74. If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABⁿ = BⁿA. Further, prove that (AB)ⁿ = AⁿBⁿ for all n Î N.

Let   $A=\left[\begin{array}{ccc}\hfill x_1\hfill & \hfill y_1\hfill & \hfill z_1\hfill \\ \hfill x_2\hfill & \hfill y_2\hfill & \hfill z_2\hfill \\ \hfill x_3\hfill & \hfill y_3\hfill & \hfill z_3\hfill \end{array}\right]$

Given  $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 0\hfill & \hfill 2\hfill \end{array}\right]$   ...(1)

$\left[\begin{array}{ccc}\hfill x_1+z_1\hfill & \hfill x_2+z_2\hfill & \hfill x_3+z_3\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 0\hfill & \hfill 2\hfill \end{array}\right]$

∴    x₁ + z₁ = 2                … (2)

x₂ + z₂ = 0               … (3)

x₃ + z₃ = 0                … (4)

Given   $A=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill -4\hfill & \hfill 0\hfill & \hfill 4\hfill \end{array}\right]$

$\left[\begin{array}{ccc}\hfill -x_1+z_1\hfill & \hfill -x_2+z_2\hfill & \hfill -x_3+z_3\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 0\hfill & \hfill 4\hfill \end{array}\right]$

⇒   – x₁ + z₁ = −4             … (5)

–x₂ + z₂ = 0                … (6)

–x₃ + z₃ = 4                … (7)

Given  $A=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \end{array}\right]$

$\left[\begin{array}{ccc}\hfill y_1\hfill & \hfill y_2\hfill & \hfill y_3\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \end{array}\right]$

∴     y₁ = 0, y₂ = 2, y₃ = 0

∴     from (2), (3), (4), (5), (6) and (7)

x₁ = 3, x₂ = 0, x₃ = –1

y₁ = 0, y₂ = 2, y₃ = 0

z₁ = –1, z₂ = 0, z₃ = 3

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

Now (A – 3I)  $\left[\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z\hfill \end{array}\right]$  = $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \end{array}\right]$  

$\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \end{array}\right]$  

$\left[\begin{array}{ccc}\hfill -z\hfill & \hfill -y\hfill & \hfill -x\hfill \end{array}\right]$  =   $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \end{array}\right]$

[z = 1], [y = –2], [x = –3]