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Question: If P is a \[3 \times 3\] matrix such that \[{P^T} = 2P + I\], where \[{P^T}\] is the transpose of P ...

If P is a 3×33 \times 3 matrix such that PT=2P+I{P^T} = 2P + I, where PT{P^T} is the transpose of P and I is the 3×33 \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

Explanation

Solution

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 PT=2P+I{P^T} = 2P + I …………….(1)
Now, taking transpose on both the sides, we will get:-
⇒(PT)T=(2P+I)T\Rightarrow {\left( {{P^T}} \right)^T} = {\left( {2P + I} \right)^T}
Now, this is a well - known fact that for any matrix A, (AT)T=A{\left( {{A^T}} \right)^T} = A and for any two matrices A and B, we have: (A+B)T=AT+BT{(A + B)^T} = {A^T} + {B^T} where AT{A^T} is transpose of matrix A.
⇒P=2PT+IT\Rightarrow P = 2{P^T} + {I^T}
We know that IT=I{I^T} = I for any identity matrix I.
⇒P=2PT+I\Rightarrow P = 2{P^T} + I
Multiplying the above equation by 2 and rewriting it by arranging terms, we will get:-
⇒4PT+2I=2P\Rightarrow 4{P^T} + 2I = 2P ……………….(2)
Now, we will subtract equation (1) from equation (2) to obtain:-
⇒4PT+2I−PT=2P−(2P+I)\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:-
⇒3PT+2I=2P−2P−I\Rightarrow 3{P^T} + 2I = 2P - 2P - I
Clubbing the like terms again to obtain the following equation:-
⇒3PT+2I=−I\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:-
⇒3PT=−2I−I\Rightarrow 3{P^T} = - 2I - I
Clubbing the like terms again in the above equation to further obtain:-
⇒3PT=−3I\Rightarrow 3{P^T} = - 3I
Cancelling the 3 from both the sides in multiplication, we will get:-
⇒PT=−I\Rightarrow {P^T} = - I
Taking transpose on both the sides, we will get:-
⇒(PT)T=−(I)T\Rightarrow {\left( {{P^T}} \right)^T} = - {\left( I \right)^T}
⇒P=−IT=−I\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.