Problem 37
Question
Determine whether the statement is true or false. Justify your answer. Cramer's Rule cannot be used to solve a system of linear equations when the determinant of the coefficient matrix is zero.
Step-by-Step Solution
Verified Answer
The statement is true. Cramer's Rule cannot be used to solve a system of linear equations when the determinant of the coefficient matrix is zero because the system may not have a unique solution in this case.
1Step 1: Analyzing the Statement
Cramer's Rule can be used to solve a system of linear equations by using determinants. According to Cramer's Rule, the system of linear equations has a unique solution if and only if the determinant of the coefficient matrix of the system of equations is not zero.
2Step 2: Verifying the Statement
The statement 'Cramer's Rule cannot be used to solve a system of linear equations when the determinant of the coefficient matrix is zero' is evaluated. It aligns with the condition of Cramer's Rule. When the determinant of the coefficient matrix is zero, the system of equations may either have infinite solutions or no solutions, hence Cramer's Rule cannot be used.
3Step 3: Statement Evaluation
Given this information, it can be concluded that the statement is true. If the determinant of the coefficient matrix of a system is zero, Cramer's Rule cannot be used to find a unique solution for that system.
Other exercises in this chapter
Problem 36
Solve the system of linear equations and check any solution algebraically. $$\left\\{\begin{array}{r} 3 x-2 y-6 z=-4 \\ -3 x+2 y+6 z=1 \\ x-y-5 z=-3 \end{array}
View solution Problem 37
Determine whether the matrix is in row-echelon form. If it is, determine if it is also in reduced row-echelon form. $$\left[\begin{array}{rrrr} 3 & 0 & 3 & 7 \\
View solution Problem 37
Use the matrix capabilities of a graphing utility to evaluate the determinant. $$\left|\begin{array}{rrrrrr} 3 & -2 & 4 & 3 & 1 & 3 \\ -1 & 0 & 2 & 1 & 0 & 0 \\
View solution Problem 37
Solve the system by the method of elimination and check any solutions using a graphing utility. $$\left\\{\begin{array}{r} \frac{x+2}{4}+\frac{y-1}{4}=1 \\ x-y=
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