Question

Show that if $A=cI$, where C is a real number and I is the identity matrix, then AB=BA whenever B is an$n\times n$ matrix.

Short answer

$A=\left(\begin{array}{ccc}a& \dots & 0\\ ⋮& \ddots & ⋮\\ 0& \cdots & a\end{array}\right)=aI$

Step-by-step solution

Step: 1

Suppose that$A=\left(\begin{array}{ccc}{a}_{11}& \dots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{n1}& \cdots & a_{nn}\end{array}\right)$

Let $B=\left(\begin{array}{ccc}0& \dots & 1\\ ⋮& \ddots & ⋮\\ 0& \cdots & 0\end{array}\right)$

Step: 2

Because AB=BA, if follows that

Hence$A=\left(\begin{array}{ccc}a& \dots & 0\\ ⋮& \ddots & ⋮\\ 0& \cdots & a\end{array}\right)=aI$