solve the following system of linear equations using matrix method 2x-y-z=7,3x+y-z=7,x+y-z=3

 

Asked by prashubh2000 | 25th Apr, 2019, 11:31: AM

Expert Answer:

2x - y - z = 7, 3x + y - z = 7,x + y - z = 3

 begin mathsize 16px style open square brackets table row 2 cell negative 1 end cell cell negative 1 end cell row 3 1 cell negative 1 end cell row 1 1 cell negative 1 end cell end table close square brackets open square brackets table row straight x row straight y row straight z end table close square brackets equals open square brackets table row 7 row 7 row 3 end table close square brackets
Use space the space steps space given space below space to space solve space this space matrix
1 right parenthesis space straight R subscript 1 space left right arrow straight R subscript 3
2 right parenthesis space space straight R subscript 2 space rightwards arrow straight R subscript 2 space minus space 3 straight R subscript 1 space and space straight R subscript 3 space rightwards arrow straight R subscript 3 space minus space 2 straight R subscript 1
3 right parenthesis space space straight R subscript 2 space rightwards arrow fraction numerator straight R subscript 2 over denominator negative 2 end fraction
4 right parenthesis space straight R subscript 1 space rightwards arrow straight R subscript 1 space minus space straight R subscript 2 space and space space straight R subscript 3 space rightwards arrow straight R subscript 3 space plus 3 straight R subscript 2
5 right parenthesis space straight R subscript 3 space rightwards arrow fraction numerator straight R subscript 3 over denominator negative 2 end fraction
6 right parenthesis space straight R subscript 2 space rightwards arrow straight R subscript 2 plus straight R subscript 3
We space get space straight x equals 2 comma space straight y equals negative 1 space and space straight z equals negative 2
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Answered by Sneha shidid | 25th Apr, 2019, 12:01: PM