$$A = \begin{bmatrix} 1 & -3 & -3 \ 1 & 5 & 1 \ 1 & 7 & 2 \end{bmatrix}, \quad b = \begin{bmatrix} 5 \ -3 \ -5 \end{bmatrix}$$ The least square solution solves the normal equation $A^TA\hat{x} = A^Tb$. --- $$A^T = \begin{bmatrix} 1 &...

Linear Transformation: Definition, Example, and Solution

STEP 1 - Given Data

$$A = \begin{bmatrix} 1 & -3 \ 3 & 5 \ -1 & 7 \end{bmatrix}, \quad v = \begin{bmatrix} -2 \ 1 \end{bmatrix}, \quad b = \begin{bmatrix} 3 \ 2 \ 1 \end{bmatrix}, \quad x = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}, \quad T(x) = Ax$$

Definition of Linear Transformation

A transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is a linear transformation if for all $\mathbf{u}, \mathbf{v} \in \mathbb{R}^n$ and all scalars $c$:

1. Additivity: $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$
2. Homogeneity: $T(c\mathbf{v}) = c,T(\mathbf{v})$

Example: For any $m \times n$ matrix $A$, the map $T(\mathbf{x}) = A\mathbf{x}$ is a linear transformation (a matrix transformation).

STEP 2 - Solve

Part (a): Find $T(v)$

$$T(v) = Av = \begin{bmatrix} 1 & -3 \ 3 & 5 \ -1 & 7 \end{bmatrix} \begin{bmatrix} -2 \ 1 \end{bmatrix}$$

• Row 1: $(1)(-2) + (-3)(1) = -2 - 3 = -5$
• Row 2: $(3)(-2) + (5)(1) = -6 + 5 = -1$
• Row 3: $(-1)(-2) + (7)(1) = 2 + 7 = 9$

$$\boxed{T(v) = \begin{bmatrix} -5 \ -1 \ 9 \end{bmatrix}}$$

Part (b): Find $x \in \mathbb{R}^2$ with $T(x) = b$

Solve $Ax = b$:

$$\begin{bmatrix} 1 & -3 \ 3 & 5 \ -1 & 7 \end{bmatrix}\begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 3 \ 2 \ 1 \end{bmatrix}$$

Augmented matrix:

$$\left[\begin{array}{cc|c} 1 & -3 & 3 \ 3 & 5 & 2 \ -1 & 7 & 1 \end{array}\right]$$

$R_2 \to R_2 - 3R_1$: $(0,\ 5+9,\ 2-9) = (0, 14, -7)$

$R_3 \to R_3 + R_1$: $(0,\ 7-3,\ 1+3) = (0, 4, 4)$

$$\left[\begin{array}{cc|c} 1 & -3 & 3 \ 0 & 14 & -7 \ 0 & 4 & 4 \end{array}\right]$$

$R_2 \to \tfrac{1}{14}R_2$: $(0, 1, -0.5)$

$R_3 \to R_3 - 4R_2$:

• Col 2: $4 - 4(1) = 0$
• Col 3 (RHS): $4 - 4(-0.5) = 4 + 2 = 6$

$$\left[\begin{array}{cc|c} 1 & -3 & 3 \ 0 & 1 & -0.5 \ 0 & 0 & 6 \end{array}\right]$$

The last row states $0 = 6$, which is impossible.

Verification of consistency using $R_3 - \frac{4}{14}R_2$ (before scaling): $$4 - \tfrac{4}{14}(-7) = 4 + 2 = 6 \neq 0$$

Same contradiction.

Conclusion: The system $Ax = b$ is inconsistent. There is no $x \in \mathbb{R}^2$ whose image under $T$ is $b = \begin{bmatrix} 3 \ 2 \ 1 \end{bmatrix}$.

Geometric reason: The columns of $A$ span at most a 2-dimensional subspace (plane) of $\mathbb{R}^3$. The vector $b$ does not lie in $\text{Col}(A)$, so $b$ is not in the range of $T$.

$$\boxed{\text{No solution: } b \text{ is not in the range of } T.}$$

State the conclusion: part (a) gives $T(v) = [-5, -1, 9]^T$, and the row reduction in part (b) reaches the contradiction $0 = 6$, so no such $x$ exists and the system is inconsistent.

