Consistency of a System of Linear Equations

Given Data

System of equations: $$x + y + z = 4 \quad \cdots (1)$$ $$x + 2y + 2z = 2 \quad \cdots (2)$$ $$2x + 2y + z = 5 \quad \cdots (3)$$

---

Definitions

Consistent System: A system of linear equations is consistent if it has at least one solution (either a unique solution or infinitely many solutions). By the rank criterion: $$\text{rank}(A) = \text{rank}([A|b]) \implies \text{consistent}$$

Inconsistent System: A system is inconsistent if it has no solution: $$\text{rank}(A) \neq \text{rank}([A|b]) \implies \text{inconsistent}$$

---

Examples

Consistent Example: $$x + y = 3, \quad x - y = 1$$ Solving gives $x = 2, ; y = 1$ (unique solution): consistent.

Inconsistent Example: $$x + y = 3, \quad x + y = 5$$ Contradictory equations, so no solution exists: inconsistent.

---

Testing Consistency and Solving

Step 1: Augmented Matrix

$$[A|b] = \begin{bmatrix} 1 & 1 & 1 & | & 4 \ 1 & 2 & 2 & | & 2 \ 2 & 2 & 1 & | & 5 \end{bmatrix}$$

Step 2: Row Reduction

$R_2 \leftarrow R_2 - R_1$: $$\begin{bmatrix} 1 & 1 & 1 & | & 4 \ 0 & 1 & 1 & | & -2 \ 2 & 2 & 1 & | & 5 \end{bmatrix}$$

$R_3 \leftarrow R_3 - 2R_1$: $$\begin{bmatrix} 1 & 1 & 1 & | & 4 \ 0 & 1 & 1 & | & -2 \ 0 & 0 & -1 & | & -3 \end{bmatrix}$$

Step 3: Consistency Check

The matrix is in echelon form with 3 non-zero rows: $$\text{rank}(A) = 3, \quad \text{rank}([A|b]) = 3, \quad \text{number of unknowns} = 3$$

Since $\text{rank}(A) = \text{rank}([A|b]) = 3 = $ number of unknowns, the system is consistent with a unique solution. No row of the form $[0;0;0,|,k],\ k \neq 0$ appears.

Step 4: Back Substitution

From Row 3: $$-z = -3 \implies z = 3$$

From Row 2: $$y + z = -2 \implies y + 3 = -2 \implies y = -5$$

From Row 1: $$x + y + z = 4 \implies x - 5 + 3 = 4 \implies x = 6$$

Step 5: Verification

---

Final Answer

The system is consistent with the unique solution: $$\boxed{x = 6, \quad y = -5, \quad z = 3}$$

Inverse of a Matrix

STEP 1 - EXTRACT: Given Data

$$A = \begin{bmatrix} 5 & 1 & 2 \ 1 & 0 & 3 \ 4 & -3 & 8 \end{bmatrix}$$

Required: (a) condition for existence of inverse, (b) find $A^{-1}$.

STEP 2 - SOLVE

Condition for a Matrix to Have an Inverse

A square matrix $A$ of order $n \times n$ has an inverse if and only if it is non-singular, that is:

$$\det(A) \neq 0$$

If $\det(A) = 0$, $A$ is singular and $A^{-1}$ does not exist. The inverse is given by:

$$A^{-1} = \frac{1}{\det(A)} , \text{adj}(A)$$

---

Step 1: Determinant

Expanding along the first row:

$$\det(A) = 5\begin{vmatrix} 0 & 3 \ -3 & 8 \end{vmatrix} - 1\begin{vmatrix} 1 & 3 \ 4 & 8 \end{vmatrix} + 2\begin{vmatrix} 1 & 0 \ 4 & -3 \end{vmatrix}$$

$$= 5(0 + 9) - 1(8 - 12) + 2(-3 - 0)$$

$$= 45 + 4 - 6 = 43$$

$$\det(A) = 43 \neq 0 \implies A^{-1} \text{ exists.}$$

---

Step 2: Cofactors

$$C_{11} = +\begin{vmatrix} 0 & 3 \ -3 & 8 \end{vmatrix} = 9$$ $$C_{12} = -\begin{vmatrix} 1 & 3 \ 4 & 8 \end{vmatrix} = -(8-12) = 4$$ $$C_{13} = +\begin{vmatrix} 1 & 0 \ 4 & -3 \end{vmatrix} = -3$$

