Question

If P is a \[3 \times 3\] matrix such that \[{P^T} = 2P + I\], where \[{P^T}\] is the transpose of P and I is the \[3 \times 3\] identity matrix, then there exists a column matrix $X = \left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right] \ne \left[ {\begin{array}{*{20}{c}}
  0 \\
  0 \\
  0
\end{array}} \right]$ such that
A. $PX = \left[ {\begin{array}{*{20}{c}}
  0 \\
  0 \\
  0
\end{array}} \right]$
B. PX = X
C. PX = 2X
D. PX = -X
E. $PX = \left[ {\begin{array}{*{20}{c}}
  2 \\
  0 \\
  0
\end{array}} \right]$
F. PX = 3X
G.PX = 5X
H. PX = -3X

Answer 1

Hint: We will use some identities of transpose of a matrix and then use the elimination and substitution method to find out the matrix P and then we will verify which option is correct.

Complete step-by-step solution:
Since we are already given that \[{P^T} = 2P + I\] …………….(1)
Now, taking transpose on both the sides, we will get:-
\[ \Rightarrow {\left( {{P^T}} \right)^T} = {\left( {2P + I} \right)^T}\]
Now, this is a well - known fact that for any matrix A, \[{\left( {{A^T}} \right)^T} = A\] and for any two matrices A and B, we have: ${(A + B)^T} = {A^T} + {B^T}$ where ${A^T}$ is transpose of matrix A.
\[ \Rightarrow P = 2{P^T} + {I^T}\]
We know that \[{I^T} = I\] for any identity matrix I.
\[ \Rightarrow P = 2{P^T} + I\]
Multiplying the above equation by 2 and rewriting it by arranging terms, we will get:-
\[ \Rightarrow 4{P^T} + 2I = 2P\] ……………….(2)
Now, we will subtract equation (1) from equation (2) to obtain:-
\[ \Rightarrow 4{P^T} + 2I - {P^T} = 2P - (2P + I)\]
Clubbing the like terms and opening the parenthesis on the right hand side to obtain:-
\[ \Rightarrow 3{P^T} + 2I = 2P - 2P - I\]
Clubbing the like terms again to obtain the following equation:-
\[ \Rightarrow 3{P^T} + 2I = - I\]
Taking the 2I from addition in left hand side to subtraction in right hand side, we will then obtain the following expression:-
\[ \Rightarrow 3{P^T} = - 2I - I\]
Clubbing the like terms again in the above equation to further obtain:-
\[ \Rightarrow 3{P^T} = - 3I\]
Cancelling the 3 from both the sides in multiplication, we will get:-
\[ \Rightarrow {P^T} = - I\]
Taking transpose on both the sides, we will get:-
\[ \Rightarrow {\left( {{P^T}} \right)^T} = - {\left( I \right)^T}\]
\[ \Rightarrow P = - {I^T} = - I\]
Now, we have obtained the matrix P that it is negative of the identity matrix.
Now, let us go through all the options one by one to see whether it satisfies them or not.
Option A: $PX = \left[ {\begin{array}{*{20}{c}}
  0 \\
  0 \\
  0
\end{array}} \right]$
Since, we have obtained that P = -I and X is not equal to zero. Hence the LHS of $PX = \left[ {\begin{array}{*{20}{c}}
  0 \\
  0 \\
  0
\end{array}} \right]$ that is PX cannot be equal to zero. Hence, it is an incorrect option.
Option B: PX = X
Since, we have obtained that P = -I. Hence the LHS of PX = X that is PX will be –X which can never be equal to X because X is never equal to zero. Hence, it is an incorrect option.
Option C: PX = 2X
Since, we have obtained that P = -I. Hence the LHS of PX = X that is PX will be –X which can never be equal to 2X because X is never equal to zero. Hence, it is an incorrect option.
Option D: PX = -X
Since, we have obtained that P = -I. Hence the LHS of PX = X that is PX will be –X is equal to RHS. Hence, it is a correct option.
Option E: $PX = \left[ {\begin{array}{*{20}{c}}
  2 \\
  0 \\
  0
\end{array}} \right]$
Since, we have obtained that P = -I and X is not equal to zero. Hence the LHS of $PX = \left[ {\begin{array}{*{20}{c}}
  2 \\
  0 \\
  0
\end{array}} \right]$ that is PX is equal to –x which can only be equal to RHS if $X = \left[ {\begin{array}{*{20}{c}}
  { - 2} \\
  0 \\
  0
\end{array}} \right]$ which necessarily need not be true. Hence, it is an incorrect option.
Option F: PX = 3X
Since, we have obtained that P = -I. Hence the LHS of PX = X that is PX will be –X which can never be equal to 3X because X is never equal to zero. Hence, it is an incorrect option.
Option G: PX = 5X
Since, we have obtained that P = -I. Hence the LHS of PX = X that is PX will be –X which can never be equal to 5X because X is never equal to zero. Hence, it is an incorrect option.
Option H: PX = -3X
Since, we have obtained that P = -I. Hence the LHS of PX = X that is PX will be –X which can never be equal to -3X because X is never equal to zero. Hence, it is an incorrect option.

$\therefore $ The correct option is (D).

Note: The students must note that we were given the condition of non – triviality in the question itself. If that would not have been given we would never have been able to discard so many of the options and this could have been a multiple correct question then.
The students must also know that we get the transpose of a matrix by exchanging rows into columns and columns into rows.