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UPSC Maths 2013 Paper 1 Q1a — Solution

10 marks · Section A

Question

Find the inverse of the matrix

A=[1312−1732−1]A=\begin{bmatrix}1 & 3 & 1\\ 2 & -1 & 7\\ 3 & 2 & -1\end{bmatrix}

by using elementary row operations. Hence solve the system of linear equations

x+3y+z=10,2x−y+7z=21,3x+2y−z=4.x+3y+z=10,\quad 2x-y+7z=21,\quad 3x+2y-z=4.

Technique

Gauss-Jordan on [A∣I][A|I] to obtain A−1A^{-1}; matrix-vector product to solve the system.

Solution

Strategy. Augment AA with I3I_3, reduce to RREF using elementary row operations; the right block becomes A−1A^{-1}. Then x=A−1b\mathbf{x}=A^{-1}\mathbf{b}.

Step 1 — Row-reduce [A∣I][A|I]

[A∣I]=[1311002−1701032−1001][A|I]=\left[\begin{array}{ccc|ccc}1&3&1&1&0&0\\2&-1&7&0&1&0\\3&2&-1&0&0&1\end{array}\right]

R2→R2−2R1R_2\to R_2-2R_1, R3→R3−3R1R_3\to R_3-3R_1:

[1311000−75−2100−7−4−301]\left[\begin{array}{ccc|ccc}1&3&1&1&0&0\\0&-7&5&-2&1&0\\0&-7&-4&-3&0&1\end{array}\right]

R3→R3−R2R_3\to R_3-R_2:

[1311000−75−21000−9−1−11]\left[\begin{array}{ccc|ccc}1&3&1&1&0&0\\0&-7&5&-2&1&0\\0&0&-9&-1&-1&1\end{array}\right]

R2→−17R2R_2\to -\tfrac{1}{7}R_2, R3→−19R3R_3\to -\tfrac{1}{9}R_3:

[13110001−5/72/7−1/700011/91/9−1/9]\left[\begin{array}{ccc|ccc}1&3&1&1&0&0\\0&1&-5/7&2/7&-1/7&0\\0&0&1&1/9&1/9&-1/9\end{array}\right]

R2→R2+57R3R_2\to R_2+\tfrac{5}{7}R_3: new R2=(0,1,0 ∣ 2/7+5/63, −1/7+5/63, −5/63)=(0,1,0 ∣ 23/63, −4/63, −5/63)R_2 = (0,1,0\,|\,2/7+5/63,\,-1/7+5/63,\,-5/63) = (0,1,0\,|\,23/63,\,-4/63,\,-5/63).

R1→R1−R3R_1\to R_1-R_3 then R1→R1−3R2R_1\to R_1-3R_2:

So

  A−1=163[−1352223−4−577−7].  \boxed{\;A^{-1}=\frac{1}{63}\begin{bmatrix}-13 & 5 & 22\\ 23 & -4 & -5\\ 7 & 7 & -7\end{bmatrix}.\;}

(Sanity: det⁡A=63\det A=63 by cofactor expansion along row 1 — 1(1−14)−3(−2−21)+1(4+3)=−13+69+7=631(1-14)-3(-2-21)+1(4+3)=-13+69+7=63 ✓.)

Step 2 — Solve Ax=(10,21,4)TA\mathbf{x}=(10,21,4)^{T}

x=A−1 ⁣(10214)=163 ⁣(−130+105+88230−84−2070+147−28)=163 ⁣(63126189)=(123).\mathbf{x}=A^{-1}\!\begin{pmatrix}10\\21\\4\end{pmatrix}=\frac{1}{63}\!\begin{pmatrix}-130+105+88\\230-84-20\\70+147-28\end{pmatrix}=\frac{1}{63}\!\begin{pmatrix}63\\126\\189\end{pmatrix}=\begin{pmatrix}1\\2\\3\end{pmatrix}.

Answer

  x=1,  y=2,  z=3.  \boxed{\;x=1,\;y=2,\;z=3.\;}

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