$$C_{21} = -\begin{vmatrix} 1 & 2 \ -3 & 8 \end{vmatrix} = -(8+6) = -14$$ $$C_{22} = +\begin{vmatrix} 5 & 2 \ 4 & 8 \end{vmatrix} = 40-8 = 32$$ $$C_{23} = -\begin{vmatrix} 5 & 1 \ 4 & -3 \end{vmatrix} = -(-15-4) = 19$$

$$C_{31} = +\begin{vmatrix} 1 & 2 \ 0 & 3 \end{vmatrix} = 3$$ $$C_{32} = -\begin{vmatrix} 5 & 2 \ 1 & 3 \end{vmatrix} = -(15-2) = -13$$ $$C_{33} = +\begin{vmatrix} 5 & 1 \ 1 & 0 \end{vmatrix} = -1$$

---

Step 3: Cofactor Matrix

$$C = \begin{bmatrix} 9 & 4 & -3 \ -14 & 32 & 19 \ 3 & -13 & -1 \end{bmatrix}$$

Step 4: Adjoint (transpose of C)

$$\text{adj}(A) = \begin{bmatrix} 9 & -14 & 3 \ 4 & 32 & -13 \ -3 & 19 & -1 \end{bmatrix}$$

Step 5: Inverse

$$A^{-1} = \frac{1}{43}\begin{bmatrix} 9 & -14 & 3 \ 4 & 32 & -13 \ -3 & 19 & -1 \end{bmatrix}$$

$$A^{-1} = \begin{bmatrix} \dfrac{9}{43} & \dfrac{-14}{43} & \dfrac{3}{43} \[6pt] \dfrac{4}{43} & \dfrac{32}{43} & \dfrac{-13}{43} \[6pt] \dfrac{-3}{43} & \dfrac{19}{43} & \dfrac{-1}{43} \end{bmatrix}$$

---

Verification (first entry of $A A^{-1}$)

Row 1 of $A$ times column 1 of $\text{adj}(A)$: $$5(9) + 1(4) + 2(-3) = 45 + 4 - 6 = 43 = \det(A) ;\checkmark$$

Dividing by 43 gives $1$, confirming the diagonal entry. The inverse is correct.

Dimensions of Null Space and Column Space of A

Step 1 - Given data

$$A = \begin{bmatrix} -3 & 6 & -1 & 1 & -7 \ 1 & -2 & 2 & 3 & -1 \ 2 & -4 & 5 & 8 & -4 \end{bmatrix}, \quad 3 \times 5 \text{ matrix}$$

Number of columns $n = 5$.

Step 2 - Row reduce to echelon form

Swap $R_1 \leftrightarrow R_2$:

$$\begin{bmatrix} 1 & -2 & 2 & 3 & -1 \ -3 & 6 & -1 & 1 & -7 \ 2 & -4 & 5 & 8 & -4 \end{bmatrix}$$

$R_2 \to R_2 + 3R_1$: $(-3+3, 6-6, -1+6, 1+9, -7-3) = (0,0,5,10,-10)$

$R_3 \to R_3 - 2R_1$: $(2-2, -4+4, 5-4, 8-6, -4+2) = (0,0,1,2,-2)$

$$\begin{bmatrix} 1 & -2 & 2 & 3 & -1 \ 0 & 0 & 5 & 10 & -10 \ 0 & 0 & 1 & 2 & -2 \end{bmatrix}$$

$R_2 \to \tfrac{1}{5}R_2$: gives $(0,0,1,2,-2)$

$R_3 \to R_3 - R_2$: gives $(0,0,0,0,0)$

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

Step 3 - Pivots and free variables

Pivots appear in columns 1 and 3.

• Rank $= 2$
• Free variables: $x_2, x_4, x_5$ (3 free variables)

Step 4 - Dimension of column space

$$\dim(\text{Col } A) = \text{rank}(A) = 2$$

Basis (pivot columns of original $A$):

$$\left{ \begin{bmatrix} -3 \ 1 \ 2 \end{bmatrix},\ \begin{bmatrix} -1 \ 2 \ 5 \end{bmatrix} \right}$$

Step 5 - Dimension of null space (Rank-Nullity)

$$\dim(\text{Nul } A) = n - \text{rank}(A) = 5 - 2 = 3$$

Final Answer

Check: $2 + 3 = 5 = n$ ✓

Eigenvalues of the Matrix

Step 1 - Extract: Given Data

$$A = \begin{bmatrix} 6 & 3 & -8 \ 0 & -2 & 0 \ 1 & 0 & -3 \end{bmatrix}$$

Step 2 - Solve

Eigenvalues satisfy $\det(A - \lambda I) = 0$.

Form $(A - \lambda I)$

$$A - \lambda I = \begin{bmatrix} 6-\lambda & 3 & -8 \ 0 & -2-\lambda & 0 \ 1 & 0 & -3-\lambda \end{bmatrix}$$

Determinant (expand along row 2)

Row 2 has entries $0,\ (-2-\lambda),\ 0$. Only the middle term survives, with cofactor sign $(-1)^{2+2}=+1$:

$$\det(A - \lambda I) = (-2-\lambda)\det\begin{bmatrix} 6-\lambda & -8 \ 1 & -3-\lambda \end{bmatrix}$$

Expand the 2x2 minor

$$(6-\lambda)(-3-\lambda) - (-8)(1)$$

Compute $(6-\lambda)(-3-\lambda)$:

$$= -18 - 6\lambda + 3\lambda + \lambda^2 = \lambda^2 - 3\lambda - 18$$

Add $8$:

$$\lambda^2 - 3\lambda - 18 + 8 = \lambda^2 - 3\lambda - 10 = (\lambda - 5)(\lambda + 2)$$

Full characteristic equation

$$\det(A - \lambda I) = (-2-\lambda)(\lambda - 5)(\lambda + 2) = 0$$

Since $-2 - \lambda = -(\lambda + 2)$:

$$-(\lambda + 2)^2(\lambda - 5) = 0$$

Solve

$$(\lambda + 2)^2 (\lambda - 5) = 0$$

Result

$$\boxed{\lambda = -2 \text{ (repeated)}, \quad \lambda = 5}$$

Check (trace): sum of eigenvalues $= -2 + (-2) + 5 = 1$; trace of $A = 6 + (-2) + (-3) = 1$. ✓

Check (determinant): product $= (-2)(-2)(5) = 20$; $\det(A) = -2\cdot\det\begin{bmatrix}6&-8\1&-3\end{bmatrix} = -2(-18+8) = -2(-10)=20$. ✓

LU Factorization

Step 1 - Given Data

$$A = \begin{bmatrix} 2 & 5 \ 6 & -7 \end{bmatrix}$$

Goal: find $L$ (lower triangular, unit diagonal) and $U$ (upper triangular) such that $A = LU$.

---

Step 2 - Row Reduction to Get U

Pivot (1,1) = 2. Eliminate entry $a_{21} = 6$.

Multiplier: $$\ell_{21} = \frac{6}{2} = 3$$

Row operation: $R_2 \rightarrow R_2 - 3R_1$

$$\begin{bmatrix} 2 & 5 \ 6 - 3(2) & -7 - 3(5) \end{bmatrix} = \begin{bmatrix} 2 & 5 \ 0 & -22 \end{bmatrix}$$

So: $$U = \begin{bmatrix} 2 & 5 \ 0 & -22 \end{bmatrix}$$

---

Step 3 - Construct L

Unit diagonal, with the multiplier $\ell_{21} = 3$ below:

$$L = \begin{bmatrix} 1 & 0 \ 3 & 1 \end{bmatrix}$$

---

Step 4 - Verify $LU = A$

$$ LU = \begin{bmatrix} 1 & 0 \ 3 & 1 \end{bmatrix}\begin{bmatrix} 2 & 5 \ 0 & -22 \end{bmatrix} = \begin{bmatrix} 2 & 5 \ 6 & 15-22 \end{bmatrix} = \begin{bmatrix} 2 & 5 \ 6 & -7 \end{bmatrix} = A \checkmark $$

---

Result

$$\boxed{A = \begin{bmatrix} 1 & 0 \ 3 & 1 \end{bmatrix}\begin{bmatrix} 2 & 5 \ 0 & -22 \end{bmatrix}}$$