Determine whether the following system has no solution, an infinite number of solutions or a unique solution.3x+ 6y +22 z =-2x5 y17 z7zx + 2ySelect AnswerSelect AnswerSelect AnswerSelect Answer1.2.3.-3 x5x5 x-3 x5x5x8x16x-24 x-II+-+++5 y3y31 y5y3 y31 y4 y8 y12 y-+1-++++--+3 zN3 z9zN3z3z9z=|| ||====|| 11=10 z20 z30 z =73|| ||051-2751-32= 2= 4-6

The given system of linear equations,the augmented matrix isHere, It can be observed thatRank of the…

Answered: Determine whether the following system…

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Related questions

Question

Determine whether the following system has no solution, an infinite number of solutions or a unique solution.
3x
+ 6y +
22 z =
-2x
5 y
17 z
7z
x + 2y
Select Answer
Select Answer
Select Answer
Select Answer
1.
2.
3.
-3 x
5x
5 x
-3 x
5x
5x
8x
16x
-24 x
-
I
I
+
-
+
+
+
5 y
3y
31 y
5y
3 y
31 y
4 y
8 y
12 y
-
+
1
-
+
+
+
+
-
-
+
3 z
N
3 z
9z
N
3z
3z
9z
=
|| ||
=
=
=
=
|| 11
=
10 z
20 z
30 z =
73
|| ||
0
5
1
-27
5
1
-32
= 2
= 4
-6

Expert Solution

Step 1: Introduce to the question 1

The given system of linear equations,

$space space space space 3 x space plus space 6 y space plus space 22 z equals 7 minus 2 x space minus space 5 y space minus space 17 z equals 3 space space space space space space x space plus space 2 y space plus space space space 7 z equals 0$

the augmented matrix is

$open square brackets right enclose table row 3 6 22 row cell negative 2 end cell cell negative 5 end cell cell negative 17 end cell row 1 2 7 end table end enclose table row 7 row 3 row 0 end table close square brackets Performed space the space row space operartions comma R subscript 1 rightwards arrow 1 third R subscript 1 open square brackets right enclose table row 1 2 cell 22 over 3 end cell row cell negative 2 end cell cell negative 5 end cell cell negative 17 end cell row 1 2 7 end table end enclose table row cell 7 over 3 end cell row 3 row 0 end table close square brackets R subscript 2 rightwards arrow R subscript 2 plus 2 R subscript 1 open square brackets right enclose table row 1 2 cell 22 over 3 end cell row 0 cell negative 1 end cell cell negative 7 over 3 end cell row 1 2 7 end table end enclose table row cell 7 over 3 end cell row cell 23 over 3 end cell row 0 end table close square brackets R subscript 3 rightwards arrow R subscript 3 minus R subscript 1 open square brackets right enclose table row 1 2 cell 22 over 3 end cell row 0 cell negative 1 end cell cell negative 7 over 3 end cell row 0 0 cell negative 1 third end cell end table end enclose table row cell 7 over 3 end cell row cell 23 over 3 end cell row cell negative 7 over 3 end cell end table close square brackets R subscript 2 rightwards arrow open parentheses negative 1 close parentheses R subscript 2 open square brackets right enclose table row 1 2 cell 22 over 3 end cell row 0 1 cell 7 over 3 end cell row 0 0 cell negative 1 third end cell end table end enclose table row cell 7 over 3 end cell row cell negative 23 over 3 end cell row cell negative 7 over 3 end cell end table close square brackets R subscript 1 rightwards arrow R subscript 1 minus 2 R subscript 2 open square brackets right enclose table row 1 0 cell 8 over 3 end cell row 0 1 cell 7 over 3 end cell row 0 0 cell negative 1 third end cell end table end enclose table row cell 53 over 3 end cell row cell negative 23 over 3 end cell row cell negative 7 over 3 end cell end table close square brackets R subscript 3 rightwards arrow open parentheses negative 3 close parentheses R subscript 3 open square brackets right enclose table row 1 0 cell 8 over 3 end cell row 0 1 cell 7 over 3 end cell row 0 0 1 end table end enclose table row cell 53 over 3 end cell row cell negative 23 over 3 end cell row 7 end table close square brackets R subscript 1 rightwards arrow R subscript 1 minus 8 over 3 R subscript 3 comma space R subscript 2 rightwards arrow R subscript 2 minus 7 over 3 R subscript 3 open square brackets right enclose table row 1 0 0 row 0 1 0 row 0 0 1 end table end enclose table row cell negative 1 end cell row cell negative 24 end cell row 7 end table close square brackets$

Here, It can be observed that

Rank of the matrix A = 3 and

Rank of the augmented matrix = 3.

$rightwards double arrow$ Rank of the matrix A = Rank of the augmented matrix = 3

Also, it is equal to a number of unknowns.

Thus, the system has unique solution.

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