Skip to content
Exercise 2.1 · Q5

Q.A=[1−13210331]A = \begin{bmatrix} 1 & -1 & 3 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{bmatrix}, apply 3R33R_3 and then C3+2C2C_3 + 2C_2.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
4% · 5/121 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: A=[1−13210331]A=\begin{bmatrix} 1 & -1 & 3 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{bmatrix}. Apply 3R33R_3 (multiply every entry of row 3 by 3): row 3 becomes (9,9,3)(9,9,3), giving A∼[1−13210993]A\sim\begin{bmatrix} 1 & -1 & 3 \\ 2 & 1 & 0 \\ 9 & 9 & 3 \end{bmatrix}.

Step 2: Now apply C3→C3+2C2C_3\to C_3+2C_2 to THIS matrix (not the original): every entry of column 3 gets (old column 3 entry) +2×+2\times(column 2 entry, same row).

Step 3: Row 1: 3+2(−1)=13+2(-1)=1. Row 2: 0+2(1)=20+2(1)=2. Row 3: 3+2(9)=213+2(9)=21. …

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.