• Vector: $u = (1, -2, 2, 0)$ $$\u\ = \sqrt{u1^2 + u2^2 + u3^2 + u4^2}$$ $$\u\ = \sqrt{(1)^2 + (-2)^2 + (2)^2 + (0)^2} = \sqrt{1 + 4 + 4 + 0} = \sqrt{9} = 3$$ The unit vector in the same direction as $u$ is: $$v = \frac{u}{\u} = \frac{1...

$$A = \begin{bmatrix} 1 & 2 & 3 & 4 \ 2 & 4 & 7 & 8 \end{bmatrix}$$ Matrix has 2 rows, 4 columns ($n = 4$ variables). We solve $A\mathbf{x} = \mathbf{0}$. $$\begin{bmatrix} 1 & 2 & 3 & 4 \ 2 & 4 & 7 & 8 \end{bmatrix}$$ $R2 \leftarrow R...

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

$$A = \begin{bmatrix} 1 & -3 & -3 \ 1 & 5 & 1 \ 1 & 7 & 2 \end{bmatrix}, \quad b = \begin{bmatrix} 5 \ -3 \ -5 \end{bmatrix}$$ The least square solution solves the normal equation $A^TA\hat{x} = A^Tb$. --- $$A^T = \begin{bmatrix} 1 &...

• Vector: $u = (1, -2, 2, 0)$ $$\u\ = \sqrt{u1^2 + u2^2 + u3^2 + u4^2}$$ $$\u\ = \sqrt{(1)^2 + (-2)^2 + (2)^2 + (0)^2} = \sqrt{1 + 4 + 4 + 0} = \sqrt{9} = 3$$ The unit vector in the same direction as $u$ is: $$v = \frac{u}{\u} = \frac{1...

$$A = \begin{bmatrix} 1 & 2 & 3 & 4 \ 2 & 4 & 7 & 8 \end{bmatrix}$$ Matrix has 2 rows, 4 columns ($n = 4$ variables). We solve $A\mathbf{x} = \mathbf{0}$. $$\begin{bmatrix} 1 & 2 & 3 & 4 \ 2 & 4 & 7 & 8 \end{bmatrix}$$ $R2 \leftarrow R...

Linear Transformation: Definition, Example, and Solution

STEP 1 - Given Data

$$A = \begin{bmatrix} 1 & -3 \ 3 & 5 \ -1 & 7 \end{bmatrix}, \quad v = \begin{bmatrix} -2 \ 1 \end{bmatrix}, \quad b = \begin{bmatrix} 3 \ 2 \ 1 \end{bmatrix}, \quad x = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}, \quad T(x) = Ax$$

Definition of Linear Transformation

A transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is a linear transformation if for all $\mathbf{u}, \mathbf{v} \in \mathbb{R}^n$ and all scalars $c$:

1. Additivity: $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$
2. Homogeneity: $T(c\mathbf{v}) = c,T(\mathbf{v})$

Example: For any $m \times n$ matrix $A$, the map $T(\mathbf{x}) = A\mathbf{x}$ is a linear transformation (a matrix transformation).

STEP 2 - Solve

Part (a): Find $T(v)$

$$T(v) = Av = \begin{bmatrix} 1 & -3 \ 3 & 5 \ -1 & 7 \end{bmatrix} \begin{bmatrix} -2 \ 1 \end{bmatrix}$$

• Row 1: $(1)(-2) + (-3)(1) = -2 - 3 = -5$
• Row 2: $(3)(-2) + (5)(1) = -6 + 5 = -1$
• Row 3: $(-1)(-2) + (7)(1) = 2 + 7 = 9$

$$\boxed{T(v) = \begin{bmatrix} -5 \ -1 \ 9 \end{bmatrix}}$$

Part (b): Find $x \in \mathbb{R}^2$ with $T(x) = b$

Solve $Ax = b$:

$$\begin{bmatrix} 1 & -3 \ 3 & 5 \ -1 & 7 \end{bmatrix}\begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 3 \ 2 \ 1 \end{bmatrix}$$

Augmented matrix:

$$\left[\begin{array}{cc|c} 1 & -3 & 3 \ 3 & 5 & 2 \ -1 & 7 & 1 \end{array}\right]$$

$R_2 \to R_2 - 3R_1$: $(0,\ 5+9,\ 2-9) = (0, 14, -7)$

$R_3 \to R_3 + R_1$: $(0,\ 7-3,\ 1+3) = (0, 4, 4)$

$$\left[\begin{array}{cc|c} 1 & -3 & 3 \ 0 & 14 & -7 \ 0 & 4 & 4 \end{array}\right]$$

$R_2 \to \tfrac{1}{14}R_2$: $(0, 1, -0.5)$

$R_3 \to R_3 - 4R_2$:

• Col 2: $4 - 4(1) = 0$
• Col 3 (RHS): $4 - 4(-0.5) = 4 + 2 = 6$

$$\left[\begin{array}{cc|c} 1 & -3 & 3 \ 0 & 1 & -0.5 \ 0 & 0 & 6 \end{array}\right]$$

The last row states $0 = 6$, which is impossible.

