Math Problem Statement

Assignment#3.pdf

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Solve question 5

Solution

The system of equations in Question 5 is:

x + 2y - 2z &= 5 \\ x - y + 3z &= -1 \\ x - 7y + kz &= -k \end{aligned}$$ ### (a) Reduce the system to echelon form: We will use Gaussian elimination to reduce the system to row echelon form. The augmented matrix of the system is: $$\begin{pmatrix} 1 & 2 & -2 & | & 5 \\ 1 & -1 & 3 & | & -1 \\ 1 & -7 & k & | & -k \end{pmatrix}$$ **Step 1:** Subtract row 1 from rows 2 and 3 to eliminate $$x$$ from rows 2 and 3. $$R2 = R2 - R1 \quad \text{and} \quad R3 = R3 - R1$$ This gives: $$\begin{pmatrix} 1 & 2 & -2 & | & 5 \\ 0 & -3 & 5 & | & -6 \\ 0 & -9 & k+2 & | & -k-5 \end{pmatrix}$$ **Step 2:** Divide row 2 by -3 to simplify: $$R2 = \frac{R2}{-3}$$ Now the matrix becomes: $$\begin{pmatrix} 1 & 2 & -2 & | & 5 \\ 0 & 1 & -\frac{5}{3} & | & 2 \\ 0 & -9 & k+2 & | & -k-5 \end{pmatrix}$$ **Step 3:** Use row 2 to eliminate $$y$$ from row 3. We add 9 times row 2 to row 3: $$R3 = R3 + 9R2$$ This gives: $$\begin{pmatrix} 1 & 2 & -2 & | & 5 \\ 0 & 1 & -\frac{5}{3} & | & 2 \\ 0 & 0 & k - 1 & | & -k + 13 \end{pmatrix}$$ This is the echelon form of the system. ### (b) Infinitely many solutions: For the system to have infinitely many solutions, the last row must represent a true statement (i.e., 0 = 0). This happens when the coefficient of $$z$$ in the last row is zero, and the constant on the right-hand side is also zero. So, set: $$k - 1 = 0 \quad \text{and} \quad -k + 13 = 0$$ From $$k - 1 = 0$$, we get $$k = 1$$. Substitute $$k = 1$$ into $$-k + 13 = 0$$: $$-1 + 13 = 12 \neq 0$$ There was an error above. Only the first condition, $$k - 1 = 0$$, needs to be checked. This

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Systems of Equations
Row Echelon Form
Gaussian Elimination

Formulas

Row operations for Gaussian Elimination
Condition for infinite solutions (k = 1)

Theorems

Gaussian Elimination Theorem
Consistent Systems Theorem

Suitable Grade Level

Undergraduate