Use matrix method to solve the following system of equations:x +

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 Multiple Choice QuestionsLong Answer Type

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291. Use matrix method to solve the following system of equations:
x + y + z = 4
2 x – y + z = - 1
2 x + y – 3 z = – 9


The given equations are
x + y + z = 4
2x – y + z = - 1
2x + y – 3z = – 9
These equations can be written as
                            open square brackets table row 1 cell space space space space space space 1 end cell cell space space space space space space 1 end cell row 2 cell space space minus 1 end cell cell space space space space space space 1 end cell row 2 cell space space space 1 end cell cell space space space minus 3 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 4 end cell row cell negative 1 end cell row cell negative 9 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 1 cell space space space space space 1 end cell cell space space space space 1 end cell row 2 cell space minus 1 end cell cell space space space space space 1 end cell row 2 cell space space space space space 1 end cell cell space minus 3 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 4 end cell row cell negative 1 end cell row cell negative 9 end cell end table close square brackets
open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space space space 1 end cell cell space space space space 1 end cell row 2 cell space minus 1 end cell cell space space space space 1 end cell row 2 cell space space space space 1 end cell cell space minus 3 end cell end table close vertical bar space equals space 1 open vertical bar table row cell negative 1 end cell cell space space space space space 1 end cell row 1 cell space space minus 3 end cell end table close vertical bar minus 1 open vertical bar table row 2 cell space space space space space 1 end cell row 2 cell space space minus 3 end cell end table close vertical bar plus 1 space open vertical bar table row 2 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
space space space space space space space equals 1 left parenthesis 3 minus 1 right parenthesis space minus space 1 left parenthesis negative 6 minus 2 right parenthesis space plus space 1 space left parenthesis 2 plus 2 right parenthesis
space space space space space space equals 2 plus 8 plus 4 space equals space 14 space not equal to space 0
space space 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 space space space space 1 end cell row 1 cell space space minus 3 end cell end table close vertical bar comma space space space space minus open vertical bar table row 2 cell space space space space space space space 1 end cell row 2 cell space space minus 3 end cell end table close vertical bar comma space space space space open vertical bar table row 2 cell space space space minus 1 end cell row 2 cell space space space space space 1 end cell end table close vertical bar

i.e. 2, 8, 4 respectively.
Co-factors of the of the elements of second row of | A | are
negative open vertical bar table row cell negative 1 end cell cell space space space space space 1 end cell row cell space 1 end cell cell space minus 3 end cell end table close vertical bar comma space space space space open vertical bar table row 2 cell space space space space space space space space 1 end cell row 2 cell space space space minus 3 end cell end table close vertical bar comma space space space space open vertical bar table row 2 cell space space space space minus 1 end cell row 2 cell space space space space space space space 1 end cell end table close vertical bar

i.e. 2, 8, 4 respectively.
Co-factors of the of the elements of second row of | A | are
negative open vertical bar table row cell negative 1 end cell cell space space space space space space 1 end cell row cell space 1 end cell cell space space minus 3 end cell end table close vertical bar comma space space space space space open vertical bar table row 2 cell space space space space space space space 1 end cell row 2 cell space space space minus 3 end cell end table close vertical bar comma space space space space minus open vertical bar table row 1 cell space space space space 1 end cell row 2 cell space space space space 1 end cell end table close vertical bar
i.e.   4, – 5, 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 1 end cell row cell negative 1 end cell cell space space space space 1 end cell end table close vertical bar comma space space space space space space minus 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 space space open vertical bar table row 1 cell space space space space space space space 1 end cell row 2 cell space space space minus 1 end cell end table close vertical bar
i.e. 2,  1, – 3 respectively.
therefore space space space space adj space straight A space equals space open square brackets table row 2 cell space space space space space space 8 end cell cell space space space 4 end cell row 4 cell space space minus 5 end cell cell space space space space 1 end cell row 2 cell space space space space space 1 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 2 cell space space space space space 4 end cell cell space space space space space space 2 end cell row 8 cell space space space minus 5 end cell cell space space space space space space 1 end cell row 4 cell space space space space 1 end cell cell space space minus 3 end cell end table close square brackets
space space space 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 over 14 open square brackets table row 2 cell space space space space space 4 end cell cell space space space space space 2 end cell row 8 cell space space minus 5 end cell cell space space space space space 1 end cell row 4 cell space space space space 1 end cell cell space minus 3 end cell end table close square brackets
Now comma space space AX space equals space straight B space space space rightwards double arrow space space space straight X space equals straight A to the power of negative 1 end exponent straight B
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 over 14 open square brackets table row 2 cell space space space space space 4 end cell cell space space space space space 2 end cell row 8 cell space minus 5 end cell cell space space space space space 1 end cell row 4 cell space space space 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 space 4 end cell row cell negative 1 end cell row cell negative 9 end cell end table close square brackets
rightwards double arrow 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 space equals space 1 over 14 open square brackets table row cell 8 minus 4 minus 18 end cell row cell 32 plus 5 minus 9 end cell row cell 16 minus 1 plus 27 end cell end table close square brackets space space space space space rightwards double arrow 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 over 14 open square brackets table row cell negative 14 end cell row cell space 28 end cell row cell space 42 end cell end table close square brackets
rightwards double arrow 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 cell negative 1 end cell row cell space space 2 end cell row cell space 3 end cell end table close square brackets
therefore space space space space space straight x space equals space minus 1 comma space space space space space space straight y space equals space 2 comma space space space straight z space equals 3 space is space required space solution. space

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

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 Multiple Choice QuestionsShort Answer Type

293. Use matrix method to solve the following system of equations:
2x – 3y + 5z = 16
3x + 2y – 4z = – 4
x + y – 2z = – 3

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 Multiple Choice QuestionsLong Answer Type

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

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295. Use matrix method to solve the following system of equations:
2x + 6y = 2
3x – z =  8
2x - y + z = – 3 
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296. Use matrix method to solve the following system of equations:
x + y + z = 6
x – y + z = 2
2x + y – z = 1
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297. Use matrix method to solve the following system of equations:
9x – 5y - 11z = 12 
x – 3y + z = 1
2x + 3y – 7z = 2
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298. Use matrix method to solve the following system of equations:
2x + y – z = 1
x – y + z = 2
3x + y – 2z = – 1


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299. Use matrix method to solve the following system of equations:
6x + y – 3z = 5
x + 3y – 2z = 5
2x + y + 4z = 8

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

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