Math Problem Statement

find the value of β€œa” such that the following system of linear

equations

𝒙 + π’š + 𝒂𝒛 = πŸ‘, 𝒙 + π’‚π’š + 𝒛 = 𝟐, 𝒂𝒙 + π’š βˆ’ 𝒛 = 𝟏

have

i) exactly one solution

ii) No solution

iii) Infinite number of solutions and find them

put it in matrix and get triangle of zeros

Solution

To solve for the values of aa such that the system of equations:

x + y + az = 3 \\ x + ay + z = 2 \\ ax + y - z = 1 \end{cases}$$ has: 1. Exactly one solution 2. No solution 3. An infinite number of solutions we can put this system in matrix form and perform row operations to get it into an upper triangular (triangular of zeros) form. ### Step 1: Write the system in matrix form The augmented matrix for the system is: $$\begin{bmatrix} 1 & 1 & a & | & 3 \\ 1 & a & 1 & | & 2 \\ a & 1 & -1 & | & 1 \\ \end{bmatrix}$$ ### Step 2: Perform Gaussian elimination to get upper triangular form We'll aim to make zeros below the main diagonal by performing row operations. #### Row Operations: 1. **Subtract $$\text{Row 1}$$ from $$\text{Row 2}$$:** $$\text{Row 2} = \text{Row 2} - \text{Row 1} \Rightarrow \begin{bmatrix} 1 & 1 & a & | & 3 \\ 0 & a-1 & 1-a & | & -1 \\ a & 1 & -1 & | & 1 \end{bmatrix}$$ 2. **Subtract $$a \times \text{Row 1}$$ from $$\text{Row 3}$$:** $$\text{Row 3} = \text{Row 3} - a \times \text{Row 1} \Rightarrow \begin{bmatrix} 1 & 1 & a & | & 3 \\ 0 & a-1 & 1-a & | & -1 \\ 0 & 1-a & -1-a^2 & | & 1 - 3a \end{bmatrix}$$ 3. **Next, eliminate the second element in the third row by using $$\text{Row 2}$$:** Subtract $$\frac{1-a}{a-1} \times \text{Row 2}$$ from $$\text{Row 3}$$: (Continue until upper triangular form is reached.)

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Systems of Equations
Gaussian Elimination

Formulas

Ax = b
Row reduction to echelon form

Theorems

Existence and uniqueness theorem for linear systems

Suitable Grade Level

Grades 10-12