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

Hint: 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:
\[ \Rightarrow {P^{50}} = \left( {\begin{array}{*{20}{c}}
1&0&0 \\
{4 \times 50}&1&0 \\
{16 \times (1 + 2 + 3 + ... + 50)}&{4 \times 50}&1
\end{array}} \right)\]
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.