 16 + 15 + 9 \end{pmatrix}\)
= \(\scriptsize \begin{pmatrix} 121 \\ 40 \end{pmatrix}\)
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