Math Problem Statement
Suppose that the system of linear equations (): 2x + y + 2z = 0 x+ (k + 1) = 0 k−y + 4z = 0 has infinitely many solutions. Note: k∈R . (a) Find the value(s) of .k (b) Solve this use matrix to solve
Solution
Let's solve the system of linear equations step by step, focusing on both parts of the problem.
Given System of Linear Equations:
We are asked to find the value(s) of such that the system has infinitely many solutions, and then solve the system using matrices.
(a) Find the value of
For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero, indicating a dependent system. First, let's represent the system in matrix form.
The system of equations can be written as:
2 & 1 & 2 \\ 1 & 0 & k+1 \\ 0 & -1 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$ Let's calculate the determinant of the coefficient matrix, which we'll denote as $$A$$: $$A = \begin{pmatrix} 2 & 1 & 2 \\ 1 & 0 & k+1 \\ 0 & -1 & 4 \end{pmatrix}$$ The determinant of a $$3 \times 3$$ matrix is given by: $$\text{det}(A) = 2 \begin{vmatrix} 0 & k+1 \\ -1 & 4 \end{vmatrix} - 1 \begin{vmatrix} 1 & k+1 \\ 0 & 4 \end{vmatrix} + 2 \begin{vmatrix} 1 & 0 \\ 0 & -1 \end{vmatrix}$$ We can now compute each of these $$2 \times 2$$ determinants: $$\text{det}\begin{pmatrix} 0 & k+1 \\ -1 & 4 \end{pmatrix} = (0)(4) - (-1)(k+1) = k+1$$ $$\text{det}\begin{pmatrix} 1 & k+1 \\ 0 & 4 \end{pmatrix} = (1)(4) - (k+1)(0) = 4$$ $$\text{det}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = (1)(-1) - (0)(0) = -1$$ Substitute these into the determinant expression: $$\text{det}(A) = 2(k+1) - 1(4) + 2(-1)$$ $$\text{det}(A) = 2k + 2 - 4 - 2 = 2k - 4$$ For the system to have infinitely many solutions, the determinant must be zero: $$2k - 4 = 0$$ $$2k = 4$$ $$k = 2$$ Thus, the value of $$k$$ that results in infinitely many solutions is $$k = 2$$. ### (b) Solve the system using matrices Substitute $$k = 2$$ into the original system of equations: 1. $$2x + y + 2z = 0$$ 2. $$x + 3z = 0$$ 3. $$2 - y + 4z = 0$$ This system can be written in matrix form as: \[ \begin{pmatrix} 2 & 1 & 2 \\ 1 & 0 & 3 \\ 0 & -1 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ -2 \end{pmatrix} toAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
System of Equations
Determinants
Formulas
Determinant of a 3x3 matrix: det(A) = a(ei − fh) − b(di − fg) + c(dh − eg)
Theorems
Infinitely many solutions condition: Determinant of coefficient matrix = 0
Suitable Grade Level
Undergraduate level or advanced high school mathematics
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