To every square matrix $A = [a_{ij}]$ of order $n$, we can associate a unique number (real or complex) called the determinant of the square matrix $A$. This function maps the set of square matrices $M$ to the set of numbers $K$ via $f(A) = k$, where $k = |A|$ or $\text{det } A$. Only square matrices have determinants.
Mathematical Problem Statement: Evaluate the determinant $\Delta = \begin{vmatrix} x & x+1 \\ x-1 & x \end{vmatrix}$.
Step-by-Step Algorithmic Solution Trace:
- Apply the $2 \times 2$ determinant formula: $\Delta = (x)(x) - (x+1)(x-1)$.
- Expand the algebraic terms: $\Delta = x^2 - (x^2 - 1)$.
- Simplify the expression: $\Delta = x^2 - x^2 + 1 = 1$.
Final Evaluation Output: $\Delta = 1$
The determinant of a $3 \times 3$ matrix is evaluated by expanding it along any one of its three rows or three columns, reducing it to a linear combination of $2 \times 2$ determinants. The sign of each term is determined by the factor $(-1)^{i+j}$, where $i$ is the row index and $j$ is the column index.
Expansion along the first row ($R_1$):
$$|A| = a_{11}(-1)^{1+1}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} + a_{12}(-1)^{1+2}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13}(-1)^{1+3}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}$$ $$|A| = a_{11}(a_{22}a_{33} - a_{32}a_{23}) - a_{12}(a_{21}a_{33} - a_{31}a_{23}) + a_{13}(a_{21}a_{32} - a_{31}a_{22})$$| Determinant Order | Exact Mathematical Domain / Sign Properties | Active Operation Formulas / Identities | Systemic Role & Evaluation Limits |
|---|---|---|---|
| Order $1 \times 1$ | Real or Complex scalar entries | |A| = a_{11} | Direct mapping of single element. |
| Order $2 \times 2$ | Real or Complex entries | |A| = a_{11}a_{22} - a_{12}a_{21} | Determines uniqueness of $2 \times 2$ linear systems. |
| Order $3 \times 3$ | Real or Complex entries; sign matrix: $\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$ | $|A| = \sum_{j=1}^3 (-1)^{i+j} a_{ij} M_{ij}$ (along row $i$) | Evaluated along the row/column containing the maximum number of zeros for computational efficiency. |
The area of a triangle whose vertices are $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ can be expressed compactly as a determinant:
$$\Delta = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$$Mathematical Problem Statement: Find the equation of the line joining $A(1, 3)$ and $B(0, 0)$ using determinants, and find $k$ if $D(k, 0)$ is a point such that the area of triangle $ABD$ is $3$ square units.
Step-by-Step Algorithmic Solution Trace:
- Let $P(x, y)$ be any point on the line $AB$. Since $A, B, P$ are collinear, the area of triangle $ABP$ is zero: $$\frac{1}{2} \begin{vmatrix} x & y & 1 \\ 1 & 3 & 1 \\ 0 & 0 & 1 \end{vmatrix} = 0$$
- Expand along $R_3$: $1(3x - y) = 0 \implies y = 3x$. This is the equation of line $AB$.
- For point $D(k, 0)$, set the area of triangle $ABD$ to $\pm 3$: $$\frac{1}{2} \begin{vmatrix} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{vmatrix} = \pm 3$$
- Expand along $R_2$: $-\frac{1}{2}(1)(0 - 3k) = \pm 3 \implies \frac{3k}{2} = \pm 3 \implies k = \pm 2$.
Final Evaluation Output: Line equation: $y = 3x$; Parameter value: $k = \pm 2$
The adjoint of a square matrix $A = [a_{ij}]_{n \times n}$ is defined as the transpose of the cofactor matrix $[A_{ij}]_{n \times n}$, denoted by $\text{adj } A$.
$$text{adj } A = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}$$Shortcut for $2 \times 2$ matrices: Interchange the diagonal elements $a_{11}$ and $a_{22}$, and change the signs of the off-diagonal elements $a_{12}$ and $a_{21}$.
Determinants and matrices are used to solve systems of linear equations in two or three variables and to check their consistency.
Consider the system: $a_1x + b_1y + c_1z = d_1$, $a_2x + b_2y + c_2z = d_2$, $a_3x + b_3y + c_3z = d_3$. This can be written as $AX = B$, where:
$$A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}$$Mathematical Problem Statement: Solve the system of equations using the matrix method: $2x + 5y = 1$, $3x + 2y = 7$.
Step-by-Step Algorithmic Solution Trace:
- Write the system as $AX = B$: $$\begin{bmatrix} 2 & 5 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 7 \end{bmatrix}$$
- Compute $|A| = (2)(2) - (5)(3) = 4 - 15 = -11$. Since $|A| \neq 0$, a unique solution exists.
- Find $\text{adj } A$ by interchanging diagonal elements and changing signs of off-diagonals: $$\text{adj } A = \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}$$
- Calculate $A^{-1} = \frac{1}{|A|} \text{adj } A = -\frac{1}{11} \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}$.
- Solve for $X = A^{-1}B$: $$X = -\frac{1}{11} \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 7 \end{bmatrix} = -\frac{1}{11} \begin{bmatrix} 2(1) - 5(7) \\ -3(1) + 2(7) \end{bmatrix} = -\frac{1}{11} \begin{bmatrix} -33 \\ 11 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$$
Final Evaluation Output: $x = 3, y = -1$
Historical Chronicle: Evolution of Determinants