Consistency of a System of Linear Equations

STEP 1 - EXTRACT: Given Data

System of equations: $$x - 2y = 5 \quad \cdots (1)$$ $$-x + y + 5z = 2 \quad \cdots (2)$$ $$y + z = 0 \quad \cdots (3)$$

Unknowns: $x, y, z$ (3 unknowns, 3 equations).

Note: equation (1) has no $z$ term (coefficient 0) and equation (3) has no $x$ term (coefficient 0).

Definitions

Consistent system: has at least one solution (either a unique solution or infinitely many solutions).

Inconsistent system: has no solution.

By the rank criterion, if $\text{rank}(A) = \text{rank}([A|b])$ the system is consistent; otherwise inconsistent.

Examples

Consistent (unique solution): $$x + y = 3, \qquad x - y = 1 \implies x = 2,\ y = 1.$$

Inconsistent (no solution): $$x + y = 3, \qquad x + y = 5 \quad \text{(contradiction)}.$$

STEP 2 - SOLVE

Augmented matrix

$$[A|b] = \begin{bmatrix} 1 & -2 & 0 & \big| & 5 \ -1 & 1 & 5 & \big| & 2 \ 0 & 1 & 1 & \big| & 0 \end{bmatrix}$$

Row reduction

$R_2 \leftarrow R_2 + R_1$:

$$\begin{bmatrix} 1 & -2 & 0 & \big| & 5 \ 0 & -1 & 5 & \big| & 7 \ 0 & 1 & 1 & \big| & 0 \end{bmatrix}$$

$R_3 \leftarrow R_3 + R_2$:

$$\begin{bmatrix} 1 & -2 & 0 & \big| & 5 \ 0 & -1 & 5 & \big| & 7 \ 0 & 0 & 6 & \big| & 7 \end{bmatrix}$$

Consistency check

The echelon form has 3 non-zero pivots and no row of the form $[0\ 0\ 0\ |\ k]$ with $k \neq 0$.

$$\text{rank}(A) = \text{rank}([A|b]) = 3 = \text{number of unknowns}$$

Hence the system is consistent with a unique solution.

Back substitution

Row 3: $$6z = 7 \implies z = \frac{7}{6}$$

Row 2: $$-y + 5z = 7 \implies -y + \frac{35}{6} = 7 \implies -y = \frac{42-35}{6} = \frac{7}{6} \implies y = -\frac{7}{6}$$

Row 1: $$x - 2y = 5 \implies x + \frac{14}{6} = 5 \implies x = \frac{30-14}{6} = \frac{16}{6} = \frac{8}{3}$$

Verification

• (1): $\frac{8}{3} - 2\left(-\frac{7}{6}\right) = \frac{8}{3} + \frac{7}{3} = 5$ ✓
• (2): $-\frac{8}{3} - \frac{7}{6} + \frac{35}{6} = -\frac{16}{6} + \frac{28}{6} = 2$ ✓
• (3): $-\frac{7}{6} + \frac{7}{6} = 0$ ✓

Final Answer

The system is consistent with a unique solution:

$$\boxed{x = \frac{8}{3}, \quad y = -\frac{7}{6}, \quad z = \frac{7}{6}}$$

Inverse of a Matrix

Given Data

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

Condition for Invertibility

A square matrix $A$ has an inverse if and only if it is non-singular, i.e. $\det(A) \neq 0$. Then:

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

where $\text{adj}(A)$ is the transpose of the cofactor matrix.

Step 1: Determinant

Expanding along the first row:

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

$$= 5(0+9) - 1(8-12) + 2(-3-0) = 45 + 4 - 6 = 43$$

Since $\det(A) = 43 \neq 0$, the inverse exists.

Step 2: Cofactors

$$C_{11} = +(0\cdot8 - 3\cdot(-3)) = 9$$ $$C_{12} = -(1\cdot8 - 3\cdot4) = -(-4) = 4$$ $$C_{13} = +(1\cdot(-3) - 0\cdot4) = -3$$ $$C_{21} = -(1\cdot8 - 2\cdot(-3)) = -(14) = -14$$ $$C_{22} = +(5\cdot8 - 2\cdot4) = 32$$ $$C_{23} = -(5\cdot(-3) - 1\cdot4) = -(-19) = 19$$ $$C_{31} = +(1\cdot3 - 2\cdot0) = 3$$ $$C_{32} = -(5\cdot3 - 2\cdot1) = -13$$ $$C_{33} = +(5\cdot0 - 1\cdot1) = -1$$

