Use matrix method to solve the following system of equations: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

271.

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


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

or    AX = B where straight A space equals space open square brackets table row 2 cell space space minus 3 end cell cell space space 3 end cell row 2 cell space space space space space 2 end cell cell space space 3 end cell row 3 cell space space minus 2 end cell cell space space 2 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 space straight B space equals space open square brackets table row 1 row 2 row 3 end table close square brackets
 
   open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space space minus 3 end cell cell space space 3 end cell row 2 cell space space space space 2 end cell cell space space 3 end cell row 3 cell space minus 2 end cell cell space space 2 end cell end table close vertical bar space equals space 2 open vertical bar table row cell space space 2 end cell cell space space space 3 end cell row cell negative 2 end cell cell space space space space 2 space end cell end table close vertical bar minus left parenthesis negative 3 right parenthesis space open vertical bar table row 2 cell space space 3 end cell row 3 cell space space 2 end cell end table close vertical bar plus 3 space open vertical bar table row 2 cell space space space space space 2 end cell row 3 cell space space minus 2 end cell end table close vertical bar
space space space space space space space equals 2 space left parenthesis 4 plus 6 right parenthesis space plus space 3 space left parenthesis 4 minus 9 right parenthesis space plus space 3 space left parenthesis negative 4 minus 6 right parenthesis
space space space space space space space space equals 20 minus 15 minus 30 space equals space minus 25 space not equal to space 0
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 2 cell space space space space 3 end cell row cell negative 2 end cell cell space space space space 2 end cell end table close vertical bar comma space space space space space minus open vertical bar table row 2 cell space space space space 3 end cell row 3 cell space space space space 2 end cell end table close vertical bar comma space space open vertical bar table row 2 cell space space space space space 2 end cell row 3 cell space minus 2 end cell end table close vertical bar

i.e., 10, 5, – 10 respectively
Co-factors of the elements of second row of | A | are
negative open vertical bar table row cell negative 3 end cell cell space space space 3 end cell row cell negative 2 end cell cell space space space 2 end cell end table close vertical bar comma space space space space open vertical bar table row 2 cell space space space 3 end cell row 3 cell space space 2 end cell end table close vertical bar comma space space space minus open vertical bar table row 2 cell space space space minus 3 end cell row 3 cell space space minus 2 end cell end table close vertical bar
i.e., 0, – 5, – 5 respectively
 Co-factors of the elements of third row of | A | are
open vertical bar table row cell negative 3 end cell cell space space space 3 end cell row 2 cell space space space 3 end cell end table close vertical bar comma space space space space minus open vertical bar table row 2 cell space space space 3 end cell row 2 cell space space space 3 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space minus 3 end cell row 2 cell space space space space 2 end cell end table close vertical bar

i.e., 15, 0, 10 respectively
therefore space space space adj space straight A space equals space open square brackets table row 10 cell space space space space space space 5 space end cell cell space space minus 10 end cell row 0 cell space minus 5 end cell cell space minus 5 end cell row cell negative 15 end cell cell space space space 0 end cell cell space space 10 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 10 cell space space space 0 end cell cell space space minus 15 end cell row 5 cell space minus 5 end cell cell space space space space space 0 end cell row cell negative 10 end cell cell space minus 5 end cell cell space space space space 10 end cell end table close square brackets
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 25 open square brackets table row 10 cell space space space 0 end cell cell space space minus 15 end cell row 5 cell space minus 5 end cell cell space space space space space space 0 end cell row cell negative 10 end cell cell space minus 5 end cell cell space space space space 10 end cell end table close square brackets
Now space space space AX space equals space straight B space space space space rightwards double arrow space space space straight X space equals space straight A to the power of negative 1 end exponent straight B space 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 25 open square brackets table row 10 cell space space space space space 0 end cell cell space space space minus 15 end cell row 5 cell space space minus 5 end cell cell space space space space space 0 end cell row cell negative 10 end cell cell space space minus 5 end cell cell space space space space space 10 end cell end table close square brackets space open square brackets table row 1 row 2 row 3 end table close square brackets
rightwards double arrow 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 1 over 25 open square brackets table row cell 10 plus 0 minus 45 end cell row cell 5 minus 10 plus 0 end cell row cell negative 10 minus 10 plus 30 end cell end table close square brackets space space space space 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 minus 1 over 25 open square brackets table row cell negative 35 end cell row cell negative 5 end cell row cell space 10 end cell end table close square brackets
rightwards double arrow 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 7 over 5 end cell row cell 1 fifth end cell row cell fraction numerator negative 2 over denominator 5 end fraction end cell end table close square brackets space space space space rightwards double arrow space space space straight x space equals space 7 over 5 comma space space space space space straight y space equals space 1 fifth comma space space space straight z space equals space minus 2 over 5

85 Views

Advertisement
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

75 Views

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

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

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

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

Advertisement
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

Advertisement