Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given that,

[tex]\rm \: \rm A=\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right] \\ [/tex]

and

[tex]\rm I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\ [/tex]

Now, Consider

[tex]\rm \: a I +b A \\ [/tex]

[tex]\rm \: = \: a \left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]+b\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right] \\ [/tex]

[tex]\rm \: = \: \left[\begin{array}{ll}a & 0 \\ 0 & a\end{array}\right]+\left[\begin{array}{ll}0 & b \\ 0 & 0\end{array}\right] \\ [/tex]

[tex]\rm \: = \: \left[\begin{array}{ll}a & b \\ 0 & a\end{array}\right]\\ \\ [/tex]

Now, Consider

[tex]\rm \: (a I +b A )^{3} \\ [/tex]

[tex]\rm \: = \: \left[\begin{array}{ll}a & b \\ 0 & a\end{array}\right]\left[\begin{array}{ll}a & b \\ 0 & a\end{array}\right]\left[\begin{array}{ll}a & b \\ 0 & a\end{array}\right] \\ [/tex]

[tex]\rm \: = \: \left[\begin{array}{ll} {a}^{2} & 2ab \\ 0 & {a}^{2} \end{array}\right]\left[\begin{array}{ll}a & b \\ 0 & a\end{array}\right] \\ [/tex]

[tex]\rm \: = \: \left[\begin{array}{ll} {a}^{3} & 3 {a}^{2} b \\ 0 & {a}^{3} \end{array}\right] \\ \\ [/tex]

Now, Consider

[tex]\rm \: a^{3} I +3 a^{2} b A \\ [/tex]

[tex]\rm \: = \: a^{3}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] +3 a^{2} b \left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right] \\ [/tex]

[tex]\rm \: = \: \left[\begin{array}{ll} {a}^{3} & 0 \\ 0 & {a}^{3} \end{array}\right] + \left[\begin{array}{ll}0 & 3 {a}^{2}b \\ 0 & 0\end{array}\right] \\ [/tex]

[tex]\rm \: = \: \left[\begin{array}{ll} {a}^{3} & 3 {a}^{2} b \\ 0 & {a}^{3} \end{array}\right] \\ \\ [/tex]

Hence, from above, we concluded that

[tex] \red{\bf\implies \:(a I +b A )^{3} \: = \: a^{3} I +3 a^{2} b A \: } \\ \\ [/tex]