Chapter 12.2, Problem 1LC

Solution

To find: The matrix $2A$ .

The required matrix is $2A=\left[\begin{array}{cc}6& -2\\ 4& 0\end{array}\right]$ .

Given information: Matrix is, $A=\left[\begin{array}{cc}3& -1\\ 2& 0\end{array}\right]$ .

Formula used: Multiplication of a scalar $c$ with the matrix $\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$ is $\left[\begin{array}{cc}c\cdot a_{11}& c\cdot a_{12}\\ c\cdot a_{21}& c\cdot a_{22}\end{array}\right]$

Calculation:

Consider the matrix as,

  $A=\left[\begin{array}{cc}3& -1\\ 2& 0\end{array}\right]$ .

Then,

  $\begin{array}{c}2A=2\times \left[\begin{array}{cc}3& -1\\ 2& 0\end{array}\right]\\ =\left[\begin{array}{cc}2\times 3& 2\times \left(-1\right)\\ 2\times 2& 2\times 0\end{array}\right]\\ =\left[\begin{array}{cc}6& -2\\ 4& 0\end{array}\right]\end{array}$

Therefore, the required matrix is $2A=\left[\begin{array}{cc}6& -2\\ 4& 0\end{array}\right]$ .