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Determinants

Long Answer Type

271.

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

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272.

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

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273.

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

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

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275. Use matrix method to solve the following system of equations:
3x – 4y + 2z = – 1
2x + 3y + 5z = 7
x + z = 2

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

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277. Use matrix method to solve the following system of equations:
x + y + z = 3
x + 2y + 3z = 4
x + 4y + 9z = 6

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

The given equations are
3x + y + z + 3
2x – y – z = 2
– x – y + z = 1
The given equation can be written as
$open square brackets table row 3 cell space space space space 1 end cell cell space space space 1 end cell row 2 cell space minus 1 end cell cell space minus 1 end cell row cell negative 1 end cell cell space minus 1 end cell cell space space space space 1 end cell end table close square brackets space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 3 row 2 row 1 end table close square brackets$

or $AX space equals space straight B space where space straight A space equals space open square brackets table row 3 cell space space space space 1 end cell cell space space space space space 1 end cell row 2 cell space minus 1 end cell cell space minus 1 end cell row cell negative 1 end cell cell negative 1 end cell cell space space space space 1 end cell end table close square brackets comma space space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space space straight B space equals space open square brackets table row 3 row 2 row 1 end table close square brackets$
$Now space straight A space equals space open vertical bar table row 3 cell space space space space 1 end cell cell space space space space 1 end cell row 2 cell space minus 1 end cell cell space minus 1 end cell row cell negative 1 end cell cell space minus 1 end cell cell space space space 1 end cell end table close vertical bar space equals space 3 open vertical bar table row cell negative 1 end cell cell space space minus 1 end cell row cell negative 1 end cell cell space space space space space space 1 end cell end table close vertical bar minus 1 open vertical bar table row 2 cell space space space minus 1 end cell row cell negative 1 end cell cell space space space space space space space 1 end cell end table close vertical bar plus 1 space open vertical bar table row 2 cell space space space minus 1 end cell row cell negative 1 end cell cell space space space minus 1 end cell end table close vertical bar space space space space space equals 3 left parenthesis negative 1 minus 1 right parenthesis minus left parenthesis 2 minus 1 right parenthesis plus left parenthesis negative 2 minus 1 right parenthesis space equals space minus 6 minus 1 minus 3 space equals space minus 10 space not equal to 0 therefore space space space space straight A to the power of negative 1 end exponent space space exists.$
Co-factors of the elements of first low of | A | are
$open vertical bar table row cell negative 1 end cell cell space space minus 1 end cell row cell negative 1 end cell cell space space space space space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space minus 1 end cell row cell negative 1 end cell cell space space space space 1 end cell end table close vertical bar comma space space space space open vertical bar table row cell space 2 end cell cell space space space minus 1 end cell row cell negative 1 end cell cell space space space minus 1 end cell end table close vertical bar$

i.e. – 2, – 1, – 3 respectively
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row 1 cell space space space 1 end cell row cell negative 1 end cell cell space space 1 end cell end table close vertical bar comma space space open vertical bar table row cell space space 3 end cell cell space space space 1 end cell row cell negative 1 end cell cell space space space 1 end cell end table close vertical bar comma space space space minus open vertical bar table row 3 cell space space space space space space 1 end cell row 1 cell space space minus 1 end cell end table close vertical bar$
i.e. –2, 4, 2 respectively
Co-factors of the elements of third row of | A | are
$open vertical bar table row cell space space 1 end cell cell space space space space space space 1 end cell row cell negative 1 end cell cell space space minus 1 end cell end table close vertical bar space minus space open vertical bar table row 3 cell space space space 1 end cell row 2 cell space minus 1 end cell end table close vertical bar comma space space space space open vertical bar table row 3 cell space space space space space 1 end cell row 2 cell space minus 1 end cell end table close vertical bar$
i.e. 0, 5, 5 respectively
$therefore space space space adj. space straight A space equals space open square brackets table row cell negative 2 end cell cell space space minus 1 end cell cell space space minus 3 end cell row cell negative 2 end cell cell space space space space 4 end cell cell space space space space space 2 end cell row cell space space 0 end cell cell space space space space 5 end cell cell space minus 5 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 2 end cell cell space space space minus 2 end cell cell space space space 0 end cell row cell negative 1 end cell cell space space space space 4 end cell cell space space space space 5 end cell row cell negative 3 end cell cell space space space 2 end cell cell space minus 5 end cell end table close square brackets 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 minus 1 over 10 open square brackets table row cell negative 2 end cell cell space minus 2 end cell cell space space 0 end cell row cell negative 1 end cell cell space space 4 end cell cell space space 5 end cell row 3 cell space space 2 end cell cell space minus 5 end cell end table close square brackets Now space space space AX space equals space straight B space space space space space space space space space rightwards double arrow space space space space space straight X space equals space 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 minus 1 over 10 open square brackets table row cell negative 2 end cell cell space space minus 2 end cell cell space space space 0 end cell row cell negative 1 end cell cell space space space 4 end cell cell space space space 5 end cell row cell negative 3 end cell cell space space space 2 end cell cell negative 5 end cell end table close square brackets space open square brackets table row 3 row 2 row 1 end table close square brackets space space space rightwards double arrow space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space minus 1 over 10 open square brackets table row cell negative 6 minus 4 plus 0 end cell row cell negative 3 plus 8 plus 5 end cell row cell negative 9 plus 4 minus 5 end cell end table close square brackets 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 minus 1 over 10 open square brackets table row cell negative 10 end cell row 10 row cell negative 10 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 open square brackets table row cell space space space space 1 end cell row cell negative 1 end cell row cell space space 1 end cell end table close square brackets therefore space space space space space solution space is space straight x space equals space 1 comma space space space space straight y space equals space minus 1 comma space space space straight z space space equals 1$

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

78 Views

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

78 Views

Previous Year Papers

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Test Yourself

Long Answer Type

278. Use matrix method to solve the following system of equations:3x + y + z = 32x – y –z = 2–x –y + z = 1

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