Step 3: Cofactor Matrix

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

Step 4: Adjugate (transpose of C)

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

Step 5: Inverse

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

$$\boxed{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 (spot check)

Row 1 of $A$ times column 1 of $A^{-1}$: $$5\cdot\tfrac{9}{43} + 1\cdot\tfrac{4}{43} + 2\cdot\tfrac{-3}{43} = \tfrac{45+4-6}{43} = \tfrac{43}{43} = 1 \checkmark$$

Row 2 times column 2: $$1\cdot\tfrac{-14}{43} + 0\cdot\tfrac{32}{43} + 3\cdot\tfrac{19}{43} = \tfrac{-14+57}{43} = \tfrac{43}{43} = 1 \checkmark$$

The result is confirmed.

Given Data

Linear transformation $T:\mathbb{R}^2 \to \mathbb{R}^2$: $$T(x) = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}\begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} -x_2 \ x_1 \end{bmatrix}$$

Vectors:

• $u = \begin{bmatrix} 4 \ 1 \end{bmatrix}$
• $v = \begin{bmatrix} 2 \ 3 \end{bmatrix}$
• $u+v = \begin{bmatrix} 6 \ 4 \end{bmatrix}$

Step 1: Image of $u$

With $x_1 = 4$, $x_2 = 1$: $$T(u) = \begin{bmatrix} -x_2 \ x_1 \end{bmatrix} = \begin{bmatrix} -1 \ 4 \end{bmatrix}$$

Step 2: Image of $v$

With $x_1 = 2$, $x_2 = 3$: $$T(v) = \begin{bmatrix} -3 \ 2 \end{bmatrix}$$

Step 3: Image of $u+v$

With $x_1 = 6$, $x_2 = 4$: $$T(u+v) = \begin{bmatrix} -4 \ 6 \end{bmatrix}$$

Verification (Linearity)

$$T(u) + T(v) = \begin{bmatrix} -1 \ 4 \end{bmatrix} + \begin{bmatrix} -3 \ 2 \end{bmatrix} = \begin{bmatrix} -4 \ 6 \end{bmatrix} = T(u+v)\ \checkmark$$

Summary

This matrix effects a $90^\circ$ counterclockwise rotation.

Determinant: Definition and Computation

Given Data

$$A = \begin{bmatrix} -2 & 8 & -9 \ -1 & 7 & 0 \ 1 & -4 & 2 \end{bmatrix}$$

Definition

A determinant is a scalar value associated with every square matrix, obtained from its entries via a well-defined rule. It is denoted $\det(A)$ or $|A|$. Geometrically it represents the scaling factor of the linear transformation described by the matrix (signed volume). A matrix is invertible iff its determinant is nonzero.

Computation Without Expanding (Row Reduction to Triangular Form)

Rules used:

• Adding a multiple of one row to another does not change the determinant.
• Swapping two rows multiplies the determinant by $-1$.
• The determinant of a triangular matrix is the product of its diagonal entries.

Step 1: Swap $R_1 \leftrightarrow R_3$ (to get pivot 1)

$$B = \begin{bmatrix} 1 & -4 & 2 \ -1 & 7 & 0 \ -2 & 8 & -9 \end{bmatrix}, \qquad \det(A) = -\det(B)$$

Step 2: Eliminate below pivot in column 1

$R_2 \leftarrow R_2 + R_1$:

$$\begin{bmatrix} 1 & -4 & 2 \ 0 & 3 & 2 \ -2 & 8 & -9 \end{bmatrix}$$

$R_3 \leftarrow R_3 + 2R_1$:

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

Column 2 already has a zero below the second pivot, so the matrix is upper triangular. These row replacements do not change the determinant.

Step 3: Product of diagonal entries

$$\det(B) = 1 \times 3 \times (-5) = -15$$

Step 4: Restore the row swap

$$\det(A) = -\det(B) = -(-15) = \boxed{15}$$

Verification (direct expansion along Row 1)

$$\det(A) = -2,(7\cdot 2 - 0\cdot(-4)) - 8,((-1)\cdot 2 - 0\cdot 1) + (-9),((-1)(-4) - 7\cdot 1)$$ $$= -2(14) - 8(-2) - 9(4 - 7)$$ $$= -28 + 16 - 9(-3) = -28 + 16 + 27 = 15 \checkmark$$

Final answer: $\det(A) = 15$

QR Factorization of $A = \begin{bmatrix} 2 & 1 \ 3 & -1 \end{bmatrix}$

Step 1 - EXTRACT: Given data

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

Columns: $$\mathbf{a_1} = \begin{bmatrix} 2 \ 3 \end{bmatrix}, \qquad \mathbf{a_2} = \begin{bmatrix} 1 \ -1 \end{bmatrix}$$

Goal: find orthogonal $Q$ and upper-triangular $R$ with $A = QR$.

