If P is a 3 × 3 real matrix such that PT=a P+(a-1) I, where a1, then

Question

Tardigrade Admin · Accepted Answer

P T = aP +( a -1) I ⇒ P =a P T +( a -1) I ⇒ P T - P = a ( P - P T ) ⇒ P = P T , text as a ≠-1 text Now, P = aP +( a -1) I ⇒ P =- I ⇒| P |=1 ⇒| textAdj P |=1

Q. If $P$ is a $3 \times 3$ real matrix such that $P^{T} = a P + (a - 1) I$ , where $a > 1$ , then

$∣ A d j P ∣ > 1$

$∣ A d j P ∣ = \frac{1}{2}$

$P$ is a singular matrix

$∣ A d j P ∣ = 1$

$P^{T} = a P + (a - 1) I$
$\Rightarrow P = a P^{T} + (a - 1) I$
$\Rightarrow P^{T} - P = a (P - P^{T})$
$\Rightarrow P = P^{T}, as a \neq = - 1$
$Now, P = a P + (a - 1) I$
$\Rightarrow P = - I \Rightarrow ∣ P ∣ = 1$
$\Rightarrow ∣ Adj P ∣ = 1$

Q. If P is a 3×3 real matrix such that PT=aP+(a−1)I, where a>1, then

∣AdjP∣>1

∣AdjP∣=21​

P is a singular matrix

∣AdjP∣=1

PT=aP+(a−1)I ⇒P=aPT+(a−1)I ⇒PT−P=a(P−PT) ⇒P=PT, as a=−1 Now, P=aP+(a−1)I ⇒P=−I⇒∣P∣=1 ⇒∣AdjP∣=1

Q. If $P$ is a $3 \times 3$ real matrix such that $P^{T} = a P + (a - 1) I$ , where $a > 1$ , then

$∣ A d j P ∣ > 1$

$∣ A d j P ∣ = \frac{1}{2}$

$P$ is a singular matrix

$∣ A d j P ∣ = 1$

$P^{T} = a P + (a - 1) I$
$\Rightarrow P = a P^{T} + (a - 1) I$
$\Rightarrow P^{T} - P = a (P - P^{T})$
$\Rightarrow P = P^{T}, as a \neq = - 1$
$Now, P = a P + (a - 1) I$
$\Rightarrow P = - I \Rightarrow ∣ P ∣ = 1$
$\Rightarrow ∣ Adj P ∣ = 1$