Question

Let P = \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 4&1&0 \\\ {16}&4&1 \end{array}} \right) and I be the identity matrix of order 3. If $Q = [{q_{ij}}]$ is a matrix such that ${P^{50}} - Q = I$, then $\dfrac{{{q_{31}} + {q_{32}}}}{{{q_{21}}}}$equals
A. 52
B. 103
C. 201
D. 205

Answer 1

This is a very interesting problem related with matrices and their properties. First compute the matrix ${P^{50}}$. Then find the difference matrix ${P^{50}} - Q$. Finally equate it with identity matrix I of the same order, element by element values. Some mathematical operations will give the result.

Complete step-by-step answer:
Given matrix in the problem is,
$P = \left( {\begin{array}{*{20}{c}}
1&0&0 \\
4&1&0 \\
{16}&4&1

\end{array}} \right)$Now, we will compute the value of matrix${P^2} = P \times P$as follows

{P^2} = \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 4&1&0 \\\ {16}&4&1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 4&1&0 \\\ {16}&4&1 \end{array}} \right) \\\ \Rightarrow {P^2} = \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 8&1&0 \\\ {16 \times (1 + 2)}&8&1 \end{array}} \right) \\\

Now, we will compute ${P^3} = {P^2} \times P$as follows

{P^3} = \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 8&1&0 \\\ {16 \times (1 + 2)}&8&1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 4&1&0 \\\ {16}&4&1 \end{array}} \right) \\\ \Rightarrow {P^3} = \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ {12}&1&0 \\\ {16 \times (1 + 2 + 3)}&{12}&1 \end{array}} \right) \\\

Similarly we can compute other matrix with higher powers. So, we can see the pattern of the values of
matrix P with some power.
Thus we can conclude with the value of ${P^{50}}$ as follows:

1&0&0 \\\ {4 \times 50}&1&0 \\\ {16 \times (1 + 2 + 3 + ... + 50)}&{4 \times 50}&1 \end{array}} \right)$$ After simplifying the above matrix as follws : $$ \Rightarrow {P^{50}} = \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ {200}&1&0 \\\ {20400}&{200}&1 \end{array}} \right)$$ Now, we need to assume the matrix Q of order 3 as: $Q = \left( {\begin{array}{*{20}{c}} {{q_{11}}}&{{q_{12}}}&{{q_{13}}} \\\ {{q_{21}}}&{{q_{22}}}&{{q_{23}}} \\\ {{q_{31}}}&{{q_{32}}}&{{q_{33}}} \end{array}} \right)$here $Q = [{q_{ij}}]$with general terms as ${q_{ij}}$with ith row and jth column. Here I is the identity matrix of order 3. So, $$I = \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right)$$ Therefore we have, ${P^{50}} - Q = I$ After substituting the terms and their values we will get $$ \Rightarrow \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ {200}&1&0 \\\ {20400}&{200}&1 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} {{q_{11}}}&{{q_{12}}}&{{q_{13}}} \\\ {{q_{21}}}&{{q_{22}}}&{{q_{23}}} \\\ {{q_{31}}}&{{q_{32}}}&{{q_{33}}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right)$$ $$$$Further simplification will give, $$ \Rightarrow \left( {\begin{array}{*{20}{c}} {1 - {q_{11}}}&{{q_{12}}}&{{q_{13}}} \\\ {200 - {q_{21}}}&{1 - {q_{22}}}&{{q_{23}}} \\\ {20400 - {q_{31}}}&{200 - {q_{32}}}&{1 - {q_{33}}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right)$$ We will compare individual corresponding elements on both the sides, we will get

20400 - {q_{31}} = 0 \\
\Rightarrow {q_{31}} = 20400 \\

Similarly we can have $ 200 - {q_{21}} = 0 \\\ \Rightarrow {q_{21}} = 200 \\\ $ And finally we will have $ 200 - {q_{32}} = 0 \\\ \Rightarrow {q_{32}} = 200 \\\ $ Now after getting required three terms, we will evaluate the following term with needed substitution and further simplification, $ \dfrac{{{q_{31}} + {q_{32}}}}{{{q_{21}}}} = \dfrac{{20400 + 200}}{{200}} \\\ \Rightarrow \dfrac{{{q_{31}} + {q_{32}}}}{{{q_{21}}}} = \dfrac{{20600}}{{200}} \\\ \Rightarrow \dfrac{{{q_{31}} + {q_{32}}}}{{{q_{21}}}} = 103 \\\ $ $\therefore $ The required value is 103. **Thus option B is the correct answer.** **Note:** Above tricky question will be lengthy, if suitable concept is not used for its solution. Also knowledge of matrices, about their terms and further Identity matrix will help a lot for finding solutions. Term by term comparisons are used here by following the principle of equivalent matrices.