System of Linear Equations: Definition, Consistency, and Condition

Given Data

System of equations: $$x_1 - 4x_2 + 7x_3 = g \quad (1)$$ $$3x_2 - 5x_3 = h \quad (2)$$ $$-2x_1 + 5x_2 - 9x_3 = k \quad (3)$$

Coefficient matrix and augmented matrix: $$A = \begin{bmatrix} 1 & -4 & 7 \ 0 & 3 & -5 \ -2 & 5 & -9 \end{bmatrix}, \qquad [A \mid b] = \begin{bmatrix} 1 & -4 & 7 & g \ 0 & 3 & -5 & h \ -2 & 5 & -9 & k \end{bmatrix}$$

1. Definition of a System of Linear Equations

A linear equation in variables $x_1, x_2, \ldots, x_n$ is an equation of the form: $$a_1x_1 + a_2x_2 + \cdots + a_nx_n = b$$ where $a_1, \ldots, a_n, b$ are constants.

A system of linear equations is a finite collection of linear equations involving the same variables. In matrix form it is written $Ax = b$.

2. When is a System Consistent?

A system is consistent if it has at least one solution (either a unique solution or infinitely many). It is inconsistent if it has no solution.

Equivalently, $Ax = b$ is consistent if and only if $\operatorname{rank}(A) = \operatorname{rank}([A\mid b])$. In row-echelon form, no row of the type $[,0\ 0\ \cdots\ 0 \mid c,]$ with $c \neq 0$ may appear.

3. Finding the Condition on $g, h, k$

Row reduction of the augmented matrix:

$$\begin{bmatrix} 1 & -4 & 7 & g \ 0 & 3 & -5 & h \ -2 & 5 & -9 & k \end{bmatrix}$$

Step 1: $R_3 \leftarrow R_3 + 2R_1$

$$R_3 = (-2+2,\ 5-8,\ -9+14,\ k+2g) = (0,\ -3,\ 5,\ k+2g)$$

$$\begin{bmatrix} 1 & -4 & 7 & g \ 0 & 3 & -5 & h \ 0 & -3 & 5 & k+2g \end{bmatrix}$$

Step 2: $R_3 \leftarrow R_3 + R_2$

$$R_3 = (0,\ -3+3,\ 5-5,\ k+2g+h) = (0,\ 0,\ 0,\ 2g+h+k)$$

$$\begin{bmatrix} 1 & -4 & 7 & g \ 0 & 3 & -5 & h \ 0 & 0 & 0 & 2g+h+k \end{bmatrix}$$

Consistency condition: The last row gives the equation $$0 = 2g + h + k.$$

For the system to be consistent, this must not be a contradiction, so:

$$\boxed{2g + h + k = 0}$$

Correction Note

In a hand solution it is easy to carry $R_3$ correctly to $k + 2g$ in Step 1 and then write the final constant as $g + h + k$. Adding $R_2$ to $R_3$ actually gives: $$(k + 2g) + h = 2g + h + k,$$ not $g + h + k$, so the condition is $2g + h + k = 0$.

Verification: Try $g=1, h=0, k=-2$ (satisfies $2g+h+k = 2+0-2 = 0$).

• Eq (2): $3x_2 - 5x_3 = 0$
• Eq (1): $x_1 - 4x_2 + 7x_3 = 1$
• Eq (3): $-2x_1 + 5x_2 - 9x_3 = -2$

Compute $2\times(1) + (3)$: $2(x_1-4x_2+7x_3) + (-2x_1+5x_2-9x_3) = -3x_2+5x_3$. LHS constant: $2(1)+(-2)=0$. So $-3x_2+5x_3 = 0 \Rightarrow 3x_2-5x_3=0$, which matches Eq (2). Consistent. ✓

This confirms the correct condition is $2g + h + k = 0$.

Summary Table

LU Factorization

Given Data

$$A = \begin{bmatrix} 2 & 3 & 4 \ 4 & 5 & 10 \ 4 & 8 & 2 \end{bmatrix}$$

Goal: find $L$ (unit lower triangular) and $U$ (upper triangular) with $A = LU$ using Doolittle's method.

Step 1: Eliminate Column 1

Pivot $= 2$.

Multiplier $m_{21} = \frac{4}{2} = 2$: $$R2 \to R2 - 2R1 = [0,\ 5-6,\ 10-8] = [0,\ -1,\ 2]$$

Multiplier $m_{31} = \frac{4}{2} = 2$: $$R3 \to R3 - 2R1 = [0,\ 8-6,\ 2-8] = [0,\ 2,\ -6]$$

$$\begin{bmatrix} 2 & 3 & 4 \ 0 & -1 & 2 \ 0 & 2 & -6 \end{bmatrix}$$

Step 2: Eliminate Column 2

Pivot $= -1$.

