Question

If $A=\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$ then $(a I+b A)^{n}$ is (where $I$ is the identity matrix of order $2$ )

• A. $a ^{2} I+a^{n-1} b \cdot A$
• B. $a^{n} I+n a^{n} b A$
• C. $a^{n} I+n \cdot a^{n-1} b \cdot A$
• D. $a^{n} I+b^{n} A$

Answer 1

Answer: (C)

$A =\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$
${[ aI + bA ]^{1}=\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}+\begin{bmatrix}0 & b \\ 0 & 0\end{bmatrix}=\begin{bmatrix} a & b \\ 0 & a \end{bmatrix} }$
${[ aI + IA ]^{2}=\begin{bmatrix} a & b \\ 0 & a \end{bmatrix}\begin{bmatrix} a & b \\ 0 & a \end{bmatrix}=\begin{bmatrix} a ^{2} & 2 ab \\ 0 & a ^{2}\end{bmatrix} }$
${[ aI + bA ]^{3}=\begin{bmatrix} a ^{2} & 2 ab \\ 0 & a ^{2}\end{bmatrix}\begin{bmatrix}a & b \\ 0 & a \end{bmatrix}=\begin{bmatrix} a ^{3} & 3^{2} b \\ 0 & a ^{3}\end{bmatrix} }$
$\therefore[ aI + bA ]^{ n }=\begin{bmatrix} a ^{ n } & na a ^{ n -1} b \\ 0 & a ^{ n }\end{bmatrix}$
$= a ^{ n } I + n \cdot a ^{ n -1} bA$