Back to Course

SS3: MATHEMATICS - 2ND TERM

0% Complete
0/0 Steps
  1. Matrices I | Week 1
    6 Topics
  2. Matrices II | Week 2
    1 Topic
    |
    1 Quiz
  3. Commercial Arithmetic | Week 3
    7 Topics
    |
    1 Quiz
  4. Coordinate Geometry | Week 4
    8 Topics
    |
    1 Quiz
  5. Differentiation of Algebraic Expressions | Week 5 & 6
    7 Topics
  6. Application of Differentiation | Week 7
    4 Topics
    |
    1 Quiz
  7. Integration | Week 8
    8 Topics
    |
    1 Quiz



  • Do you like this content?

  • Follow us

Lesson Progress
0% Complete

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:

\(\scriptsize \begin{pmatrix}4 & 7 & 6 \\ 2 & 3 & 1 \end{pmatrix}. \scriptsize \begin{pmatrix} 8 \\ 5 \\ 9 \end{pmatrix}\)

=\(\scriptsize \begin{pmatrix}4 \times 8 & 7 \times 5 & 6 \times 9 \\ 2 \times 8 & 3 \times 5 & 1 \times 9 \end{pmatrix}\)

= \(\scriptsize \begin{pmatrix} 32 + 35 + 54 \\ 16 + 15 + 9 \end{pmatrix}\)

= \(\scriptsize \begin{pmatrix} 121 \\ 40 \end{pmatrix}\)

Responses

Your email address will not be published. Required fields are marked *

back-to-top
error: