Question

Let P be a square matrix of order 4 such that, $16P^4 - 96P^3 + 216P^2 + 81I_4 = O$, and det(P) = 16 then which of following is/are true? (Where det(P) represent determinant value of matrix P, O is null matrix and $P^{-1}$ represents inverse of matrix P and adj(P) represents Adjoint of matrix P)

• A. det(adj($2I-3P^{-1}$)) = $3^6$
• B. adj(adjP) = $2^8$P
• C. adj(adj(adjP)) = $2^{28}P^{-1}$
• D. det(adj($2I-3P^{-1}$)) = $\pm 3^9$

Answer 1

Answer: (B), (C), (D)

(B), (C), (D)

Answer 2

Let the given matrix equation be

$16P^4 - 96P^3 + 216P^2 + 81I_4 = O$

This equation can be rewritten as:

$16P^4 - 96P^3 + 216P^2 + 0P + 81I = O$

Since $\det(P) = 16 \neq 0$, P is invertible, so $P^{-1}$ exists.

From $(2P - 3I)^4 = -216P$, take the determinant:

$\det((2P - 3I)^4) = \det(-216P)$

$(\det(2P - 3I))^4 = \det(-216I \cdot P) = \det(-216I) \det(P)$

$(\det(2P - 3I))^4 = (-216)^4 \det(P)$ (since order is 4)

$(\det(2P - 3I))^4 = (-(2^3 \times 3^3))^4 \times 16 = (2^3 \times 3^3)^4 \times 2^4 = 2^{12} \times 3^{12} \times 2^4 = 2^{16} \times 3^{12}$.

$\det(2P - 3I) = \pm (2^{16} \times 3^{12})^{1/4} = \pm 2^4 \times 3^3 = \pm 16 \times 27 = \pm 432$.

Let's consider the expression $2I - 3P^{-1}$.

We need to find $\det(\text{adj}(2I - 3P^{-1})) = (\det(2I - 3P^{-1}))^3$.

Let $A = 2I - 3P^{-1}$.

$AP = (2I - 3P^{-1})P = 2P - 3I$.

$A = (2P - 3I)P^{-1}$.

$\det(A) = \det((2P - 3I)P^{-1}) = \det(2P - 3I) \det(P^{-1}) = \det(2P - 3I) / \det(P)$.

We know $\det(P) = 16$. So $\det(A) = \det(2P - 3I) / 16$.

$\det(A) = \det(2I - 3P^{-1}) = \det(2P - 3I) / 16 = (\pm 432) / 16 = \pm 27$.

Now, $\det(\text{adj}(2I - 3P^{-1})) = (\det(2I - 3P^{-1}))^3 = (\pm 27)^3 = (\pm 3^3)^3 = \pm (3^3)^3 = \pm 3^9$.

So, option (D) $\det(\text{adj}(2I-3P^{-1})) = \pm 3^9$ is true.

(B) adj(adjP) = $2^8$P.

For an $n \times n$ matrix P, adj(adjP) = $(\det(P))^{n-2} P$.

Here $n=4$.

adj(adjP) = $(\det(P))^{4-2} P = (\det(P))^2 P$.

Given $\det(P) = 16$.

adj(adjP) = $(16)^2 P = (2^4)^2 P = 2^8 P$.

So, option (B) adj(adjP) = $2^8$P is true.

(C) adj(adj(adjP)) = $2^{28}P^{-1}$.

Let $Q = \text{adj}(\text{adjP})$. We found $Q = (\det(P))^2 P = 16^2 P$.

Now we need to find adj(Q) = adj($16^2 P$).

adj($kM$) = $k^{n-1}$ adj(M) for an $n \times n$ matrix M and scalar k.

Here $k = 16^2 = (2^4)^2 = 2^8$, $M=P$, $n=4$.

adj($16^2 P$) = $(16^2)^{4-1}$ adj(P) = $(16^2)^3$ adj(P) = $16^6$ adj(P).

We know that adj(P) = $\det(P) P^{-1}$.

adj(adj(adjP)) = $16^6 (\det(P) P^{-1}) = 16^6 (16 P^{-1}) = 16^7 P^{-1}$.

$16^7 = (2^4)^7 = 2^{28}$.

So, adj(adj(adjP)) = $2^{28}P^{-1}$.

Option (C) adj(adj(adjP)) = $2^{28}P^{-1}$ is true.

The question asks which of the following is/are true. Options (B), (C), and (D) are true.