Matrices and Determinants 1 Question 4
4. Let $P=\left[\begin{array}{lll}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1\end{array}\right]$ and $Q=\left[q_{i j}\right]$ be two $3 \times 3$ matrices such that $Q-P^5=I_3$. Then, $\frac{q_{21}+q_{31}}{q_{32}} $ equal to
(a) 10
(b) 135
(c) 9
(d) 15
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Answer:
Correct Answer: 4. (a)
Solution:
- Given matrix $ P=\left[\begin{array}{lll} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{array}\right]+\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] $
$\Rightarrow \quad P=X+I$ (let)
Now, $P^5=(I+X)^5$
$ \begin{array}{r} =I+{ }^5 C_1(X)+{ }^5 C_2\left(X^2\right)+{ }^5 C_3\left(X^3\right)+\ldots \\ {\left[\because I^n=I, I \cdot A=A \text { and }(a+x)^n={ }^n C_0 a^n+\right.} \\ \left.{ }^n C_1 a^{n-1} x+\ldots+{ }^n C_n x^n\right] \end{array} $
Here,
$ X^2=\left[\begin{array}{lll} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{array}\right]\left[\begin{array}{lll} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{array}\right] $
and
$ X^3=X^2 \cdot X=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{array}\right]\left[\begin{array}{lll} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] $
$ \Rightarrow X^4=X^5=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] $
So,
$ \begin{aligned} P^5 & =I+5\left[\begin{array}{lll} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{array}\right]+10\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{array}\right] \\ & =\left[\begin{array}{ccc} 1 & 0 & 0 \\ 15 & 1 & 0 \\ 135 & 15 & 1 \end{array}\right] \end{aligned} $
and $Q=I+P^5=\left[\begin{array}{ccc}2 & 0 & 0 \\ 15 & 2 & 0 \\ 135 & 15 & 2\end{array}\right]=\left[q_{i j}\right]$
$\Rightarrow q_{21}=15, q_{31}=135$ and $q_{32}=15$
Hence, $\frac{q_{21}+q_{31}}{q_{32}}=\frac{15+135}{15}=\frac{150}{15}=10$
