PYQs / GATE CS / 2022 / Set 1 / Q45 GATE CS 2022 Set 1 — Question 45 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +2 / -0.67 Medium LU Decomposition Linear Algebra Engineering Mathematics
Engineering Mathematics → Linear Algebra → LU Decomposition
Last updated 5 September 2026
Question Consider solving the following system of simultaneous equations using LU decomposition.
x 1 + x 2 − 2 x 3 = 4 x_1 + x_2 - 2x_3 = 4 x 1 + x 2 − 2 x 3 = 4 x 1 + 3 x 2 − x 3 = 7 x_1 + 3x_2 - x_3 = 7 x 1 + 3 x 2 − x 3 = 7 2 x 1 + x 2 − 5 x 3 = 7 2x_1 + x_2 - 5x_3 = 7 2 x 1 + x 2 − 5 x 3 = 7 where
L L L and
U U U are denoted as
L = ( L 11 0 0 L 21 L 22 0 L 31 L 32 L 33 ) , U = ( U 11 U 12 U 13 0 U 22 U 23 0 0 U 33 ) L = \begin{pmatrix} L_{11} & 0 & 0 \\ L_{21} & L_{22} & 0 \\ L_{31} & L_{32} & L_{33} \end{pmatrix}, \quad U = \begin{pmatrix} U_{11} & U_{12} & U_{13} \\ 0 & U_{22} & U_{23} \\ 0 & 0 & U_{33} \end{pmatrix} L = L 11 L 21 L 31 0 L 22 L 32 0 0 L 33 , U = U 11 0 0 U 12 U 22 0 U 13 U 23 U 33 Which one of the following is the correct combination of values for
L 32 L_{32} L 32 ,
U 33 U_{33} U 33 , and
x 1 x_1 x 1 ?
Correct answer (D) L₃₂ = -(1)/(2), U₃₃ = -(1)/(2), x₁ = 0
Solution We perform LU decomposition (
A = L U A = LU A = LU ) using the Doolittle algorithm (where
L i i = 1 L_{ii} = 1 L ii = 1 ).
Matrix
A = ( 1 1 − 2 1 3 − 1 2 1 − 5 ) A = \begin{pmatrix} 1 & 1 & -2 \\ 1 & 3 & -1 \\ 2 & 1 & -5 \end{pmatrix} A = 1 1 2 1 3 1 − 2 − 1 − 5 .
1. First row of U (same as A):
U 11 = 1 , U 12 = 1 , U 13 = − 2 U_{11}=1, U_{12}=1, U_{13}=-2 U 11 = 1 , U 12 = 1 , U 13 = − 2 .
2. First column of L :
L 11 = 1 L_{11}=1 L 11 = 1 .
L 21 = A 21 / U 11 = 1 / 1 = 1 L_{21} = A_{21}/U_{11} = 1/1 = 1 L 21 = A 21 / U 11 = 1/1 = 1 .
L 31 = A 31 / U 11 = 2 / 1 = 2 L_{31} = A_{31}/U_{11} = 2/1 = 2 L 31 = A 31 / U 11 = 2/1 = 2 .
3. Second row of U :
U 22 = A 22 − L 21 U 12 = 3 − 1 ( 1 ) = 2 U_{22} = A_{22} - L_{21}U_{12} = 3 - 1(1) = 2 U 22 = A 22 − L 21 U 12 = 3 − 1 ( 1 ) = 2 .
U 23 = A 23 − L 21 U 13 = − 1 − 1 ( − 2 ) = 1 U_{23} = A_{23} - L_{21}U_{13} = -1 - 1(-2) = 1 U 23 = A 23 − L 21 U 13 = − 1 − 1 ( − 2 ) = 1 .
4. Second column of L :
L 22 = 1 L_{22}=1 L 22 = 1 .
L 32 = ( A 32 − L 31 U 12 ) / U 22 = ( 1 − 2 ( 1 ) ) / 2 = − 1 / 2 L_{32} = (A_{32} - L_{31}U_{12}) / U_{22} = (1 - 2(1)) / 2 = -1/2 L 32 = ( A 32 − L 31 U 12 ) / U 22 = ( 1 − 2 ( 1 )) /2 = − 1/2 .
