Let A = [ cases 2 & 1 & 2 6 & 2 & 11 3 & 3 & 2 cases ] and P = [ cases 1 & 2 & 0 5 & 0 & 2 7 & 1 & 5 cases ] . The sum of the prime factors of | P ^ - 1 AP - 2I| is equal to

Correct answer: A. 26 Let, y = | P ^ - 1 AP - 2I| ⇒ y = | P ^ - 1 AP - 2 P ^ - 1 P| ⇒ y = |( P ^ - 1 I) ( A - 2I ) P| ⇒ y = | P ^ - 1 I| | A - 2I | | P | We know that, |A| = 1 | A ^ - 1 | . ⇒ y = | A - 2I | ⇒ y = | cases 2 & 1 & 2 6 & 2 & 11 3 & 3 & 2 cases | - | cases 2 & 0 & 0 0 & 2 & 0 0 & 0 & 2 cases | ⇒ y = | cases 0 & 1 & 2 6 & 0 & 11 3 & 3 & 0 cases | ⇒ y = 0 - (0 - 33) + 2(18 - 0) ⇒ y = 69 Now, prime factor of 69 is 3 and 23 . So, sum = 26

JEE Main 202429 JanuaryEvening Shift

Exam: JEE Main
Year: 2024
Date: 29 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2024 · 29 January · Evening Shift · Mathematics · Matrices

Let A = [ cases 2 & 1 & 2 6 & 2 & 11 3 & 3 & 2 cases ] and P = [...

Let

A = \left[

\begin{cases}2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2\end{cases}

\right] and

P = \left[

\begin{cases}1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5\end{cases}

\right]

. The sum of the prime factors of \left|{P}^{- 1}AP - 2I\right|

is equal to

Options

A $26$
B $27$
C $66$
D $23$

Correct answer

A. $26$

Step-by-step solution

Let, $y = \left|{P}^{- 1}AP - 2I\right|$

$⇒ y = \left|{P}^{- 1}AP - 2{P}^{- 1}P\right|$

$⇒ y = \left|\left({P}^{- 1}I\right) ( A - 2I ) P\right|$

$⇒ y = \left|{P}^{- 1}I\right| | A - 2I | | P |$

We know that, \left|A\right| = \dfrac{1}{\left|{A}^{- 1}\right|} $.$ ⇒ y = | A - 2I | $$ ⇒ y = \left| $\begin{cases}2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2\end{cases}$ \right| - \left| $\begin{cases}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{cases}$ \right| $$ ⇒ y = \left| $\begin{cases}0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0\end{cases}$ \right| $$ ⇒ y = 0 - \left(0 - 33\right) + 2\left(18 - 0\right) $$ ⇒ y = 69 $Now, prime factor of$ 69 $is$ 3 and $23$ .

So, sum $= 26$

Practice Matrices on Quantrex Academy →

More from Matrices