Question

Find the value of determinant of $({P^2} + {Q^2})$ such that P and Q be $3 \times 3 $ matrices and $P \ne Q$. If ${P^3} = {Q^3}$ and ${P^2}Q = P{Q^2}$.Choose from the correct answer given below.
(A) $0$
(B) $1$
(C) $ - 2$
(D) $ - 1$

Answer 1

Hint:For solving this type of question, we consider the given data ${P^3} = {Q^3}$, ${P^2}Q = P{Q^2}$ as equation 1 and 2 respectively. Then we subtract both the equations with each other and we get the value of ${P^2} + {Q^2}$ After this we find the determinant of ${P^2} + {Q^2}$.

Complete step-by-step answer:
According to the question it is given that P and Q are the matrices of order $3 \times 3$. And $P \ne Q$
It is also given that
$ \Rightarrow {P^3} = {Q^3} - - - - - (1)$
and
$ \Rightarrow {P^2}Q = P{Q^2} - - - - - (2)$
We find the value of $\left| {{P^2} + {Q^2}} \right|$ for this we subtract equation (2) from (1), we get
$ \Rightarrow {P^3} - {P^2}Q = {Q^3} - {Q^2}P$
Now taking ${P^2}$ common from L.H.S and -${Q^2}$ common from R.H.S, we get
$ \Rightarrow {P^2}(P - Q) = - {Q^2}(P - Q)$
Now , we take the R.H.S terms to the L.H.S side we get,
$ \Rightarrow {P^2}(P - Q) + {Q^2}(P - Q) = 0$
Now we take $(P - Q)$ common, we get
$ \Rightarrow (P - Q)({P^2} + {Q^2}) = 0$
Now it is given that $P \ne Q$
So, $(P - Q) \ne 0$
Therefore from equation we get
$\left| {{P^2} + {Q^2}} \right| = 0$
The value of determinant of $({P^2} + {Q^2})$ is 0

So, the correct answer is “Option A”.

Additional Information:Matrix: It is an array of many numbers. A square matrix is a matrix which has the number of rows and columns equal.
Determinant of a matrix: It is a special number that can be calculated from a square matrix. The determinant of a matrix P is denoted by det(P) or $\left| P \right|$.
Note:For solving these kinds of questions we have to always remember the conditions given to us and compute these two equations and the matrices should be of the same order.