To solve The system of linear equations by using the Cramer’s rule.
The given system of equations has No solution .
Given information:
The system of equations are { 4 x − y + z = − 5 2 x + 2 y + 3 z = 10 6 x + y + 4 z = − 5 .
Formula used:
If { a 1 x + b 1 y + + c 1 z = d 1 a 2 x + b 2 y + c 2 z = d 2 a 3 x + b 3 y + c 3 z = d 3 represents the system of equations then the value of x and y are calculated as x = | d 1 b 1 c 1 d 2 b 2 c 2 d 3 b 3 c 3 | | a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 | = D x D , y = | a 1 d 1 c 1 a 2 d 2 c 2 a 3 d 3 c 3 | | a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 | = D y D and z = | a 1 b 1 d 1 a 2 b 2 d 2 a 3 b 3 d 3 | | a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 | = D z D
Calculation:
Compute the determinant of the matrix as follows,
D = [ 4 − 1 1 2 2 3 6 1 4 ] = 4 | 2 3 1 4 | − ( − 1 ) | 2 3 6 4 | + 1 | 2 2 6 1 | = 4 ⋅ 5 − 10 − 10 = 0
As the determinant of the equation is nonzero, so the solution of the equations exist.
Therefore, the given system has no solution.
The given system of equations has
Given information:
The system of equations are
Formula used:
If
Calculation:
Compute the determinant of the matrix as follows,
As the determinant of the equation is nonzero, so the solution of the equations exist.
Therefore, the given system has no solution.
Answer to Problem 29E
The given system of equations has
Explanation of Solution
Given information:
The system of equations are
Formula used:
If
Calculation:
Compute the determinant of the matrix as follows,
As the determinant of the equation is nonzero, so the solution of the equations exist.
Therefore, the given system has no solution.
Chapter 7 Solutions
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
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