5. Third row of U :
U 33 = A 33 − L 31 U 13 − L 32 U 23 = − 5 − 2 ( − 2 ) − ( − 0.5 ) ( 1 ) = − 5 + 4 + 0.5 = − 0.5 U_{33} = A_{33} - L_{31}U_{13} - L_{32}U_{23} = -5 - 2(-2) - (-0.5)(1) = -5 + 4 + 0.5 = -0.5 U 33 = A 33 − L 31 U 13 − L 32 U 23 = − 5 − 2 ( − 2 ) − ( − 0.5 ) ( 1 ) = − 5 + 4 + 0.5 = − 0.5 .
So,
L 32 = − 1 / 2 L_{32} = -1/2 L 32 = − 1/2 and
U 33 = − 1 / 2 U_{33} = -1/2 U 33 = − 1/2 .
Now solve for
x x x using
L y = b Ly = b L y = b and
U x = y Ux = y U x = y , where
b = ( 4 7 7 ) b = \begin{pmatrix} 4 \\ 7 \\ 7 \end{pmatrix} b = 4 7 7 .
Forward substitution (L y = b Ly=b L y = b y 1 = 4 y_1 = 4 y 1 = 4 .
y 2 = 7 − y 1 = 3 y_2 = 7 - y_1 = 3 y 2 = 7 − y 1 = 3 .
y 3 = 7 − 2 y 1 − ( − 0.5 ) y 2 = 7 − 8 + 1.5 = 0.5 y_3 = 7 - 2y_1 - (-0.5)y_2 = 7 - 8 + 1.5 = 0.5 y 3 = 7 − 2 y 1 − ( − 0.5 ) y 2 = 7 − 8 + 1.5 = 0.5 .
Backward substitution (U x = y Ux=y U x = y − 0.5 x 3 = 0.5 ⟹ x 3 = − 1 -0.5x_3 = 0.5 \implies x_3 = -1 − 0.5 x 3 = 0.5 ⟹ x 3 = − 1 .
2 x 2 + x 3 = 3 ⟹ 2 x 2 − 1 = 3 ⟹ x 2 = 2 2x_2 + x_3 = 3 \implies 2x_2 - 1 = 3 \implies x_2 = 2 2 x 2 + x 3 = 3 ⟹ 2 x 2 − 1 = 3 ⟹ x 2 = 2 .
x 1 + x 2 − 2 x 3 = 4 ⟹ x 1 + 2 − 2 ( − 1 ) = 4 ⟹ x 1 + 4 = 4 ⟹ x 1 = 0 x_1 + x_2 - 2x_3 = 4 \implies x_1 + 2 - 2(-1) = 4 \implies x_1 + 4 = 4 \implies x_1 = 0 x 1 + x 2 − 2 x 3 = 4 ⟹ x 1 + 2 − 2 ( − 1 ) = 4 ⟹ x 1 + 4 = 4 ⟹ x 1 = 0 .
Thus,
L 32 = − 1 / 2 L_{32} = -1/2 L 32 = − 1/2 ,
U 33 = − 1 / 2 U_{33} = -1/2 U 33 = − 1/2 , and
x 1 = 0 x_1 = 0 x 1 = 0 .
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Correct answer (D) L₃₂ = -(1)/(2), U₃₃ = -(1)/(2), x₁ = 0
Solution We perform LU decomposition (
A = L U A = LU A = LU ) using the Doolittle algorithm (where
L i i = 1 L_{ii} = 1 L ii = 1 ).
Matrix
A = ( 1 1 − 2 1 3 − 1 2 1 − 5 ) A = \begin{pmatrix} 1 & 1 & -2 \\ 1 & 3 & -1 \\ 2 & 1 & -5 \end{pmatrix} A = 1 1 2 1 3 1 − 2 − 1 − 5 .
1. First row of U (same as A):
U 11 = 1 , U 12 = 1 , U 13 = − 2 U_{11}=1, U_{12}=1, U_{13}=-2 U 11 = 1 , U 12 = 1 , U 13 = − 2 .
2. First column of L :
L 11 = 1 L_{11}=1 L 11 = 1 .
L 21 = A 21 / U 11 = 1 / 1 = 1 L_{21} = A_{21}/U_{11} = 1/1 = 1 L 21 = A 21 / U 11 = 1/1 = 1 .
L 31 = A 31 / U 11 = 2 / 1 = 2 L_{31} = A_{31}/U_{11} = 2/1 = 2 L 31 = A 31 / U 11 = 2/1 = 2 .
