Algebraic Operations on Matrices - Multiplication of a Matrix by a Scalar

For example , ${\displaystyle 3\times \left[\begin{array}{cc}2& 4\\ -5& 1\end{array}\right]}$
To get the result of matrix each element of  the matrix  multiply by 3.
The order of the above matrix is same as the order of the original matrix. The original matrix is  of the order ${\displaystyle 2\times 2}$  so the resulting matrix 
will also be of the same order.
${\displaystyle 3\times \left[\begin{array}{cc}2& 4\\ -5& 1\end{array}\right]=\left[\begin{array}{cc}3\times 2& 3\times 4\\ 3\times -5& 3\times 1\end{array}\right]}$
=${\displaystyle \left[\begin{array}{cc}6& 12\\ -15& 3\end{array}\right]}$
Thus ${\displaystyle k\times \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}$ = ${\displaystyle \left[\begin{array}{cc}ka& kb\\ kc& kd\end{array}\right]}$

Multiplying a matrix to a number is called scalar Multiplication.

Negative of matrix :
Z = ${\displaystyle \left[\begin{array}{cc}1& 3\\ -5& 2\end{array}\right]}$  -Z = ?
-Z = `-1 xx Z

=${\displaystyle -1\times \left[\begin{array}{cc}1& 3\\ -5& 2\end{array}\right]}$

=${\displaystyle \left[\begin{array}{cc}1\cdot -1& 3\cdot -1\\ -5\cdot -1& 2\cdot -1\end{array}\right]}$

=${\displaystyle \left[\begin{array}{cc}-1& -3\\ 5& -2\end{array}\right]}$

The negative of a matrix is denoted by –Z. We define –Z = (–1) Z.

Difference of matrices:
 If A = ${\displaystyle \left[a_{ij}\right]}$, B = ${\displaystyle \left[b_{ij}\right]}$ are two matrices of the same order, say m × n, then difference A – B is defined as a matrix D = ${\displaystyle \left[d_{ij}\right]}$, where ${\displaystyle d_{ij}=a_{ij}–b_{ij}}$, for all value of i and j. In other words, D = A – B = A + (–1) B, that is sum of the matrix A and the matrix – B.
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