← 2020 Paper 1
UPSC 2020 Maths Optional Paper 1 Q4a — Step-by-Step Solution 15 marks · Section A
Solution of system of linear equations · Linear Algebra · asked 7× in 13 yrs · Read the full method →
Question
Let A = [ 1 0 2 2 − 1 3 4 1 8 ] A=\begin{bmatrix}1&0&2\\2&-1&3\\4&1&8\end{bmatrix} A = 1 2 4 0 − 1 1 2 3 8 B = [ − 11 2 2 − 4 0 1 6 − 1 − 1 ] B=\begin{bmatrix}-11&2&2\\-4&0&1\\6&-1&-1\end{bmatrix} B = − 11 − 4 6 2 0 − 1 2 1 − 1 A B AB A B det ( A ) \det(A) det ( A ) det ( B ) \det(B) det ( B ) x + 2 z = 3 , 2 x − y + 3 z = 3 , 4 x + y + 8 z = 14 x+2z=3,\ 2x-y+3z=3,\ 4x+y+8z=14 x + 2 z = 3 , 2 x − y + 3 z = 3 , 4 x + y + 8 z = 14
Technique
Direct matrix multiplication shows B = A − 1 B=A^{-1} B = A − 1 det ( A B ) = det A det B \det(AB)=\det A\det B det ( A B ) = det A det B A x = b A\mathbf{x}=\mathbf{b} A x = b x = A − 1 b = B b \mathbf{x}=A^{-1}\mathbf{b}=B\mathbf{b} x = A − 1 b = B b
Solution
(i) Compute A B AB A B
( A B ) i j = ∑ k A i k B k j (AB)_{ij}=\sum_k A_{ik}B_{kj} ( A B ) ij = ∑ k A ik B k j
Row 1 ([ 1 , 0 , 2 ] [1,0,2] [ 1 , 0 , 2 ] 1 ( − 11 ) + 0 ( − 4 ) + 2 ( 6 ) = 1 ; 1 ( 2 ) + 0 ( 0 ) + 2 ( − 1 ) = 0 ; 1 ( 2 ) + 0 ( 1 ) + 2 ( − 1 ) = 0. \;1(-11)+0(-4)+2(6)=1;\quad 1(2)+0(0)+2(-1)=0;\quad 1(2)+0(1)+2(-1)=0. 1 ( − 11 ) + 0 ( − 4 ) + 2 ( 6 ) = 1 ; 1 ( 2 ) + 0 ( 0 ) + 2 ( − 1 ) = 0 ; 1 ( 2 ) + 0 ( 1 ) + 2 ( − 1 ) = 0. [ 2 , − 1 , 3 ] [2,-1,3] [ 2 , − 1 , 3 ] 2 ( − 11 ) + ( − 1 ) ( − 4 ) + 3 ( 6 ) = 0 ; 2 ( 2 ) + ( − 1 ) ( 0 ) + 3 ( − 1 ) = 1 ; 2 ( 2 ) + ( − 1 ) ( 1 ) + 3 ( − 1 ) = 0. \;2(-11)+(-1)(-4)+3(6)=0;\quad 2(2)+(-1)(0)+3(-1)=1;\quad 2(2)+(-1)(1)+3(-1)=0. 2 ( − 11 ) + ( − 1 ) ( − 4 ) + 3 ( 6 ) = 0 ; 2 ( 2 ) + ( − 1 ) ( 0 ) + 3 ( − 1 ) = 1 ; 2 ( 2 ) + ( − 1 ) ( 1 ) + 3 ( − 1 ) = 0. [ 4 , 1 , 8 ] [4,1,8] [ 4 , 1 , 8 ] 4 ( − 11 ) + 1 ( − 4 ) + 8 ( 6 ) = 0 ; 4 ( 2 ) + 1 ( 0 ) + 8 ( − 1 ) = 0 ; 4 ( 2 ) + 1 ( 1 ) + 8 ( − 1 ) = 1. \;4(-11)+1(-4)+8(6)=0;\quad 4(2)+1(0)+8(-1)=0;\quad 4(2)+1(1)+8(-1)=1. 4 ( − 11 ) + 1 ( − 4 ) + 8 ( 6 ) = 0 ; 4 ( 2 ) + 1 ( 0 ) + 8 ( − 1 ) = 0 ; 4 ( 2 ) + 1 ( 1 ) + 8 ( − 1 ) = 1.
