Solve the following system of equations by matrix method:2x –

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsLong Answer Type

Advertisement

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


The given equations are
2x – 3 y + 5 z =11
3x + 2 y – 4 z = – 5
x + y – 2 z = –3
These equations can be written as
                              open square brackets table row 2 cell space space minus 3 end cell cell space space space space space space 5 end cell row 3 cell space space space space space 2 end cell cell space minus 4 end cell row 1 cell space space 1 end cell cell space space minus 2 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 11 end cell row cell negative 5 end cell row cell negative 3 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 2 cell space space minus 3 end cell cell space space space space space space 5 end cell row 3 cell space space space 2 end cell cell space space minus 4 end cell row 1 cell space space 1 end cell cell space space minus 2 end cell end table close square brackets comma space space space space straight X 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 cell space space 11 end cell row cell negative 5 end cell row cell negative 3 end cell end table close square brackets
space space space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space minus 3 end cell cell space space space space 5 end cell row 3 cell space space space 2 end cell cell space minus 4 end cell row 1 cell space space 1 end cell cell space space minus 2 end cell end table close vertical bar space equals space 2 open vertical bar table row 2 cell space space space minus 4 end cell row 1 cell space space minus 2 end cell end table close vertical bar minus left parenthesis negative 3 right parenthesis open vertical bar table row 3 cell space space minus 4 end cell row 1 cell space space minus 2 end cell end table close vertical bar plus 5 open vertical bar table row 3 cell space space 2 end cell row 1 cell space space 1 end cell end table close vertical bar
space space space space space space space space space space space space space space equals space 2 left parenthesis negative 4 plus 4 right parenthesis plus 3 left parenthesis negative 6 plus 4 right parenthesis plus 5 left parenthesis 3 minus 2 right parenthesis space equals space 0 minus 6 plus 5 space equals space minus 1 not equal to space 0
therefore space space space space space straight A to the power of negative 1 end exponent space exists
space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight A subscript 11 space equals space open vertical bar table row 2 cell space space space minus 4 end cell row 1 cell space space minus 2 end cell end table close vertical bar space equals space minus 4 plus 4 space equals space 0
space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight A subscript 12 space equals space minus open vertical bar table row 3 cell space space space minus 4 end cell row 1 cell space space minus 2 end cell end table close vertical bar space equals space minus left parenthesis negative 6 plus 4 right parenthesis space equals space 2
space space space space space space space space space space space space space space space space space space space space space space space space space space straight A subscript 13 space equals space open vertical bar table row 3 cell space space space space 2 end cell row 1 cell space space space space 1 end cell end table close vertical bar space equals space 3 minus 2 space equals space 1
space space space

                     straight A subscript 21 space equals space minus open vertical bar table row cell negative 3 end cell cell space space space space 5 end cell row 1 cell space minus 2 end cell end table close vertical bar space equals space minus left parenthesis 6 minus 5 right parenthesis space equals space minus 1
straight A subscript 22 space equals space open vertical bar table row 2 cell space space space space 5 end cell row 1 cell space space minus 2 end cell end table close vertical bar space equals space minus 4 minus 5 space equals space minus 9
straight A subscript 23 space equals space minus open vertical bar table row 2 cell space space space minus 3 end cell row 1 cell space space space space space 1 end cell end table close vertical bar space equals space minus left parenthesis 2 plus 3 right parenthesis space equals space minus 5
straight A subscript 31 space equals space open vertical bar table row cell negative 3 end cell cell space space space space 5 end cell row 2 cell space space minus 4 end cell end table close vertical bar space equals space 12 minus 10 space equals space 2
straight A subscript 32 space equals space minus open vertical bar table row 2 cell space space space space space 5 end cell row 3 cell space space minus 4 end cell end table close vertical bar space equals space minus left parenthesis negative 8 minus 15 right parenthesis space equals space 23
straight A subscript 33 space equals space open vertical bar table row 2 cell space space space space minus 3 end cell row 3 cell space space space space space space 2 end cell end table close vertical bar space equals space 4 plus 9 space equals space 13

therefore space space space space adj. space straight A space equals space open square brackets table row 0 cell space space space 2 end cell cell space space space space space 1 end cell row cell negative 1 end cell cell space minus 9 end cell cell space space minus 5 end cell row 2 cell space space space 23 end cell cell space space space 13 end cell end table close square brackets to the power of apostrophe space space space equals open square brackets table row 0 cell space space minus 1 end cell cell space space 2 end cell row 2 cell space minus 9 end cell cell space 23 end cell row 1 cell negative 5 end cell cell space 13 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 open square brackets table row 0 cell space space minus 1 end cell cell space space space space 2 end cell row 2 cell space space minus 9 end cell cell space space 23 end cell row 1 cell space space minus 5 end cell cell space space space 13 end cell end table close square brackets
Now space space space space AX space equals space straight B 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
space space rightwards double arrow 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 negative open square brackets table row 0 cell space space minus 1 end cell cell space space space space 2 end cell row 2 cell space minus 9 end cell cell space space 23 end cell row 1 cell space space minus 5 end cell cell space 13 end cell end table close square brackets space open square brackets table row 11 row cell negative 5 end cell row cell negative 3 end cell end table close square brackets
rightwards double arrow space space 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 minus open square brackets table row cell 0 plus 5 minus 6 end cell row cell 24 plus 45 minus 69 end cell row cell 11 plus 25 minus 39 end cell end table close square brackets space equals space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space minus open square brackets table row cell negative 1 end cell row cell negative 2 end cell row cell negative 3 end cell end table close square brackets
rightwards double arrow 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 1 row 2 row 3 end table close square brackets
therefore space space space space space space space space space space space space straight x space equals space 1 comma space space space straight y space equals space 2 comma space space space straight z space equals space 3

78 Views

Advertisement
262.

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

74 Views

263.

Solve (Use matrix method):
x + y = 0
y + z = 1
z + x = 3   

74 Views

264. Using matrices, solve the following system of linear equations:
2x + y + z = 7
x – y – z = – 4
3x + 2y + z = 10  
84 Views

Advertisement
265.

Using matrices,  following system of linear equations:
x – y + 2 z = 1
2 y – 3z = 1
3x – 2y + 4z = 2  

76 Views

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

72 Views

Advertisement
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

77 Views

Advertisement