Question

lf $A=$ diagonal $(3,3,3)$ , then $A^4=$

• A. $12A$
• B. $81A$
• C. $684A$
• D. $27A$

Answer 1

The correct option is D $27A$

$A=$ diagonal $(3,3,3)$, (diagonal $({\lambda }_1.....{\lambda }_n)$ is a diagonal matrix of size $n\times n$ with elements $a_{ii}={\lambda }_i$)

So $A$ is represented as $\left[\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]$

A square matrix is called a diagonal matrix if non-diagonal entries are all zero. The main diagonal can be constants or zeros.

For calculating the powers of diagonal matrices

$D^n=\left[\begin{array}{ccccc}{\left(d_{11}\right)}^n& 0& 0& ......& 0\\ 0& {\left(d_{22}\right)}^n& 0& ......& 0\\ .& .& .& .......& 0\\ 0& 0& 0& ......& {\left(d_{nn}\right)}^n\end{array}\right]$ ($D$ is a diagonal matrix)

$\therefore A.T.PA=\left[\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]\therefore A^4=\left[\begin{array}{ccc}{\left(3\right)}^4& 0& 0\\ 0& {\left(3\right)}^4& 0\\ 0& 0& {\left(3\right)}^4\end{array}\right]$

$=3^3\left[\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]$ ($\because$ Taking $3^3$ as common)

$=27A$