Matrix Algebra - Addition & Subtraction, Scalar Multiplication - SS3 Mathematics Lesson Note

ADDITION AND SUBTRACTION OF MATRICES

The sum (or difference) of two matrices \(A\) and \(B\) is a matrix whose elements are the result of the sum (or difference) of the corresponding entries of \(A\) and \(B\).

Given two matrices, \(A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix}\) and \(B = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ \end{bmatrix}\),

The matrix \(A + B = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix} + \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ \end{bmatrix} = \begin{bmatrix} a_{11} + b_{11} & a_{12} + b_{12} \\ a_{21} + b_{21} & a_{22} + b_{22} \\ \end{bmatrix}\)

Likewise, the matrix \(A - B = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix} - \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ \end{bmatrix} = \begin{bmatrix} a_{11} - b_{11} & a_{12} - b_{12} \\ a_{21} - b_{21} & a_{22} - b_{22} \\ \end{bmatrix}\)

Example 1 Given \(A = \begin{bmatrix} 4 & 1 \\ 0 & 2 \\ \end{bmatrix}\) and \(B = \begin{bmatrix} 8 & - 5 \\ 13 & 6 \\ \end{bmatrix}\). Solve:

1. \(A + B\)

2. \(A - B\)

3. \(B - A\)

Solution

1. \(A + B = \begin{bmatrix} 4 + 8 & 1 + ( - 5) \\ 0 + 13 & 2 + 6 \\ \end{bmatrix} = \begin{bmatrix} 12 & - 4 \\ 13 & 8 \\ \end{bmatrix}\)

2. \(A - B = \begin{bmatrix} 4 - 8 & 1 - ( - 5) \\ 0 - 13 & 2 - 6 \\ \end{bmatrix} = \begin{bmatrix} - 4 & 6 \\ - 13 & - 4 \\ \end{bmatrix}\)

3. \(B - A = \ \begin{bmatrix} 8 - 4 & - 5 - 1 \\ 13 - 0 & 6 - 2 \\ \end{bmatrix} = \begin{bmatrix} 4 & - 6 \\ 13 & 4 \\ \end{bmatrix}\)

SCALAR MULTIPLICATION OF MATRICES

The product of a matrix \(A\) and a scalar (number) \(K\) is called the scalar product of a matrix \(A\) and the scalar \(K\) and is defined by the matrix \(KA\) whose elements are \(K\) times the elements of \(A\).

If \(A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix}\), then \(KA = \begin{bmatrix} Ka_{11} & Ka_{12} \\ Ka_{21} & Ka_{22} \\ \end{bmatrix}\).

Example 2 Given that \(A = \begin{bmatrix} 5 & 3 & 4 \\ 2 & - 3 & 1 \\ \end{bmatrix}\) and \(B = \begin{bmatrix} 0 & 3 & - 1 \\ 5 & 3 & 6 \\ \end{bmatrix}\). Evaluate \(4(2A + 3B)\)

Solution

\[4(2A + 3B) = \ \ 4\left\lbrack 2\begin{bmatrix} 5 & 3 & 4 \\ 2 & - 3 & 1 \\ \end{bmatrix} + 3\begin{bmatrix} 0 & 3 & - 1 \\ 5 & 3 & 6 \\ \end{bmatrix} \right\rbrack\]

\[= 4\left\lbrack \begin{bmatrix} 2(5) & 2(3) & 2(4) \\ 2(2) & 2( - 3) & 2(1) \\ \end{bmatrix} + \begin{bmatrix} 3(0) & 3(3) & 3( - 1) \\ 3(5) & 3(3) & 3(6) \\ \end{bmatrix} \right\rbrack = 4\left\lbrack \begin{bmatrix} 10 & 6 & 8 \\ 4 & - 6 & 2 \\ \end{bmatrix} + \begin{bmatrix} 0 & 9 & - 3 \\ 15 & 9 & 18 \\ \end{bmatrix} \right\rbrack\]

\[= 4\left\lbrack \begin{bmatrix} 10 + 0 & 6 + 9 & 8 + ( - 3) \\ 4 + 15 & - 6 + 9 & 2 + 18 \\ \end{bmatrix} \right\rbrack = 4\begin{bmatrix} 10 & 15 & 5 \\ 19 & 3 & 20 \\ \end{bmatrix}\]

\[= \begin{bmatrix} 4(10) & 4(15) & 4(5) \\ 4(19) & 4(3) & 4(20) \\ \end{bmatrix} = \begin{bmatrix} 40 & 60 & 20 \\ 76 & 12 & 80 \\ \end{bmatrix}\]

MATRIX MULTIPLICATION

Two matrices \(A\) and \(B\) can be multiplied together only when the number of columns of the first matrix \(A\) equals the number of rows of the second matrix \(B\). A \(3 \times 2\) matrix multiplies a \(2 \times 3\) matrix to produce a \(3 \times 3\) matrix.

Generally, the product of two matrices A and B defined as AB is a matrix C such that:

\[AB = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \\ \end{bmatrix}\begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ \end{bmatrix} = C = \begin{bmatrix} c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \\ \end{bmatrix}\]

Where \(c_{11} = a_{11}b_{11} + a_{12}b_{21}\)

\(c_{12} = a_{11}b_{12} + a_{12}b_{22}\)

\(c_{13} = a_{11}b_{13} + a_{12}b_{23}\)

\[c_{21} = a_{21}b_{11} + a_{22}b_{21}\]

\(c_{22} = a_{21}b_{12} + a_{22}b_{22}\)

\(c_{23} = a_{21}b_{13} + a_{22}b_{23}\)

\[c_{31} = a_{31}b_{11} + a_{32}b_{21}\]

\(c_{32} = a_{31}b_{12} + a_{32}b_{22}\)

\(c_{33} = a_{31}b_{13} + a_{32}b_{23}\)

Example 3 Let \(A = \begin{bmatrix} 2 & 3 \\ 0 & - 2 \\ 4 & 5 \\ \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 3 & 2 \\ - 1 & 0 & 6 \\ \end{bmatrix}\). Find \(AB\)

Solution

\(A = \begin{bmatrix} 2 & 3 \\ 0 & - 2 \\ 4 & 5 \\ \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 3 & 2 \\ - 1 & 0 & 6 \\ \end{bmatrix}\)

\[c_{11} = 2(1) + 3( - 1)\]

\(c_{12} = 2(3) + 3(0)\)

\(c_{13} = 2(2) + 3(6)\)

\[c_{21} = 0(1) + ( - 2)( - 1)\]

\(c_{22} = 0(3) + ( - 2)(0)\)

\(c_{23} = 0(2) + ( - 2)(6)\)

\[c_{31} = 4(1) + 5( - 1)\]

\(c_{32} = 4(3) + 5(0)\)

\(c_{33} = 4(2) + 5(6)\)

\[c_{11} = 2 - 3\]

\(c_{12} = 6 + 0\)

\(c_{13} = 4 + 18\)

\[c_{21} = 0 + 2\]

\(c_{22} = 0 + 0\)

\(c_{23} = 0 - 12\)

\[c_{31} = 4 - 5\]

\(c_{32} = 12 + 0\)

\[c_{33} = 8 + 30\]

\[AB = C = \begin{bmatrix} - 1 & 6 & 22 \\ 2 & 0 & - 12 \\ - 1 & 12 & 38 \\ \end{bmatrix}\]