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Concept 03 / 09Matrices and Determinants

Determinants: Definition, Expansion and Geometric Meaning

The 2 x 2 and 3 x 3 determinants, expansion along any row or column with the checkerboard of signs, the determinant as a signed area or volume scale factor, and why det(AB) = det A det B while det(A + B) has no such rule.

Figure 1
To every square matrix AA we attach a number, its determinant, written det⁡A\det A or ∣A∣|A|. For a 2×22 \times 2 matrix, ∣abcd∣=ad−bc\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc. For a 3×33 \times 3 matrix, expand along the first row, multiplying each entry by the 2×22 \times 2 determinant left after deleting its row and column, with alternating signs: ∣a1b1c1a2b2c2a3b3c3∣=a1(b2c3−b3c2)−b1(a2c3−a3c2)+c1(a2b3−a3b2)\begin{vmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{vmatrix} = a_{1}(b_{2}c_{3} - b_{3}c_{2}) - b_{1}(a_{2}c_{3} - a_{3}c_{2}) + c_{1}(a_{2}b_{3} - a_{3}b_{2}). You may expand along any row or any column and get the same value, provided you use the checkerboard of signs: the sign attached to position (i,j)(i, j) is (−1)i+j(-1)^{i+j}, giving +,−,++, -, + in the first row, −,+,−-, +, - in the second, and +,−,++, -, + in the third. In practice, expand along the row or column with the most zeros. Minors and cofactors. The minor MijM_{ij} is the determinant obtained by deleting row ii and column jj. The cofactor is Cij=(−1)i+jMijC_{ij} = (-1)^{i+j}M_{ij}. Expansion along row ii reads det⁡A=∑jaijCij\det A = \sum_{j} a_{ij}C_{ij}. A striking fact used to derive the inverse: if you multiply the entries of one row by the cofactors of a different row, the sum is 00, because that sum is the determinant of a matrix with two equal rows. Geometric meaning. The two columns (a,c)(a, c) and (b,d)(b, d) of a 2×22 \times 2 matrix span a parallelogram, and ad−bcad - bc is its area with a sign: positive if the second column is anticlockwise from the first, negative if clockwise. The figure shows columns (4,1)(4, 1) and (1,3)(1, 3), with area 4⋅3−1⋅1=114 \cdot 3 - 1 \cdot 1 = 11. In three dimensions, the three columns span a parallelepiped, and det⁡A\det A is its signed volume, which is the scalar triple product of the columns. Viewed as a transformation, AA multiplies every area (or volume) by ∣det⁡A∣|\det A|, and a negative determinant means the transformation reverses orientation, like a reflection. This single idea explains the main facts. det⁡A=0\det A = 0 exactly when the columns are linearly dependent (parallel in 2D, coplanar in 3D): the transformation squashes space into a lower dimension, so no inverse can exist. det⁡(AB)=det⁡A⋅det⁡B\det(AB) = \det A \cdot \det B: scaling by one factor and then another multiplies the factors. Consequently det⁡(An)=(det⁡A)n\det(A^{n}) = (\det A)^{n} and det⁡(A−1)=1det⁡A\det(A^{-1}) = \frac{1}{\det A}. det⁡(kA)=kndet⁡A\det(kA) = k^{n}\det A for an n×nn \times n matrix, because each of the nn directions is stretched by kk. So det⁡(2A)=8det⁡A\det(2A) = 8\det A for 3×33 \times 3 matrices, not 2det⁡A2\det A. det⁡(A+B)≠det⁡A+det⁡B\det(A + B) \neq \det A + \det B in general. There is no addition rule, since sums of transformations do not correspond to anything simple about volumes. Triangular matrices. The determinant of an upper or lower triangular matrix (including a diagonal matrix) is the product of its diagonal entries. Expanding down the first column repeatedly shows this. It is also why row-reduction to triangular form is an efficient way to evaluate large determinants. Area of a triangle. With vertices (x1,y1),(x2,y2),(x3,y3)(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}), the