Skip to content
Miscellaneous Exercise 2(A) · Q71

Q.Show with usual notations that for any matrix A=[aij]3×3A = [a_{ij}]_{3\times3}: a11A21+a12A22+a13A23=0a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
59% · 71/121 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Step 1: Recall that expanding a determinant along row 2 using row 2's OWN cofactors reproduces ∣A∣|A|: a21A21+a22A22+a23A23=∣A∣a_{21}A_{21}+a_{22}A_{22}+a_{23}A_{23}=|A|.

Step 2: The cofactors A21,A22,A23A_{21},A_{22},A_{23} depend only on rows 1 and 3 of AA (since each is formed by deleting row 2 and one column) -- they do NOT depend on what row 2's own entries actually are.

Step 3: So consider a new matrix A′A', identical to AA except that row 2 is replaced by a second copy of row 1 (i.e. A′A' has row 1 == row 2 =(a11,a12,a13)=(a_{11},a_{12},a_{13}), and row 3 unchanged). Since A21,A22,A23A_{21},A_{22},A_{23} don't depend on row 2's entries, they are exactly the same cofactors for A′A' as for AA.

Step 4: Expanding ∣A′∣|A'| along its (new) row 2 using these SAME cofactors: ∣A′∣=a11A21+a12A22+a13A23|A'| = a_{11}A_{21}+a_{12}A_{22}+a_{13}A_{23} (using row 1's entries a11,a12,a13a_{11},a_{12},a_{13} in place of row 2's, since that is what row 2 of A′A' now equals). …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.