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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

75 Views

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

These equations can be written as

open square brackets table row 1 cell space space space 1 end cell cell space space space 1 end cell row 1 cell space space space 2 end cell cell space space space 3 end cell row 1 cell space space 4 end cell cell space space space 9 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 3 row 4 row 6 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 1 end cell cell space space space 1 end cell row 1 cell space space space 2 end cell cell space space space 3 end cell row 1 cell space space space 4 end cell cell space space space 9 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 3 row 4 row 6 end table close square brackets space space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space 1 end cell cell space space space space 1 end cell row 1 cell space space space 2 end cell cell space space space space 3 end cell row 1 cell space space space space 4 end cell cell space space space space 9 end cell end table close vertical bar space equals space 1 open vertical bar table row 2 cell space space space 3 end cell row 4 cell space space space 9 end cell end table close vertical bar comma space space space minus 1 space open vertical bar table row 1 cell space space space 3 end cell row 1 cell space space space 9 end cell end table close vertical bar space plus space 1 space open vertical bar table row 1 cell space space space 2 end cell row 1 cell space space space 4 end cell end table close vertical bar space space space space space space space space space space space equals space 1 left parenthesis 18 minus 12 right parenthesis minus 1 left parenthesis 9 minus 3 right parenthesis space plus space 1 space left parenthesis 4 minus 2 right parenthesis space equals space 6 minus 6 plus 2 space equals space 2

Co-factors of the elements of first row of | A | are

open vertical bar table row 2 cell space space space 3 end cell row 4 cell space space space 9 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space 3 end cell row 1 cell space space 9 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space space 2 end cell row 1 cell space space space 4 end cell end table close vertical bar

i.e., 6, – 6, 2 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 4 cell space space 9 end cell end table close vertical bar comma space space space space open vertical bar table row 1 cell space space space 1 end cell row 1 cell space space 9 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space 1 end cell row 1 cell space space space 4 end cell end table close vertical bar

i.e., – 5, 8, – 3 respectively
Co-factors of the elements of third row of |A| are

open vertical bar table row 1 cell space space space 1 end cell row 2 cell space space space 3 end cell end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space space space space 1 end cell row 1 cell space space space space 3 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space 1 end cell row 1 cell space space space 2 end cell end table close vertical bar straight i. straight e. space space 1 comma space – space 2 comma space space 1 space space space respectively.

adj space straight A space equals space space open square brackets table row 6 cell space space minus 6 end cell cell space space space space 2 end cell row cell negative 5 end cell cell space space space space 8 end cell cell space minus 3 end cell row 1 cell space space minus 2 end cell cell space space space 1 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell space space space 6 end cell cell space space minus 5 end cell cell space space space space 1 end cell row cell negative 6 end cell cell space space space space space 8 end cell cell space minus 2 end cell row cell space space 2 end cell cell space minus 3 end cell cell space space space 1 end cell end table close square brackets

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 half open square brackets table row cell space space space 6 end cell cell space space minus 5 end cell cell space space space 1 end cell row cell negative 6 end cell cell space space space space 8 end cell cell negative 2 end cell row cell space space 2 end cell cell space minus 3 end cell cell space space space 1 end cell end table close square brackets Now space space AX space equals space straight B space space rightwards double arrow space 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 half open square brackets table row cell space space space 6 end cell cell space space space minus 5 end cell cell space space space space 1 end cell row cell negative 6 end cell cell space space space space space 8 end cell cell space minus 2 end cell row cell space space space 2 end cell cell space space minus 3 end cell cell space space 1 end cell end table close square brackets space open square brackets table row 3 row 4 row 6 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 1 half open square brackets table row cell 18 minus 20 plus 6 end cell row cell negative 18 plus 32 minus 12 end cell row cell 6 minus 12 plus 6 end cell end table close square brackets 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 1 half open square brackets table row 4 row 2 row 0 end table close square brackets 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 2 row 1 row 0 end table close square brackets therefore space space space space space straight x space equals 2 comma space space space straight y space equals space 1 comma space space space straight z space space equals 0

76 Views

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

77 Views

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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Long Answer Type

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

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