Using matrices,  following system of linear equations:x – y +

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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  
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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  


The given equations are
x - y + 2z = 1
2y – 3z = 1
3x – 2y + 4z = 2
These equations can be written as
open square brackets table row 1 cell space space minus 1 end cell cell space space space space space space 2 end cell row 0 cell space space space space space space 2 end cell cell space minus 3 end cell row 3 cell space minus 2 end cell cell space space space space 4 end cell end table close square brackets 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 1 row 2 end table close square brackets

or     AX space space equals space straight B space space space space where space straight A space equals space open square brackets table row 1 cell space space minus 1 end cell cell space space space space space 2 end cell row 0 cell space space space space space 2 end cell cell space space minus 3 end cell row 3 cell space space space minus 2 end cell cell space space space space space 4 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 straight B space equals space open square brackets table row 1 row 1 row 2 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 minus 1 end cell cell space space space space space 2 end cell row 0 cell space space space space space 2 end cell cell space space minus 3 end cell row 3 cell space space minus 2 end cell cell space space space 4 end cell end table close vertical bar space equals space 1 open vertical bar table row 2 cell space space space minus 3 end cell row cell negative 2 end cell cell space space space space space space space 4 end cell end table close vertical bar minus left parenthesis negative 1 right parenthesis space open vertical bar table row 0 cell space space space minus 3 end cell row 3 cell space space space space space 4 end cell end table close vertical bar plus 2 space open vertical bar table row 0 cell space space space space space 2 end cell row 3 cell space minus 2 end cell end table close vertical bar
space space space space space equals 1 space left parenthesis 8 minus 6 right parenthesis space plus space 1 left parenthesis 0 plus 9 right parenthesis space plus space 2 left parenthesis 0 minus 6 right parenthesis space equals space 1 left parenthesis 2 right parenthesis space plus space 1 left parenthesis 9 right parenthesis space plus space 2 space left parenthesis negative 6 right parenthesis
space space space space space equals space 2 plus 9 minus 12 space equals space minus 1 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 space 2 end cell cell space space space minus 3 end cell row cell negative 2 end cell cell space space space space space 4 end cell end table close vertical bar comma space space space space minus open vertical bar table row 0 cell space space space minus 3 end cell row 3 cell space space space space space space space 4 end cell end table close vertical bar comma space space open vertical bar table row 0 cell space space space space space space 2 end cell row 3 cell space space minus 2 end cell end table close vertical bar

i.e. 8 – 6, – (0 + 9), 0 – 6 i.e. 2, – 9, – 6 respectively.
Co-factors 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 2 end cell row cell negative 2 end cell cell space space space space 4 end cell end table close vertical bar comma space space space space open vertical bar table row 1 cell space space space 2 end cell row 3 cell space space 4 end cell end table close vertical bar comma space space space space minus open vertical bar table row 1 cell space space minus 1 end cell row 3 cell space minus 2 end cell end table close vertical bar
i.e.   – (– 4 + 4), 4 – 6, – (– 2 + 7) i.e. 0, – 2, – 1 respectively.
Co-factors of the elements of third row of | A | are
open vertical bar table row cell negative 1 end cell cell space space space space 2 end cell row 2 cell space minus 3 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space space 2 end cell row 0 cell space minus 3 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space minus 1 end cell row 0 cell space space space 2 end cell end table close vertical bar
i.e. 3 – 4, – (– 3 – 0), 2+ 0 i.e. – 1, 3, 2 respectively.
therefore space space space space space space space space space space space adj. space straight A space equals space open square brackets table row 2 cell space space minus 9 end cell cell space space space minus 6 end cell row 0 cell space space minus 2 end cell cell space space minus 1 end cell row cell negative 1 end cell cell space space space space 3 end cell cell space space space space 2 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell space 2 end cell cell space space space space 0 end cell cell space space minus 1 end cell row cell negative 9 end cell cell space space minus 2 end cell cell space space space space space 3 end cell row cell negative 6 end cell cell space space minus 1 end cell cell space space space space space 2 end cell end table close square brackets
therefore space space space space space 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 open square brackets table row cell space space space space 2 end cell cell space space space space space 0 end cell cell space space space minus 1 end cell row cell negative 9 end cell cell space space space minus 2 end cell cell space space space space space 3 end cell row cell negative 6 end cell cell space space space minus 1 end cell cell space space space space 2 end cell end table close square brackets space equals space open square brackets table row 2 cell space space space 0 end cell cell space space space space space 1 end cell row 9 cell space space 2 end cell cell space space minus 3 end cell row 6 cell space space 1 end cell cell space space minus 2 end cell end table close square brackets
Now comma space space space space space space space AX space equals space straight B space space space space space space space space space space space space space rightwards double arrow space space space space space straight A to the power of negative 1 end exponent straight B
therefore 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 cell negative 2 end cell cell space space space 0 end cell cell space space space space space 1 end cell row 9 cell space 2 end cell cell space minus 3 end cell row 6 1 cell space minus 2 end cell end table close square brackets space open square brackets table row 1 row 1 row 2 end table close square brackets
therefore 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 cell negative 2 plus 0 plus 2 end cell row cell 9 plus 2 minus 6 end cell row cell 6 plus 1 minus 4 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 open square brackets table row 0 row 5 row 3 end table close square brackets
therefore space space space space space straight x space equals space 0 comma space space space space straight y space equals space 5 comma space space space space straight z space equals space 3.

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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

81 Views

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

110 Views

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

75 Views

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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