Verification of consistency using $R_3 - \frac{4}{14}R_2$ (before scaling): $$4 - \tfrac{4}{14}(-7) = 4 + 2 = 6 \neq 0$$

Same contradiction.

Conclusion: The system $Ax = b$ is inconsistent. There is no $x \in \mathbb{R}^2$ whose image under $T$ is $b = \begin{bmatrix} 3 \ 2 \ 1 \end{bmatrix}$.

Geometric reason: The columns of $A$ span at most a 2-dimensional subspace (plane) of $\mathbb{R}^3$. The vector $b$ does not lie in $\text{Col}(A)$, so $b$ is not in the range of $T$.

$$\boxed{\text{No solution: } b \text{ is not in the range of } T.}$$

State the conclusion: part (a) gives $T(v) = [-5, -1, 9]^T$, and the row reduction in part (b) reaches the contradiction $0 = 6$, so no such $x$ exists and the system is inconsistent.

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

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.

STEP 1 - EXTRACT

Given data:

$$\mathbf{u} = \begin{bmatrix} 1 \ -3 \ 2 \end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix} 1 \ -1 \ 3 \end{bmatrix}$$

Tasks:

1. State the condition for equality of two column vectors in $\mathbb{R}^2$, with an example.
2. Compute $\mathbf{u} + 3\mathbf{v}$ and $\mathbf{u} - 2\mathbf{v}$.

Note: The vectors given are actually in $\mathbb{R}^3$ (three components), though the theory question refers to $\mathbb{R}^2$. The equality definition generalizes; I answer the theory for $\mathbb{R}^2$ as asked and compute using the given $\mathbb{R}^3$ vectors.

STEP 2 - SOLVE

Part 1: Equality of two column vectors in $\mathbb{R}^2$

Two column vectors in $\mathbb{R}^2$ are equal if and only if their corresponding components are equal.

If $$\mathbf{u} = \begin{bmatrix} u_1 \ u_2 \end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix} v_1 \ v_2 \end{bmatrix}$$ then $$\mathbf{u} = \mathbf{v} \iff u_1 = v_1 \ \text{and}\ u_2 = v_2.$$

Example: $$\begin{bmatrix} 3 \ -2 \end{bmatrix} = \begin{bmatrix} 3 \ -2 \end{bmatrix} \quad \text{(equal, matching entries)}$$ but $$\begin{bmatrix} 3 \ -2 \end{bmatrix} \neq \begin{bmatrix} -2 \ 3 \end{bmatrix} \quad \text{(order matters, entries differ)}.$$

Part 2: Compute $\mathbf{u} + 3\mathbf{v}$

$$3\mathbf{v} = 3\begin{bmatrix} 1 \ -1 \ 3 \end{bmatrix} = \begin{bmatrix} 3 \ -3 \ 9 \end{bmatrix}$$

$$\mathbf{u} + 3\mathbf{v} = \begin{bmatrix} 1 \ -3 \ 2 \end{bmatrix} + \begin{bmatrix} 3 \ -3 \ 9 \end{bmatrix} = \begin{bmatrix} 4 \ -6 \ 11 \end{bmatrix}$$

Compute $\mathbf{u} - 2\mathbf{v}$

$$2\mathbf{v} = \begin{bmatrix} 2 \ -2 \ 6 \end{bmatrix}$$

$$\mathbf{u} - 2\mathbf{v} = \begin{bmatrix} 1 \ -3 \ 2 \end{bmatrix} - \begin{bmatrix} 2 \ -2 \ 6 \end{bmatrix} = \begin{bmatrix} -1 \ -1 \ -4 \end{bmatrix}$$

Final Results

$$\mathbf{u} + 3\mathbf{v} = \begin{bmatrix} 4 \ -6 \ 11 \end{bmatrix}, \qquad \mathbf{u} - 2\mathbf{v} = \begin{bmatrix} -1 \ -1 \ -4 \end{bmatrix}$$

Definition of Ring and Proof that (Z⁺, +, ×) is Not a Ring

Definition of Ring

An algebraic structure (R, +, ×) with two binary operations, addition (+) and multiplication (×), is called a ring if it satisfies the following conditions:

Proof: (Z⁺, +, ×) is NOT a Ring

Let Z⁺ = {1, 2, 3, 4, ...} be the set of positive integers.

