JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023Let A=(mpamp;namp;q),d=∣A∣=0 and ∣A−d(Adj A)∣=0. Then:
Choose the correct answer:
- A.
1+d2=m2+q2
- B.
1+d2=(m+q)2
(1+d)2=(m+q)2
Explanation
Solution:
-
Matrix Setup: d=mq−np. Adj A=(q−pamp;−namp;m).
-
Determinant Condition:
∣A−d(Adj A)∣=m−dqp+dpamp;n+dnamp;q−dm=0(m−dq)(q−dm)−np(1+d)2=0 -
Expansion:
mq−dm2−dq2+d2mq−np(1+d)2=0Substitute np=mq−d:
mq(1+d2)−d(m2+q2)−(mq−d)(1+d)2=0 -
Simplify: After expansion and canceling terms, we get:
(1+d)2=(m+q)2
Explanation
Solution:
-
Matrix Setup: d=mq−np. Adj A=(q−pamp;−namp;m).
-
Determinant Condition:
∣A−d(Adj A)∣=m−dqp+dpamp;n+dnamp;q−dm=0(m−dq)(q−dm)−np(1+d)2=0 -
Expansion:
mq−dm2−dq2+d2mq−np(1+d)2=0Substitute np=mq−d:
mq(1+d2)−d(m2+q2)−(mq−d)(1+d)2=0 -
Simplify: After expansion and canceling terms, we get:
(1+d)2=(m+q)2