Using matrices, solve the following system of linear equations:2

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 Multiple Choice QuestionsLong Answer Type

261.

Solve the following system of equations by matrix method:
2x – 3y + 5z =11
3 x + 2y – 4 z = – 5
x + y – 2 z = –3

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262.

Use matrix method to solve the equations:
x + y + z = 3
2x – y + z = 2
x – 2y + 3z = 2

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263.

Solve (Use matrix method):
x + y = 0
y + z = 1
z + x = 3   

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264. Using matrices, solve the following system of linear equations:
2x + y + z = 7
x – y – z = – 4
3x + 2y + z = 10  


The given equations are
2x + y + z = 7
x – y – z = – 4
3x + 2y + z = 10
These equations can by written as
                           open square brackets table row 2 cell space space space space 1 end cell cell space space space space space 1 end cell row 1 cell space minus 1 end cell cell space minus 1 end cell row 3 cell space space space space 2 end cell cell space space space space space 1 end cell end table close square brackets space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row cell space space space space 7 end cell row cell negative 4 end cell row cell space space 10 end cell end table close square brackets

or space space space AX space equals space straight B space where space straight A space equals space open square brackets table row 2 cell space space 1 end cell cell space space space space 1 end cell row 1 cell negative 1 end cell cell negative 1 end cell row 3 cell space space space 2 end cell cell space space 1 end cell end table close square brackets comma space space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space space straight B space equals space open square brackets table row cell space space space space 7 end cell row cell negative 4 end cell row cell space space 10 end cell end table close square brackets
open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space space space space 1 end cell cell space space space space space space space 1 end cell row 1 cell space minus 1 end cell cell space space space minus 1 end cell row 3 cell space space space 2 end cell cell space space space space space 1 end cell end table close vertical bar space equals space 2 space open vertical bar table row cell negative 1 end cell cell space space space minus 1 end cell row 2 cell space space space space space space 1 end cell end table close vertical bar minus space 1 open vertical bar table row 1 cell space space space minus 1 end cell row 3 cell space space space space space 1 end cell end table close vertical bar plus 1 open vertical bar table row 1 cell space space minus 1 end cell row 3 cell space space space space space 2 end cell end table close vertical bar
space space space space space space equals 2 space left parenthesis negative 1 plus 2 right parenthesis minus 1 left parenthesis 1 plus 3 right parenthesis space plus space 1 left parenthesis 2 plus 3 right parenthesis space equals space 2 left parenthesis 1 right parenthesis space minus space 1 left parenthesis 4 right parenthesis space plus space 1 left parenthesis 5 right parenthesis
space space space space space space space equals space 2 space minus 4 plus 5 space equals space 3 space not equal to 0
therefore space space space space straight A to the power of negative 1 end exponent space exists.

Co-factors of the elements of first row of | A | are
open vertical bar table row cell negative 1 end cell cell space space minus 1 end cell row 2 cell space space space space space 1 end cell end table close vertical bar comma space space space space space minus open vertical bar table row 1 cell space space minus 1 end cell row 3 cell space space space space space space 1 end cell end table close vertical bar comma space space space space open vertical bar table row 1 cell space space space space minus 1 end cell row 3 cell space space space space space space 2 end cell end table close vertical bar
i.e.    – 1 + 2, – (1 + 3), 2 + 3  i.e. 1, – 4, 5 respectively.
Co-factors of the elements of second row of | A | are
negative open vertical bar table row 1 cell space space space 1 end cell row 2 cell space space space 1 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space space 1 end cell row 3 cell space space 1 end cell end table close vertical bar comma space space space minus open vertical bar table row 2 cell space space space 1 end cell row 3 cell space space 2 end cell end table close vertical bar
i.e. – (1 – 2), 2 – 3, – (4 – 3) i.e. 1, – 1, –1 respectively.
Co-factors of the elements of third row of | A | are
open vertical bar table row cell space space 1 end cell cell space space space space space 1 end cell row cell negative 1 end cell cell space minus 1 end cell end table close vertical bar comma space space space minus open vertical bar table row 2 cell space space space space space space 1 end cell row 1 cell space space minus 1 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space space space space space 1 end cell row 1 cell space space minus 1 end cell end table close vertical bar
i.e. – 1 + 1, – (2 – 1), – 2 – 1 i.e. 0, 3, – 3 respectively.
therefore space space adj. space straight A space equals space open square brackets table row 1 cell space space space minus 4 end cell cell space space space space space 5 end cell row 1 cell space space minus 1 end cell cell space minus 1 end cell row 0 cell space space 3 end cell cell space minus 3 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 1 cell space space space space 1 end cell cell space space space space 0 end cell row cell negative 4 end cell cell space minus 1 end cell cell space space space space space 3 end cell row 5 cell negative 1 end cell cell space minus 3 end cell end table close square brackets
therefore space space space space straight A to the power of negative 1 end exponent space equals space fraction numerator adj space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space 1 third open square brackets table row 1 cell space space space space space 1 end cell cell space space space space 0 end cell row cell negative 4 end cell cell space space minus 1 end cell cell space space space space space 3 end cell row 5 cell space minus 1 end cell cell space minus 3 end cell end table close square brackets

Now comma space space space space space space space space space space space space space space AX space space equals space straight B space space space space space space space rightwards double arrow space space space space straight X space equals space straight A to the power of negative 1 end exponent straight B
space therefore space space space space space space space space space space space space space space space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 third open square brackets table row 1 cell space space space space space 1 end cell cell space space space 0 end cell row cell negative 4 end cell cell space space minus 1 end cell cell space space space space 3 end cell row 5 cell space space minus 1 end cell cell space minus 3 end cell end table close square brackets space space open square brackets table row cell space space space 7 end cell row cell negative 4 end cell row 10 end table close square brackets
therefore space space space space space space space space space space space space space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 third open square brackets table row cell 7 minus 4 plus 0 end cell row cell negative 28 plus 4 plus 30 end cell row cell 35 plus 4 minus 30 end cell end table close square brackets space space space space space rightwards double arrow space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 third open square brackets table row 3 row 6 row 9 end table close square brackets
therefore space space space space space space space space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 1 row 2 row 3 end table close square brackets
therefore space space space space space straight x space equals space 1 comma space space space straight y space equals space 2 comma space space space straight z space equals space 3. space space space space space space space space space space space space space space space

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265.

Using matrices,  following system of linear equations:
x – y + 2 z = 1
2 y – 3z = 1
3x – 2y + 4z = 2  

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266.

Use matrix method to solve the following system of equations:
x – y + 2z = 7  
3x + 4 y – 5 z = – 5
2x – y + 3z = 12

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267.

Use matrix method to solve the following system of equations:
x – y + 2z = 7  
3x + 4 y – 5 z = – 5
2x – y + 3z = 12

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268.

Use matrix method to solve the following system of equations:
5x – y + z = 4
3x + 2y – 5z = 2
x + 3 y – 2 z = 5

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269.

Use matrix method to solve the following system of equations:
4x + 2y + 3z = 2
x + y + z = 1
3x + y – 2z = 5

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270.

Use matrix method to solve the following system of equations:
2x + 3y + 3z = 5
x – 2 y + z = – 4
3x – y – 2z = 3

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