To show that (Z⁺, +, ×) is not a ring, it is sufficient to show that at least one condition of a ring is violated.

Checking Condition 3: Existence of Additive Identity

For (Z⁺, +, ×) to be a ring, there must exist an element 0 ∈ Z⁺ such that:

a + 0 = 0 + a = a, for all a ∈ Z⁺

But 0 ∉ Z⁺, since Z⁺ = {1, 2, 3, ...} contains only positive integers.

Therefore, no additive identity exists in Z⁺.

Checking Condition 4: Existence of Additive Inverse

For (Z⁺, +, ×) to be a ring, for every a ∈ Z⁺, there must exist −a ∈ Z⁺ such that:

a + (−a) = 0

Example: Take a = 3 ∈ Z⁺. Then we need −3 ∈ Z⁺.

But −3 ∉ Z⁺, since Z⁺ contains no negative integers.

Therefore, additive inverses do not exist in Z⁺.

Conclusion

Since the set of positive integers Z⁺ fails to satisfy:

• the existence of an additive identity (0 ∉ Z⁺), and
• the existence of additive inverses (−a ∉ Z⁺ for any a ∈ Z⁺),

the algebraic structure (Z⁺, +, ×) is NOT a ring. $\blacksquare$

$$A = \begin{bmatrix} 1 & -5 & -7 \ -3 & 7 & 5 \end{bmatrix}, \quad u = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}, \quad b = \begin{bmatrix} -2 \ -2 \end{bmatrix}, \quad T(x) = Ax$$ --- $$T(u) = Au = \begin{bmatrix} 1 & -5 & -7 \ -3 &...

$$v1 = \begin{bmatrix} 1 \ 4 \ 0 \end{bmatrix},\quad v2 = \begin{bmatrix} 10 \ 2 \ 1 \end{bmatrix},\quad v3 = \begin{bmatrix} -5 \ 0 \ 6 \end{bmatrix}$$ The vectors are linearly independent iff the only solution to $c1 v1 + c2 v2 +...

$$A = \begin{bmatrix} 2 & -8 & 6 & 8 \ 3 & -9 & 5 & 10 \ -3 & 0 & 1 & -2 \ 1 & -4 & 0 & 6 \end{bmatrix}$$ Rules used: - Adding a multiple of one row to another does not change the determinant. - Factoring scalar $k$ out of a row means...

System of 3 equations in 3 unknowns: $$-2a - 3b + 4c = 5 \quad \cdots (1)$$ $$b - 2c = 4 \quad \cdots (2)$$ $$a + 3b - c = 2 \quad \cdots (3)$$ --- A linear equation in $n$ variables has the form: $$a1x1 + a2x2 + \cdots + anxn = b$$ wher...

$$A = \begin{bmatrix} 1 & 5 \ -3 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 4 & -5 \ 3 & k \end{bmatrix}$$ We must find $k$ (if any) so that $AB = BA$. --- $$AB = \begin{bmatrix} 1 & 5 \ -3 & 1 \end{bmatrix} \begin{bmatrix} 4 & -5 ...

$$A = \begin{bmatrix} 7 & 3 \ 3 & -1 \end{bmatrix}$$ --- $$\det(A - \lambda I) = 0$$ $$A - \lambda I = \begin{bmatrix} 7 - \lambda & 3 \ 3 & -1 - \lambda \end{bmatrix}$$ $$\det(A - \lambda I) = (7 - \lambda)(-1 - \lambda) - (3)(3)$$ Ex...

Homogeneous linear difference equation: $$y{k+2} - 4y{k+1} + 3yk = 0$$ Task: show the two fundamental solutions are linearly independent. Assume a solution of the form $yk = m^k$. Substituting: $$m^{k+2} - 4m^{k+1} + 3m^k = 0$$ Dividing ...