The inverse of the matrix [\table 2 , 3 , 4 ; 4 , 3 , 1 ; 1 , 2 , 4 ] is

Correct Answer is : [−245953−45−145−11565]

[\table 2 , -\frac{4}{5} , -\frac{9}{5} ; -3 , \frac{4}{5} , \frac{14}{5} ; 1 , -\frac{1}{5} , -\frac{6}{5}]

[\table -2 , \frac{4}{5} , \frac{9}{5} ; 3 , -\frac{4}{5} , -\frac{14}{5} ; -1 , \frac{1}{5} , \frac{6}{5}]

GATE Civil Engineering (CE) 2019 Shift 2 Solved Paper

Question : 55 of 65

Marks: +1, -0

The inverse of the matrix

[\table 2 , 3 , 4 ; 4 , 3 , 1 ; 1 , 2 , 4 ]

$[\table 2 , -\;{4}/{5} , -\;{9}/{5} ; -3 , \;{4}/{5} , \;{14}/{5} ; 1 , -\;{1}/{5} , -\;{6}/{5}]$
$[\table -2 , \;{4}/{5} , \;{9}/{5} ; 3 , -\;{4}/{5} , -\;{14}/{5} ; -1 , \;{1}/{5} , \;{6}/{5}]$
$[\table 10 , -4 , -9 ; -15 , 4 , 14 ; 5 , -1 , -6]$
$[\table -10 , 4 , 9 ; 15 , -4 , -14 ; -5 , 1 , 6]$

Solution:

A=[\table 2 , 3 , 4 ; 4 , 3 , 1 ; 1 , 2 , 4]

Det. of matrix,

| {A}|=2(12-2)-3(16-1)+4(8-3)

=20-45+20

=-5

{A}^{-1}=\;{1}/{| {A}|}(

adj.

{A})

(adj A)=(cof A)'

cof of

{A}_{11}=12-2=10

cof of

{A}_{12}=-(16-1)=-15

cof of

{A}_{13}=8-3=5

cof of

{A}_{21}=-(12-8)=-4

cof of

{A}_{22}=(8-4)=4

cof of

{A}_{23}=-(4-3)=-1

cof of

{A}_{31}=(3-12)=-9

cof of

{A}_{32}=-(2-16)=14

cof of

{A}_{33}=(6-12)=-6

cof

A=[\table 10 , -15 , 5 ; -4 , 4 , -1 ; -9 , 14 , -6]

\adj \A=(\cof \A){'}

=[\table 10 , -4 , -9 ; -15 , 4 , 14 ; 5 , -1 , -6]

A^{-1}=-\;{1}/{5}[\table 10 , -4 , -9 ; -15 , 4 , 14 ; 5 , -1 , -6]

=[\table -2 , 4 {/} 5 , 9{/}5 ; 3 , -4{/}5 , -14{/}5 ; -1 , 1{/}5 , 6{/}5]

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