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UPSC 2017 Maths Optional Paper 2 Q5b — Step-by-Step Solution

10 marks · Section B

Gauss-Jordan method · Numerical Analysis · asked 2× in 13 yrs · Read the full method →

Question

Explain the main steps of the Gauss–Jordan method and apply this method to find the inverse of the matrix [266286268]\begin{bmatrix}2 & 6 & 6\\ 2 & 8 & 6\\ 2 & 6 & 8\end{bmatrix}.

Technique

Gauss–Jordan elimination on [AI][A\mid I]; reduce left block to II, right block becomes A1A^{-1}.

Solution

Main steps of Gauss–Jordan inversion.

  1. Form the augmented matrix [AI][A\mid I] by appending the identity II to AA.
  2. Apply elementary row operations to reduce the left block AA to the identity II (forward elimination to upper-triangular, then back-elimination, with each pivot normalised to 11).
  3. Whatever row operations turn AA into II turn II into A1A^{-1}. When the left block is II, the right block is A1A^{-1}, i.e. [AI][IA1][A\mid I]\to[I\mid A^{-1}].

Step 1 — Augment

[266100286010268001]\left[\begin{array}{ccc|ccc}2&6&6&1&0&0\\2&8&6&0&1&0\\2&6&8&0&0&1\end{array}\right]

Step 2 — Create zeros below the first pivot

R2R2R1R_2\to R_2-R_1 and R3R3R1R_3\to R_3-R_1:

[266100020110002101]\left[\begin{array}{ccc|ccc}2&6&6&1&0&0\\0&2&0&-1&1&0\\0&0&2&-1&0&1\end{array}\right]

(The matrix is already nearly diagonal — the off-diagonals in rows 2,3 cleared at once.)

Step 3 — Eliminate the entries in row 1 above the pivots

Column 2: R1R13R2R_1\to R_1-3R_2 (since the (1,2)(1,2) entry is 6=326=3\cdot2):

[206430020110002101]\left[\begin{array}{ccc|ccc}2&0&6&4&-3&0\\0&2&0&-1&1&0\\0&0&2&-1&0&1\end{array}\right]

Column 3: R1R13R3R_1\to R_1-3R_3 (the (1,3)(1,3) entry is 6=326=3\cdot2):

[200733020110002101]\left[\begin{array}{ccc|ccc}2&0&0&7&-3&-3\\0&2&0&-1&1&0\\0&0&2&-1&0&1\end{array}\right]

Step 4 — Normalise pivots (×12\times\tfrac12 each row)

[1007232320101212000112012]\left[\begin{array}{ccc|ccc}1&0&0&\tfrac72&-\tfrac32&-\tfrac32\\[2pt]0&1&0&-\tfrac12&\tfrac12&0\\[2pt]0&0&1&-\tfrac12&0&\tfrac12\end{array}\right]

Step 5 — Read off the inverse

Answer

  A1=12[733110101]=[7232321212012012].\boxed{\;A^{-1}=\frac12\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}=\begin{bmatrix}\tfrac72&-\tfrac32&-\tfrac32\\[2pt]-\tfrac12&\tfrac12&0\\[2pt]-\tfrac12&0&\tfrac12\end{bmatrix}.}
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