3. Second row of U :
U 22 = A 22 − L 21 U 12 = 3 − 1 ( 1 ) = 2 U_{22} = A_{22} - L_{21}U_{12} = 3 - 1(1) = 2 U 22 = A 22 − L 21 U 12 = 3 − 1 ( 1 ) = 2 .
U 23 = A 23 − L 21 U 13 = − 1 − 1 ( − 2 ) = 1 U_{23} = A_{23} - L_{21}U_{13} = -1 - 1(-2) = 1 U 23 = A 23 − L 21 U 13 = − 1 − 1 ( − 2 ) = 1 .
4. Second column of L :
L 22 = 1 L_{22}=1 L 22 = 1 .
L 32 = ( A 32 − L 31 U 12 ) / U 22 = ( 1 − 2 ( 1 ) ) / 2 = − 1 / 2 L_{32} = (A_{32} - L_{31}U_{12}) / U_{22} = (1 - 2(1)) / 2 = -1/2 L 32 = ( A 32 − L 31 U 12 ) / U 22 = ( 1 − 2 ( 1 )) /2 = − 1/2 .
5. Third row of U :
U 33 = A 33 − L 31 U 13 − L 32 U 23 = − 5 − 2 ( − 2 ) − ( − 0.5 ) ( 1 ) = − 5 + 4 + 0.5 = − 0.5 U_{33} = A_{33} - L_{31}U_{13} - L_{32}U_{23} = -5 - 2(-2) - (-0.5)(1) = -5 + 4 + 0.5 = -0.5 U 33 = A 33 − L 31 U 13 − L 32 U 23 = − 5 − 2 ( − 2 ) − ( − 0.5 ) ( 1 ) = − 5 + 4 + 0.5 = − 0.5 .
So,
L 32 = − 1 / 2 L_{32} = -1/2 L 32 = − 1/2 and
U 33 = − 1 / 2 U_{33} = -1/2 U 33 = − 1/2 .
Now solve for
x x x using
L y = b Ly = b L y = b and
U x = y Ux = y U x = y , where
b = ( 4 7 7 ) b = \begin{pmatrix} 4 \\ 7 \\ 7 \end{pmatrix} b = 4 7 7 .
Forward substitution (L y = b Ly=b L y = b y 1 = 4 y_1 = 4 y 1 = 4 .
y 2 = 7 − y 1 = 3 y_2 = 7 - y_1 = 3 y 2 = 7 − y 1 = 3 .
y 3 = 7 − 2 y 1 − ( − 0.5 ) y 2 = 7 − 8 + 1.5 = 0.5 y_3 = 7 - 2y_1 - (-0.5)y_2 = 7 - 8 + 1.5 = 0.5 y 3 = 7 − 2 y 1 − ( − 0.5 ) y 2 = 7 − 8 + 1.5 = 0.5 .
Backward substitution (U x = y Ux=y U x = y − 0.5 x 3 = 0.5 ⟹ x 3 = − 1 -0.5x_3 = 0.5 \implies x_3 = -1 − 0.5 x 3 = 0.5 ⟹ x 3 = − 1 .
2 x 2 + x 3 = 3 ⟹ 2 x 2 − 1 = 3 ⟹ x 2 = 2 2x_2 + x_3 = 3 \implies 2x_2 - 1 = 3 \implies x_2 = 2 2 x 2 + x 3 = 3 ⟹ 2 x 2 − 1 = 3 ⟹ x 2 = 2 .
x 1 + x 2 − 2 x 3 = 4 ⟹ x 1 + 2 − 2 ( − 1 ) = 4 ⟹ x 1 + 4 = 4 ⟹ x 1 = 0 x_1 + x_2 - 2x_3 = 4 \implies x_1 + 2 - 2(-1) = 4 \implies x_1 + 4 = 4 \implies x_1 = 0 x 1 + x 2 − 2 x 3 = 4 ⟹ x 1 + 2 − 2 ( − 1 ) = 4 ⟹ x 1 + 4 = 4 ⟹ x 1 = 0 .
Thus,
L 32 = − 1 / 2 L_{32} = -1/2 L 32 = − 1/2 ,
U 33 = − 1 / 2 U_{33} = -1/2 U 33 = − 1/2 , and
x 1 = 0 x_1 = 0 x 1 = 0 .
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