area is 12∣∣x1y11x2y21x3y31∣∣\frac{1}{2}\left|\begin{vmatrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{vmatrix}\right|, and the three points are collinear exactly when this determinant is 00. When a question gives the area and asks for an unknown coordinate, remember the absolute value: the determinant equals ±2×\pm 2 \times area, so both signs must be considered. Key Takeaways: - 2×22 \times 2: ad−bcad - bc. 3×33 \times 3: expand along any row or column with signs (−1)i+j(-1)^{i+j}. - Cij=(−1)i+jMijC_{ij} = (-1)^{i+j}M_{ij}; row ×\times own cofactors =det⁡A= \det A; row ×\times another row's cofactors =0= 0. - ∣det⁡A∣|\det A| is the area or volume scale factor; its sign records orientation. - det⁡(AB)=det⁡Adet⁡B\det(AB) = \det A\det B; det⁡(AT)=det⁡A\det(A^{T}) = \det A; det⁡(kA)=kndet⁡A\det(kA) = k^{n}\det A. - det⁡(A+B)\det(A + B) has no general formula. - Triangular matrices: the determinant is the product of the diagonal entries. - Triangle area =12∣det⁡∣= \frac{1}{2}|\det|; both signs must be considered when solving for unknowns. - Pitfall: det⁡(3A)=3det⁡A\det(3A) = 3\det A for a 3×33 \times 3 matrix. The correct value is 27det⁡A27\det A. Brain-Teaser 1: AA is 3×33 \times 3 with det⁡A=4\det A = 4. Find det⁡(2A)\det(2A), det⁡(A2AT)\det(A^{2}A^{T}), det⁡(−A)\det(-A) and det⁡(A−1)\det(A^{-1}). Solution: det⁡(2A)=23⋅4=32\det(2A) = 2^{3} \cdot 4 = 32. det⁡(A2AT)=(det⁡A)2det⁡A=64\det(A^{2}A^{T}) = (\det A)^{2}\det A = 64. det⁡(−A)=(−1)3⋅4=−4\det(-A) = (-1)^{3} \cdot 4 = -4. det⁡(A−1)=14\det(A^{-1}) = \frac{1}{4}. Each follows from 'determinant = volume scale factor': doubling every side of a box multiplies its volume by 88, and reversing all three axes flips the orientation once overall (three flips). Brain-Teaser 2: Without expanding fully, show that ∣1aa21bb21cc2∣=(a−b)(b−c)(c−a)\begin{vmatrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{vmatrix} = (a - b)(b - c)(c - a). Solution: View the determinant DD as a polynomial in aa. If a=ba = b, two rows are equal and D=0D = 0, so (a−b)(a - b) is a factor. By the same argument, (b−c)(b - c) and (c−a)(c - a) are factors. DD has total degree 0+1+2=30 + 1 + 2 = 3 in a,b,ca, b, c, and so does the product of the three factors, so D=k(a−b)(b−c)(c−a)D = k(a - b)(b - c)(c - a) for a constant kk. To find kk, substitute convenient values: set a=0,b=1,c=2a = 0, b = 1, c = 2: D=∣100111124∣=4−2=2D = \begin{vmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \end{vmatrix} = 4 - 2 = 2 and (0−1)(1−2)(2−0)=2(0 - 1)(1 - 2)(2 - 0) = 2, so k=1k = 1. This is the Vandermonde determinant, and the 'equal rows give zero, so find the factors' method is the most efficient way to evaluate symmetric determinants. Brain-Teaser 3: The points (k,2−2k)(k, 2 - 2k), (−k+1,2k)(-k + 1, 2k) and (−4−k,6−2k)(-4 - k, 6 - 2k) are collinear. Find kk. Solution: Set the area determinant to zero: ∣k2−2k11−k2k1−4−k6−2k1∣=0\begin{vmatrix} k & 2 - 2k & 1 \\ 1 - k & 2k & 1 \\ -4 - k & 6 - 2k & 1 \end{vmatrix} = 0. Subtract row 1 from rows 2 and 3 to create zeros in the last column: row 2 becomes (1−2k,4k−2,0)(1 - 2k, 4k - 2, 0) and row 3 becomes (−4−2k,4,0)(-4 - 2k, 4, 0). Expanding along the last column, the determinant equals (1−2k)⋅4−(4k−2)(−4−2k)(1 - 2k) \cdot 4 - (4k - 2)(-4 - 2k). The second product is (4k−2)(−4−2k)=−16k−8k2+8+4k=−8k2−12k+8(4k - 2)(-4 - 2k) = -16k - 8k^{2} + 8 + 4k = -8k^{2} - 12k + 8. So the determinant is 4−8k+8k2+12k−8=8k2+4k−4=4(2k2+k−1)=4(2k−1)(k+1)4 - 8k + 8k^{2} + 12k - 8 = 8k^{2} + 4k - 4 = 4(2k^{2} + k - 1) = 4(2k - 1)(k + 1). Setting it to 00 gives k=12k = \frac{1}{2} or k=−1k = -1. Row operations before expanding keep the arithmetic manageable.