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Question

Use matrix to solve the following system of equations
x+y+z=3
x+2y+3z=4
2x+3y+4z=7

A
x=2+k,y=12k,z=k where kR
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B
x=2+k,y=12k,z=k where kR
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C
x=2k,y=12k,z=k where kR
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D
x=2+k,y=1+2k,z=k where kR
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Solution

The correct option is B x=2+k,y=12k,z=k where kR
Given system of equations can be written as
AX=B
where A=111123234
X=xyz ;B=347
Here, |A|=0
Now, we will find (adjA)B
adjA=CT=121121121T
adjA=111222111
Now, (adjA)B=111222111347
(adjA)B=000
(adjA)B=O
Hence,the system of equations has infinitely many solutions.
Let z=k where kR
Then
x+y=3k
x+2y=43k
Solving these eqns, we get
y=12k;x=2+k

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