← 2019 Paper 1

UPSC 2019 Maths Optional Paper 1 Q1d — Step-by-Step Solution

10 marks · Section A

Solution of system of linear equations · Linear Algebra · asked 7× in 13 yrs · Read the full method →

Question

If A=[1211−4130−3]A=\begin{bmatrix}1&2&1\\1&-4&1\\3&0&-3\end{bmatrix} and B=[2111−1021−1]B=\begin{bmatrix}2&1&1\\1&-1&0\\2&1&-1\end{bmatrix} then show that AB=6I3AB=6I_3. Use this result to solve the following system of equations: 2x+y+z=5, x−y=0, 2x+y−z=12x+y+z=5,\ x-y=0,\ 2x+y-z=1.

Technique

Verify AB=6I3AB=6I_3 by hand, deduce B−1=16AB^{-1}=\tfrac16 A, recognize BB as the system’s coefficient matrix, apply x=16Ac\mathbf x=\tfrac16 A\mathbf c.

Solution

Step 1 — Compute ABAB

AB=[1211−4130−3][2111−1021−1].AB=\begin{bmatrix}1&2&1\\1&-4&1\\3&0&-3\end{bmatrix}\begin{bmatrix}2&1&1\\1&-1&0\\2&1&-1\end{bmatrix}.

Row 1: (1⋅2+2⋅1+1⋅2, 1⋅1+2(−1)+1⋅1, 1⋅1+2⋅0+1(−1))=(6,0,0)(1\cdot2+2\cdot1+1\cdot2,\ 1\cdot1+2(-1)+1\cdot1,\ 1\cdot1+2\cdot0+1(-1))=(6,0,0). Row 2: (1⋅2−4⋅1+1⋅2, 1⋅1−4(−1)+1⋅1, 1⋅1−4⋅0+1(−1))=(0,6,0)(1\cdot2-4\cdot1+1\cdot2,\ 1\cdot1-4(-1)+1\cdot1,\ 1\cdot1-4\cdot0+1(-1))=(0,6,0). Row 3: (3⋅2+0−3⋅2, 3⋅1+0−3⋅1, 3⋅1+0−3(−1))=(0,0,6)(3\cdot2+0-3\cdot2,\ 3\cdot1+0-3\cdot1,\ 3\cdot1+0-3(-1))=(0,0,6).

 AB=[600060006]=6I3. \boxed{\,AB=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}=6I_3.\,}

Step 2 — Consequence: B−1B^{-1}

From AB=6I3AB=6I_3, B⋅A6=16BAB\cdot\dfrac{A}{6}=\dfrac{1}{6}BA… more directly, AB=6I⇒B−1=16AA B=6I\Rightarrow B^{-1}=\dfrac{1}{6}A (a square matrix with a left inverse, in finite dimension, has that as its two-sided inverse). Thus

B−1=16A=16[1211−4130−3].B^{-1}=\frac{1}{6}A=\frac{1}{6}\begin{bmatrix}1&2&1\\1&-4&1\\3&0&-3\end{bmatrix}.

Step 3 — Cast the system as Bx=cB\mathbf x=\mathbf c

The system

2x+y+z=5,x−y+0 z=0,2x+y−z=12x+y+z=5,\qquad x-y+0\,z=0,\qquad 2x+y-z=1

has coefficient matrix exactly B=[2111−1021−1]B=\begin{bmatrix}2&1&1\\1&-1&0\\2&1&-1\end{bmatrix} and right side c=[501]\mathbf c=\begin{bmatrix}5\\0\\1\end{bmatrix}. Hence

x=B−1c=16A c=16[1211−4130−3][501].\mathbf x=B^{-1}\mathbf c=\frac{1}{6}A\,\mathbf c=\frac{1}{6}\begin{bmatrix}1&2&1\\1&-4&1\\3&0&-3\end{bmatrix}\begin{bmatrix}5\\0\\1\end{bmatrix}.

Step 4 — Multiply out

Ac=[1⋅5+2⋅0+1⋅11⋅5−4⋅0+1⋅13⋅5+0⋅0−3⋅1]=[6612],x=16[6612]=[112].A\mathbf c=\begin{bmatrix}1\cdot5+2\cdot0+1\cdot1\\1\cdot5-4\cdot0+1\cdot1\\3\cdot5+0\cdot0-3\cdot1\end{bmatrix}=\begin{bmatrix}6\\6\\12\end{bmatrix},\qquad \mathbf x=\frac{1}{6}\begin{bmatrix}6\\6\\12\end{bmatrix}=\begin{bmatrix}1\\1\\2\end{bmatrix}.

Answer

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