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Answer: Hence proved,$A^{3}=p I+q A+r A^{2}$

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given: $A=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r\end{array}\right]$  and I is the identity matrix of order 3

So,

$I=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

Consider, LHS side

$A^{2}=A A=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r\end{array}\right]\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q &r \end{array}\right] \\\\\\ A^{2}=\left[\begin{array}{ccc}0+0+0 & 0+0+0 & 0+1+0 \\ 0+0+p & 0+0+q & 0+0+r \\ 0+0+p r & p+0+q r & 0+q+r^{2}\end{array}\right] \\\\\\ A^{2}=\left[\begin{array}{ccc} 0 & 0 & 1 \\ p & q & r \\ p r & p+q r & q+r^{2} \end{array}\right]\\$

Now, $A^3 = A^2 A$

$A^{3}=\left[\begin{array}{ccc} 0 & 0 & 1 \\ p & q & r \\ p r & p+q r & q+r^{2} \end{array}\right]\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{array}\right]\\\\\\ A^{3}=\left[\begin{array}{ccc} 0+0+p & 0+0+q & 0+0+r \\ 0+0+p r & p+0+q r & 0+q+r^{2} \\ 0+0+p q+p r^{2} & p r+0+q^{2}+q r^{2} & 0+p+q r+q r+r^{3} \end{array}\right]\\\\\\\ A^{3}=\left[\begin{array}{ccc} p & q & r \\ p r & p+q r & q+r^{2} \\ p q+p r^{2} & p r+q^{2}+q r^{2} & p+2 q r+r^{3} \end{array}\right] ....(i)$

Now consider RHS  $p I+q A+r A^{2}$

Then put the values in the equation, we get.

$p I+q A+r A^{2}$

$=p\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]+q\left[\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{array}\right]+r\left[\begin{array}{ccc} 0 & 0 & 1 \\ p & q & r \\ p r & p+q r & q+r^{2} \end{array}\right]\\\\ \\ =\left[\begin{array}{ccc} p & 0 & 0 \\ 0 & p & 0 \\ 0 & 0 & p \end{array}\right]+\left[\begin{array}{ccc} 0 & q & 0 \\ 0 & 0 & q \\ q p & q^{2} & q r \end{array}\right]+\left[\begin{array}{ccc} 0 & 0 & r \\ r p & r q & r^{2} \\ p r^{2} & r p+q r^{2} & r q+r^{3} \end{array}\right]\\ \\\\\\ =\left[\begin{array}{ccc} p+0+0 & 0+q+0 & 0+0+r \\ 0+0+r p & p+0+q r & 0+q+r^{2} \\ 0+p q+p r^{2} & r p+q^{2}+q r^{2} & p+2 q r+r^{3} \end{array}\right]\\ \\\\ =\left[\begin{array}{ccc} p & q & r \\ r p & p+q r & q+r^{2} \\ p q+p r^{2} & r p+q^{2}+q r^{2} & p+2 q r+r^{3} \end{array}\right]\\$$-----ii$

From equation i & equation ii

$A^{3}=p I+q A+r A^{2}$

Hence, proved.

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