Multiplier $m_{32} = \frac{2}{-1} = -2$: $$R3 \to R3 - (-2)R2 = R3 + 2R2 = [0,\ 2-2,\ -6+4] = [0,\ 0,\ -2]$$

$$\begin{bmatrix} 2 & 3 & 4 \ 0 & -1 & 2 \ 0 & 0 & -2 \end{bmatrix}$$

Result

$$U = \begin{bmatrix} 2 & 3 & 4 \ 0 & -1 & 2 \ 0 & 0 & -2 \end{bmatrix}, \qquad L = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 2 & -2 & 1 \end{bmatrix}$$

Verification $LU = A$

• Row 1: $[2,\ 3,\ 4]$ ✓
• Row 2: $[2(2),\ 2(3)+(-1),\ 2(4)+2] = [4,\ 5,\ 10]$ ✓
• Row 3: $[2(2),\ 2(3)+(-2)(-1),\ 2(4)+(-2)(2)+(-2)] = [4,\ 8,\ 2]$ ✓

Final Answer

$$\boxed{A = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 2 & -2 & 1 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \ 0 & -1 & 2 \ 0 & 0 & -2 \end{bmatrix}}$$

Given Data

$$A = \begin{bmatrix} 5 & -3 \ -4 & 9 \end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix} 3 \ 2 \end{bmatrix}$$

Requirement: Determine whether $\mathbf{v}$ is an eigenvector of $A$; if so, find the eigenvalue.

Definition

A non-zero vector $\mathbf{v}$ is an eigenvector of $A$ if $A\mathbf{v} = \lambda\mathbf{v}$ for some scalar $\lambda$ (the eigenvalue).

Step 1: Compute $A\mathbf{v}$

$$A\mathbf{v} = \begin{bmatrix} 5 & -3 \ -4 & 9 \end{bmatrix} \begin{bmatrix} 3 \ 2 \end{bmatrix}$$

Row 1: $$5(3) + (-3)(2) = 15 - 6 = 9$$

Row 2: $$(-4)(3) + 9(2) = -12 + 18 = 6$$

$$A\mathbf{v} = \begin{bmatrix} 9 \ 6 \end{bmatrix}$$

Step 2: Check $A\mathbf{v} = \lambda\mathbf{v}$

$$\begin{bmatrix} 9 \ 6 \end{bmatrix} = \lambda \begin{bmatrix} 3 \ 2 \end{bmatrix}$$

Component 1: $9 = 3\lambda \Rightarrow \lambda = 3$

mth168-eigenvector-check-3-2

Component 2: $6 = 2\lambda \Rightarrow \lambda = 3$

Both components yield the same scalar $\lambda = 3$.

Step 3: Verify

$$A\mathbf{v} = \begin{bmatrix} 9 \ 6 \end{bmatrix} = 3\begin{bmatrix} 3 \ 2 \end{bmatrix} = 3\mathbf{v} \quad \checkmark$$

Conclusion

Yes, $\begin{bmatrix} 3 \ 2 \end{bmatrix}$ is an eigenvector of $A$, with corresponding eigenvalue $\lambda = 3$.

Definition of a Group and Proof that (Z, .) is Not a Group

Definition of a Group

A group is an algebraic structure consisting of a non-empty set G together with a binary operation * such that the following four axioms are satisfied:

If additionally the operation is commutative (a * b = b * a for all a, b in G), then G is called an abelian group.

Proof that (Z, .) Does Not Form a Group

Here Z = {..., -2, -1, 0, 1, 2, 3, ...} is the set of all integers and (.) denotes ordinary multiplication.

We verify each group axiom:

Axiom G1 - Closure: Satisfied.

For any a, b in Z, a . b is also in Z. Example: 3 . 4 = 12, which is in Z. So closure holds.

Axiom G2 - Associativity: Satisfied.

For any a, b, c in Z, (a . b) . c = a . (b . c). This is a standard property of integer multiplication. So associativity holds.

Axiom G3 - Identity: Satisfied.

The element 1 in Z acts as the multiplicative identity since: a . 1 = 1 . a = a for all a in Z. So the identity element exists.

Axiom G4 - Inverse: NOT Satisfied.

For every a in Z, we need an element a' in Z such that: a . a' = 1

Consider a = 2 in Z. We need a' such that: 2 . a' = 1 This gives a' = 1/2

But 1/2 is not in Z (it is not an integer).

More generally, for any integer a where |a| > 1, the multiplicative inverse 1/a is not an integer, so 1/a does not belong to Z.

Also, for a = 0, there is no element a' in Z such that 0 . a' = 1, since 0 . a' = 0 for all a' in Z.

Therefore, the inverse axiom fails for (Z, .).

Conclusion

Since the inverse axiom (G4) is not satisfied in (Z, .) -- most integers do not have a multiplicative inverse within Z -- the algebraic structure (Z, .) does not form a group.