Question

Let P = [(1, 0, 0), (3, 1, 0), (9, 3, 1)]  and Q = [q_ij] be two 3 × 3 matrices such that Q – P⁵ = I₃. Then (q₂₁ + q₃₁)/q₃₂  is equal to : 

(1) 135 

(2) 15 

(3) 10 

(4) 9

Answer 1

Correct option is (3) 10

Given matrix

\(P = \begin{bmatrix}1&0&0\\3&1&0\\9&3&1\end{bmatrix}\)

\(= \begin{bmatrix}1&0&0\\3&1&0\\9&3&1\end{bmatrix} + \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)

⇒ P = X + I(let)

Now, P⁵ = (I + X)⁵

= I + ⁵C₁(X) + ⁵C₂(X²) + ⁵C₃(X³) + ...

[∵ Iⁿ = I, I⋅A and (a + x)ⁿ = ⁿC₀αⁿ + ⁿC₁αⁿ⁻¹x + ... + ⁿC_nxⁿ]

Here,

\(X^2= \begin{bmatrix}0&0&0\\3&0&0\\9&3&0\end{bmatrix} \begin{bmatrix}0&0&0\\3&0&0\\9&3&0\end{bmatrix} = \begin{bmatrix}0&0&0\\0&0&0\\9&0&0\end{bmatrix} \)

and

\(X^3 = X^2\),

\(X= \begin{bmatrix}0&0&0\\0&0&0\\9&0&0\end{bmatrix} \begin{bmatrix}0&0&0\\3&0&0\\9&3&0\end{bmatrix} = \begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix} \)

⇒ \( X^4 = X^5 = \begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix} \)

So, \(P^5 = I + 5 \begin{bmatrix}0&0&0\\3&0&0\\9&3&0\end{bmatrix} + 10\begin{bmatrix}0&0&0\\0&0&0\\9&0&0\end{bmatrix} \)

\(= \begin{bmatrix}1&0&0\\15&1&0\\135&15&1\end{bmatrix} \)

and 

\(Q = I + P^5\)

\(= \begin{bmatrix}2&0&0\\15&2&0\\135&15&2\end{bmatrix} \)

\(= [q_{ij}]\)

⇒ q₂₁ = 15, q₃₁ = 135 and q₃₂ = 15

Hence,

\(\frac{q_{21} + q_{31}}{q_{32}} = \frac{15 + 135}{15}\)

\(= \frac{150}{15}\)

\(= 10\)

Answer 2

The correct option (3) 10   

Explanation: