Linear algebra essentials
導入済み| English | 意味 |
|---|---|
| linear system/ˈlɪnɪə ˈsɪstəm/ | 連立一次方程式 |
| determinant/dɪˈtɜːmɪnənt/ | 行列式 |
| matrix inverse/ˈmeɪtrɪks ɪnˈvɜːs/ | The matrix that, multiplied by the original, gives the identity matrix; only non-singular square matrices have one. |
| singular/ˈsɪŋɡjʊlə/ | 特異 |
Three meshes, three equations
- Three coupled mesh currents give three equations in three unknowns. Clever substitution works once; it does not scale to the hundreds of equations a simulator solves.
- The structure behind every such system is a matrix. This lesson shows how to read one, solve it, and spot when it has no single answer.
A circuit as $Ax = b$
- A linear system 线性方程组 is written $Ax = b$: $A$ holds the coefficients, $x$ the unknowns, $b$ the known right-hand sides.
- For mesh analysis the diagonal of $A$ is each mesh's total resistance, the off-diagonals are the shared resistors (with a minus sign), $x$ the mesh currents and $b$ the source voltages. Coupling lives off the diagonal.
In a mesh equation Ax = b, match each part to its circuit meaning.
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Writing KVL for every mesh and collecting terms gives exactly this structure.
Elimination solves it
- Subtract a multiple of one equation from another to remove an unknown, then substitute back.
- $x + 2y = 8$ and $3x + y = 9$: subtract 3 × the first from the second to get $-5y = -15$, so $y = 3$ and $x = 2$.
Solve 2x + y = 7 and x − y = 2. What is x?
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Add the equations to remove $y$: $3x = 9$, so $x = 3$ and $y = 1$. Check: $2 \cdot 3 + 1 = 7$.
The determinant and the 2 × 2 inverse
- For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant 行列式 is $ad - bc$, and the matrix inverse 逆矩阵 is $A^{-1} = \dfrac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.
- Swap the diagonal, negate the off-diagonal, divide by the determinant. Then $x = A^{-1}b$.
For the system above, $\det A = 1 \cdot 1 - 2 \cdot 3 = -5$. So $x = \dfrac{1 \cdot 8 - 2 \cdot 9}{-5} = 2$ and $y = \dfrac{-3 \cdot 8 + 1 \cdot 9}{-5} = 3$. Elimination and the inverse agree; each checks the other.
What is the determinant of the matrix with rows (3, 2) and (4, 5)?
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$ad - bc = 3 \times 5 - 2 \times 4 = 7$. It is non-zero, so the system has one solution.
A two-mesh circuit as one matrix equation
- The two-mesh circuit from course 1 closes as $\begin{pmatrix} 6 & -4 \\ -4 & 10 \end{pmatrix}\begin{pmatrix} i_{A} \\ i_{B} \end{pmatrix} = \begin{pmatrix} 12 \\ -3 \end{pmatrix}$.
- $\det = 60 - 16 = 44$, so there is one solution: $i_{A} \approx 2.455\,\text{A}$ and $i_{B} \approx 0.682\,\text{A}$. Substituting back into both equations returns zero.
The mesh system with rows (6, −4) and (−4, 10) and right-hand side (12, −3) solves to
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$\det = 60 - 16 = 44$, so there is one solution; elimination gives 2.455 A and 0.682 A, and substitution confirms both equations. The −4 entries are the shared resistor.
When the determinant is zero
- $\det A = 0$ means the system is singular 奇异: the equations are dependent, and there is no single solution. In a circuit this usually means a floating node or a missing ground.
- $x + 2y = 8$ and $2x + 4y = 16$: the second is twice the first, so a whole line of points solves both. Change 16 to 15 and no point solves both.
How many solutions does the system x + 2y = 8, 2x + 4y = 16 have?
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$\det = 1 \cdot 4 - 2 \cdot 2 = 0$ and the equations agree, so a whole line of points solves both. With 15 instead of 16 there would be none.
A nodal analysis gives a coefficient determinant of zero. What does this most likely mean?
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A zero determinant means the equations are dependent, which in circuits usually means a node with no path to the reference. Fix the model, not the arithmetic.
Three common slips
- Rows out of step. Keep each row of $A$ with its own entry of $b$.
- "Non-zero determinant, so easy." It guarantees one solution, not a convenient one.
- "Zero determinant, so no solution." It may mean none or infinitely many; checking consistency decides which.
A system whose determinant is zero always has no solution.
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It has either no solution or infinitely many. Check whether the equations are consistent to decide which.
Recap
- $Ax = b$: diagonal = own terms, off-diagonal = coupling.
- Solve by elimination; check with $A^{-1} = \dfrac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ or by substitution.
- $\det A = 0$: singular, so no unique solution; in circuits, look for a floating node.