PDF chapter test TRY NOW
We can multiply the elements of the given matrix \(A\) by a non-zero number \(k\) to obtain a new matrix \(kA\) whose elements are multiplied by \(k\).
The matrix \(kA\) is called the scalar multiplication of \(A\).
If \(A = (a_{ij})_{m×n}\), then, \(kA = (ka_{ij})_{m×n}\) for all \(i = 1, 2,...m\), and such that \(j = 1, 2, ….n\)
Example:
Determine 5\(A + B\), if \(A =\begin{bmatrix}
2 & 4 & 6\\
7 & 5 & -4\\
-2 & 1 & 7
\end{bmatrix}, B = \begin{bmatrix}
2 & 4 & 6\\
7 & 5 & 3\\
7 & 1 & 7
\end{bmatrix}\)
2 & 4 & 6\\
7 & 5 & -4\\
-2 & 1 & 7
\end{bmatrix}, B = \begin{bmatrix}
2 & 4 & 6\\
7 & 5 & 3\\
7 & 1 & 7
\end{bmatrix}\)
Both the matrices \(A\) and \(B\) have same orders as \( 3 × 3\), so 5\(A + B\) is defined.
Therefore, we have 5\(A + B = 5\begin{bmatrix}
2 & 4 & 6\\
7 & 5 & -4\\
-2 & 1 & 7
\end{bmatrix} + \begin{bmatrix}
2 & 4 & 6\\
7 & 5 & 3\\
7 & 1 & 7
\end{bmatrix}\)
2 & 4 & 6\\
7 & 5 & -4\\
-2 & 1 & 7
\end{bmatrix} + \begin{bmatrix}
2 & 4 & 6\\
7 & 5 & 3\\
7 & 1 & 7
\end{bmatrix}\)
\(= \begin{bmatrix}
2 × 5 & 4 × 5 & 6 × 5\\
7 × 5 & 5 × 5& -4 × 5\\
-2 × 5 & 1 × 5 & 7 × 5
\end{bmatrix} + \begin{bmatrix}
2 & 4 & 6\\
7 & 5 & 3\\
7 & 1 & 7
\end{bmatrix}\)
2 × 5 & 4 × 5 & 6 × 5\\
7 × 5 & 5 × 5& -4 × 5\\
-2 × 5 & 1 × 5 & 7 × 5
\end{bmatrix} + \begin{bmatrix}
2 & 4 & 6\\
7 & 5 & 3\\
7 & 1 & 7
\end{bmatrix}\)
\(= \begin{bmatrix}
10 & 20 & 30 \\
25 & 20 & -20\\
-10 & 5 & 35
\end{bmatrix} + \begin{bmatrix}
2 & 4 & 6\\
7 & 5 & 3\\
7 & 1 & 7
\end{bmatrix}\)
10 & 20 & 30 \\
25 & 20 & -20\\
-10 & 5 & 35
\end{bmatrix} + \begin{bmatrix}
2 & 4 & 6\\
7 & 5 & 3\\
7 & 1 & 7
\end{bmatrix}\)
Now we add the two matrices.
\(= \begin{bmatrix}
12 & 24 & 36 \\
32 & 25 & -17\\
-3 & 6 & 42
\end{bmatrix}\)
12 & 24 & 36 \\
32 & 25 & -17\\
-3 & 6 & 42
\end{bmatrix}\)
Register for free to see more content