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Determinants

Long Answer Type

231.

Examine the consistency of the system of equations:
3x – y – 2z = 2
2y – z = – 1
3x – 5y = 3

80 Views

Short Answer Type

232.

Examine the consistency of the system of equations:
5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = – 1

90 Views

Long Answer Type

233.

Examine the consistencies of the system of equations:
3x – y + 2z = 3
2x + y + 3z =5
x - 2y - z = 1

91 Views

234.

Examine the consistencies of the system of equations:
x - y+ z = 3
2x - y – z = 2
– x – 2y + 2z = 1

74 Views

Short Answer Type

235. Solve the following system of equations using matrix method:
2x + 5y = 1
1x + 2y = 7

72 Views

Long Answer Type

236. Solve system of linear equations, using matrix method:
5x + 2y = 4
7x + 3y = 5

74 Views

Short Answer Type

237. Solve system of linear equations, using matrix method:
2x - y = -2
3x + 4y = 3

93 Views

238. Solve system of linear equations, using matrix method:
4x – 3y = 3
3x – 5y = 7

The given equations are
4x – 3y = 3
3x – 5y = 7
These equations can be written as
$open square brackets table row 4 cell space space minus 3 end cell row 3 cell space space space minus 5 end cell end table close square brackets space open square brackets table row straight x row straight y end table close square brackets space equals space open square brackets table row 3 row 7 end table close square brackets$

$or space space AX space equals space straight B space where space straight A space equals space open square brackets table row 4 cell space space minus 3 end cell row 3 cell space space minus 5 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 end table close square brackets comma space space space straight B space equals space open square brackets table row 3 row 7 end table close square brackets space space space space space space space space space space space space space space space space space space space space open vertical bar straight A close vertical bar space space equals space open vertical bar table row 4 cell space space minus 3 end cell row 3 cell space space minus 5 end cell end table close vertical bar space equals space minus 20 plus 9 space equals space minus 11 space not equal to space 0 therefore space space space straight A to the power of negative 1 end exponent space exists. space space space space space straight A subscript 11 space equals space minus 5 comma space space space space straight A subscript 12 space equals space minus 3 comma space space space straight A subscript 21 space equals space 3 comma space space space space straight A subscript 22 space equals space 4 space space space space space space adj. space straight A space space equals space open square brackets table row cell negative 5 end cell cell space space minus 3 end cell row 3 cell space space space space 4 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 5 end cell cell space space space space 3 end cell row cell negative 3 end cell cell space space space 4 end cell end table close square brackets 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 11 open square brackets table row cell negative 5 end cell cell space space space 3 end cell row cell negative 3 end cell cell space space space 4 end cell end table close square brackets Now comma space space AX space equals space straight B 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 rightwards double arrow space space space space space open square brackets table row straight x row straight y end table close square brackets space equals space minus 1 over 11 open square brackets table row cell negative 5 end cell cell space space space 3 end cell row cell negative 3 end cell cell space space 4 end cell end table close square brackets space open square brackets table row 3 row 7 end table close square brackets space space rightwards double arrow space space space open square brackets table row straight x row straight y end table close square brackets space equals negative 1 over 11 open square brackets table row cell negative 15 plus 21 end cell row cell negative 9 plus 28 end cell end table close square brackets space rightwards double arrow space space space space open square brackets table row straight x row straight y end table close square brackets space equals space minus 1 over 11 open square brackets table row 6 row 19 end table close square brackets space space space rightwards double arrow space space space space space open square brackets table row straight x row straight y end table close square brackets space equals space open square brackets table row cell negative 6 over 11 end cell row cell negative 19 over 11 end cell end table close square brackets therefore space space space space straight x equals negative 6 over 11 comma space space space straight y space equals space minus 19 over 11$