Consider the following system of linear equations (E): 3x-y+z=q px+3y-z=5 x-4y+2z=-1, where p and q are real numbers 
(a) Assume that (E) has a unique solution.
(i) Find the range of values of p. 
(ii) Solve (E).
(b) Assume that p=2 and (E) is consistent.
(i) Find the value of q.
(ii) Solve (E).
(c) Hence, or otherwise, determine whether the system of equations xz=y 2x+3y-z=5 3x-y+z=4 x-4y+2z=-1 is consistent

We are given a system of linear equations:

\[
\begin{aligned}
&\text{(E)}: 
\\ 
&3x - y + z = q \\
&px + 3y - z = 5 \\
&x - 4y + 2z = -1
\end{aligned}
\]
where \( p \) and \( q \) are real numbers. Let's approach the problem step by step.

### (a) Assume that (E) has a unique solution.
#### (i) Find the range of values of \( p \).

For a system of three linear equations to have a unique solution, the determinant of the coefficient matrix must be non-zero. The coefficient matrix \( A \) for this system is:

\[
A = \begin{pmatrix}
3 & -1 & 1 \\
p & 3 & -1 \\
1 & -4 & 2
\end{pmatrix}
\]

We will calculate the determinant of \( A \):

\[
\text{det}(A) = \begin{vmatrix}
3 & -1 & 1 \\
p & 3 & -1 \\
1 & -4 & 2
\end{vmatrix}
\]

We can expand this determinant using cofactor expansion along the first row:

\[
\text{det}(A) = 3 \begin{vmatrix}
3 & -1 \\
-4 & 2
\end{vmatrix}
- (-1) \begin{vmatrix}
p & -1 \\
1 & 2
\end{vmatrix}
+ 1 \begin{vmatrix}
p & 3 \\
1 & -4
\end{vmatrix}
\]

Now, calculate each 2x2 determinant:

\[
\begin{vmatrix}
3 & -1 \\
-4 & 2
\end{vmatrix} = (3)(2) - (-1)(-4) = 6 - 4 = 2
\]

\[
\begin{vmatrix}
p & -1 \\
1 & 2
\end{vmatrix} = (p)(2) - (-1)(1) = 2p + 1
\]

\[
\begin{vmatrix}
p & 3 \\
1 & -4
\end{vmatrix} = (p)(-4) - (3)(1) = -4p - 3 = -4p - 3
\]

Now substitute these back into the determinant expression:

\[
\text{det}(A) = 3(2) - (-1)(2p + 1) + 1(-4p - 3)
\]
\[
\text{det}(A) = 6 + (2p + 1) - (4p + 3)
\]
\[
\text{det}(A) = 6 + 2p + 1 - 4p - 3
\]
\[
\text{det}(A) = 4 - 2p
\]

For the system to have a unique solution, we need \( \text{det}(A) \neq 0 \):

\[
4 - 2p \neq 0 \quad \Rightarrow \quad p \neq 2
\]

Thus, the system has a unique solution for all values of \( p \) except \( p = 2 \).

#### (ii) Solve (E).

To solve the system, we can use Cramer's rule or Gaussian elimination. Since we know that \( p \neq 2 \), the system is consistent and has a unique solution. Let's leave the numerical solution for later after handling the next sub-question, which gives specific values for \( p \) and \( q \).

---

### (b) Assume that \( p = 2 \) and (E) is consistent.
#### (i) Find the value of \( q \).

If \( p = 2 \), substitute this into the system of equations:

\[
\begin{aligned}
&3

Consider the following system of linear equations (E): 3x - y + z = q, px + 3y - z = 5, x - 4y + 2z = -1, where p and q are real numbers. (a) Assume that (E) has a unique solution. (i) Find the range of values of p. (ii) Solve (E). (b) Assume that p = 2 and (E) is consistent. (i) Find the value of q. (ii) Solve (E). (c) Hence, or otherwise, determine whether the system of equations xz = y, 2x + 3y - z = 5, 3x - y + z = 4, x - 4y + 2z = -1 is consistent.

Explore how to solve a system of linear equations using determinants, including finding the range of values of p for a unique solution and solving for specific cases. This includes step-by-step methods like Cramer's rule and consistency analysis of linear systems. Ideal for undergraduate students studying Linear Algebra.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation