SATHEE: Matrices and Determinants 1 Question 4

Matrices and Determinants 1 Question 4

4. Let $P = 3 1 0$ and $Q = [q_{i j}]$ be two $3 \times 3$ matrices $\begin{array}{lll} 9 & 3 & 1 \end{array}$ such that $Q - P^{5} = I_{3}$ . Then, $\frac{q_{21} + q_{31}}{q_{32}}$ is equal to

(a) 10

(b) 135

(d) 15

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Answer:

Correct Answer: 4. (a)

Solution:

Given matrix $P = [\begin{array}{lll} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{array}] = [\begin{array}{lll} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{array}] + [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$

$\Rightarrow P = X + I$ (let)

Now, $P^{5} = (I + X)^{5}$

$\begin{array}{r} = I +^{5} C_{1} (X) +^{5} C_{2} (X^{2}) +^{5} C_{3} (X^{3}) + \dots \\ [∵ I^{n} = I, I \cdot A = A and (a + x)^{n} =^{n} C_{0} a^{n} + \\ ^{n} C_{1} a^{n - 1} x + \dots +^{n} C_{n} x^{n}] \end{array}$

Here,

$X^{2} = [\begin{array}{lll} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{array}] [\begin{array}{lll} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{array}] = [\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{array}]$

and

$X^{3} = X^{2} \cdot X = [\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{array}] [\begin{array}{lll} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{array}] = [\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$

$\Rightarrow X^{4} = X^{5} = [\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$

So,

$\begin{aligned} P^{5} & = I + 5 [\begin{array}{lll} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{array}] + 10 [\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{array}] \\ = [\begin{array}{ccc} 1 & 0 & 0 \\ 15 & 1 & 0 \\ 135 & 15 & 1 \end{array}] \end{aligned}$