SOLUTION: Consider the following system of linear equations 𝑥 + 3𝑦 + 2𝑧 − 𝑤 = −1 −3𝑥 − 7𝑦 + (𝑝 − 6)𝑧 + 2𝑤 = 1 2𝑥 + 𝑝^2𝑧 + 𝑝𝑤 = 𝑞^2

Algebra ->  Equations -> SOLUTION: Consider the following system of linear equations 𝑥 + 3𝑦 + 2𝑧 − 𝑤 = −1 −3𝑥 − 7𝑦 + (𝑝 − 6)𝑧 + 2𝑤 = 1 2𝑥 + 𝑝^2𝑧 + 𝑝𝑤 = 𝑞^2       Log On


   



Question 1205173: Consider the following system of linear equations
𝑥 + 3𝑦 + 2𝑧 − 𝑤 = −1
−3𝑥 − 7𝑦 + (𝑝 − 6)𝑧 + 2𝑤 = 1
2𝑥 + 𝑝^2𝑧 + 𝑝𝑤 = 𝑞^2
where 𝑝 and 𝑞 are real numbers.
Using Gaussian elimination, determine all possible values of 𝑝 and 𝑞 such that the system has infinitely many solutions with two free variables and solve the systems

Answer by ElectricPavlov(122) About Me  (Show Source):
You can put this solution on YOUR website!
**1. Set up the Augmented Matrix**
The given system of linear equations can be represented by the following augmented matrix:
```
[ 1 3 2 -1 | -1 ]
[-3 -7 p-6 2 | 1 ]
[ 2 0 p^2 p | q^2 ]
```
**2. Perform Row Operations**
* **R2 = R2 + 3R1:**
```
[ 1 3 2 -1 | -1 ]
[ 0 2 p -1 | -2 ]
[ 2 0 p^2 p | q^2 ]
```
* **R3 = R3 - 2R1:**
```
[ 1 3 2 -1 | -1 ]
[ 0 2 p -1 | -2 ]
[ 0 -6 p^2-4 p+2 | q^2+2 ]
```
* **R3 = R3 + 3R2:**
```
[ 1 3 2 -1 | -1 ]
[ 0 2 p -1 | -2 ]
[ 0 0 p^2+3p-4 p-1 | q^2-4 ]
```
**3. Analyze for Infinitely Many Solutions**
For the system to have infinitely many solutions with two free variables, the last row of the augmented matrix must be a row of zeros. This implies:
* p^2 + 3p - 4 = 0
* p - 1 = 0
* q^2 - 4 = 0
**4. Solve for p and q**
* **p^2 + 3p - 4 = 0**
* (p + 4)(p - 1) = 0
* p = -4 or p = 1
* **p - 1 = 0**
* p = 1
* **q^2 - 4 = 0**
* q = ±2
Since we need all conditions to be satisfied simultaneously, the only possible values for p and q are:
* **p = 1**
* **q = 2 or q = -2**
**5. Solve the System for p = 1, q = 2 (or q = -2)**
* Substitute p = 1 into the row-echelon form of the augmented matrix:
```
[ 1 3 2 -1 | -1 ]
[ 0 2 1 -1 | -2 ]
[ 0 0 0 0 | 0 ]
```
* Let z = s and w = t (where s and t are free variables).
* From the second row: 2y + s - t = -2
=> y = (-s + t - 2)/2
* From the first row: x + 3y + 2z - w = -1
=> x + 3[(-s + t - 2)/2] + 2s - t = -1
=> x = (s - t - 1)/2
**Therefore, the general solution for p = 1 and q = ±2 is:**
* x = (s - t - 1)/2
* y = (-s + t - 2)/2
* z = s
* w = t
where s and t are arbitrary real numbers.
**In summary:**
* The system of linear equations has infinitely many solutions with two free variables only when p = 1 and q = 2 or q = -2.
* The general solution for these values of p and q is given above.