A B = [ 1 0 0 0 1 0 0 0 1 ] = I 3 . \boxed{\,AB=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I_3.\,} A B = 1 0 0 0 1 0 0 0 1 = I 3 . Hence B = A − 1 B=A^{-1} B = A − 1
(ii) Determinants
Expand det A \det A det A
det A = 1 det [ − 1 3 1 8 ] − 0 + 2 det [ 2 − 1 4 1 ] = 1 ( − 8 − 3 ) + 2 ( 2 + 4 ) = − 11 + 12 = 1. \det A=1\det\begin{bmatrix}-1&3\\1&8\end{bmatrix}-0+2\det\begin{bmatrix}2&-1\\4&1\end{bmatrix}=1(-8-3)+2(2+4)=-11+12=1. det A = 1 det [ − 1 1 3 8 ] − 0 + 2 det [ 2 4 − 1 1 ] = 1 ( − 8 − 3 ) + 2 ( 2 + 4 ) = − 11 + 12 = 1. From A B = I AB=I A B = I det A ⋅ det B = det I = 1 \det A\cdot\det B=\det I=1 det A ⋅ det B = det I = 1 det B = 1 / det A = 1 \det B=1/\det A=1 det B = 1/ det A = 1
det A = 1 , det B = 1. \boxed{\det A=1,\qquad \det B=1.} det A = 1 , det B = 1. (iii) Solve the system
In matrix form the system is
[ 1 0 2 2 − 1 3 4 1 8 ] [ x y z ] = [ 3 3 14 ] , \begin{bmatrix}1&0&2\\2&-1&3\\4&1&8\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\3\\14\end{bmatrix}, 1 2 4 0 − 1 1 2 3 8 x y z = 3 3 14 , i.e. A x = b A\mathbf{x}=\mathbf{b} A x = b same matrix A A A A − 1 = B A^{-1}=B A − 1 = B
x = B b = [ − 11 2 2 − 4 0 1 6 − 1 − 1 ] [ 3 3 14 ] . \mathbf{x}=B\mathbf{b}=\begin{bmatrix}-11&2&2\\-4&0&1\\6&-1&-1\end{bmatrix}\begin{bmatrix}3\\3\\14\end{bmatrix}. x = B b = − 11 − 4 6 2 0 − 1 2 1 − 1 3 3 14 . Compute:
x = − 11 ( 3 ) + 2 ( 3 ) + 2 ( 14 ) = − 33 + 6 + 28 = 1 , x=-11(3)+2(3)+2(14)=-33+6+28=1, x = − 11 ( 3 ) + 2 ( 3 ) + 2 ( 14 ) = − 33 + 6 + 28 = 1 , y = − 4 ( 3 ) + 0 ( 3 ) + 1 ( 14 ) = − 12 + 14 = 2 , y=-4(3)+0(3)+1(14)=-12+14=2, y = − 4 ( 3 ) + 0 ( 3 ) + 1 ( 14 ) = − 12 + 14 = 2 , z = 6 ( 3 ) − 1 ( 3 ) − 1 ( 14 ) = 18 − 3 − 14 = 1. z=6(3)-1(3)-1(14)=18-3-14=1. z = 6 ( 3 ) − 1 ( 3 ) − 1 ( 14 ) = 18 − 3 − 14 = 1. Answer
( x , y , z ) = ( 1 , 2 , 1 ) . \boxed{\;(x,y,z)=(1,\,2,\,1).\;} ( x , y , z ) = ( 1 , 2 , 1 ) .