Step 2 - SOLVE (Gram-Schmidt)

First orthogonal vector

$$\mathbf{u_1} = \mathbf{a_1} = \begin{bmatrix} 2 \ 3 \end{bmatrix}, \qquad |\mathbf{u_1}| = \sqrt{4+9} = \sqrt{13}$$

$$\mathbf{q_1} = \frac{1}{\sqrt{13}}\begin{bmatrix} 2 \ 3 \end{bmatrix}$$

Second orthogonal vector

$$\mathbf{a_2}\cdot\mathbf{u_1} = (1)(2)+(-1)(3) = -1, \qquad \mathbf{u_1}\cdot\mathbf{u_1} = 13$$

$$\mathbf{u_2} = \mathbf{a_2} - \frac{-1}{13}\mathbf{u_1} = \begin{bmatrix} 1 \ -1 \end{bmatrix} + \frac{1}{13}\begin{bmatrix} 2 \ 3 \end{bmatrix} = \begin{bmatrix} \tfrac{15}{13} \ -\tfrac{10}{13} \end{bmatrix}$$

mth168-gram-schmidt-qr-2x2

$$|\mathbf{u_2}| = \frac{1}{13}\sqrt{225+100} = \frac{\sqrt{325}}{13} = \frac{5\sqrt{13}}{13} = \frac{5}{\sqrt{13}}$$

$$\mathbf{q_2} = \frac{\mathbf{u_2}}{|\mathbf{u_2}|} = \frac{1}{\sqrt{13}}\begin{bmatrix} 3 \ -2 \end{bmatrix}$$

Matrix Q

$$Q = \frac{1}{\sqrt{13}}\begin{bmatrix} 2 & 3 \ 3 & -2 \end{bmatrix}$$

Matrix R

$$R = Q^{T}A = \frac{1}{\sqrt{13}}\begin{bmatrix} 2 & 3 \ 3 & -2 \end{bmatrix}\begin{bmatrix} 2 & 1 \ 3 & -1 \end{bmatrix}$$

Compute the entries:

• $(1,1): \frac{1}{\sqrt{13}}(2\cdot2 + 3\cdot3) = \frac{13}{\sqrt{13}} = \sqrt{13}$
• $(1,2): \frac{1}{\sqrt{13}}(2\cdot1 + 3\cdot(-1)) = \frac{-1}{\sqrt{13}}$
• $(2,1): \frac{1}{\sqrt{13}}(3\cdot2 + (-2)\cdot3) = 0$
• $(2,2): \frac{1}{\sqrt{13}}(3\cdot1 + (-2)(-1)) = \frac{5}{\sqrt{13}}$

$$\boxed{R = \begin{bmatrix} \sqrt{13} & -\dfrac{1}{\sqrt{13}} \[2mm] 0 & \dfrac{5}{\sqrt{13}} \end{bmatrix}}$$

Final Answer

$$Q = \frac{1}{\sqrt{13}}\begin{bmatrix} 2 & 3 \ 3 & -2 \end{bmatrix}, \qquad R = \begin{bmatrix} \sqrt{13} & -\tfrac{1}{\sqrt{13}} \ 0 & \tfrac{5}{\sqrt{13}} \end{bmatrix}$$

Verification

$$QR = \frac{1}{\sqrt{13}}\begin{bmatrix} 2 & 3 \ 3 & -2 \end{bmatrix}\cdot\frac{1}{\sqrt{13}}\begin{bmatrix} 13 & -1 \ 0 & 5 \end{bmatrix} = \frac{1}{13}\begin{bmatrix} 26 & 13 \ 39 & -13 \end{bmatrix} = \begin{bmatrix} 2 & 1 \ 3 & -1 \end{bmatrix} = A \checkmark$$

The completed values are $R_{11}=\sqrt{13}$, $R_{12}=-1/\sqrt{13}$, $R_{21}=0$ and $R_{22}=5/\sqrt{13}$.