Determinants & Inverses

Study Sheet

Determinants & Inverses

One number that decides invertibility

The Determinant

Tip
2×2 and 3×3 by hand

det⁡(abcd)=ad−bc\det\begin{pmatrix}a&b\\c&d\end{pmatrix} = ad - bc; for 3×33\times3, expand along a row: each entry times the 2×22\times2 determinant of what remains, with alternating signs. Geometrically, ∣det⁡∣|\det| is the area (or volume) scale factor of the map.

Tip
Invertible ⇔ nonzero determinant

A−1A^{-1} exists exactly when det⁡A≠0\det A \ne 0, and for 2×22\times2: A−1=1det⁡A(d−b−ca)A^{-1} = \tfrac{1}{\det A}\begin{pmatrix} d & -b \\ -c & a\end{pmatrix} — swap the diagonal, negate the off-diagonal, divide by the determinant. Also det⁡(AB)=det⁡A det⁡B\det(AB) = \det A\,\det B, so det⁡(An)=(det⁡A)n\det(A^n) = (\det A)^n.

Example
Singular tuning

(243k)\begin{pmatrix}2&4\\3&k\end{pmatrix} fails to invert when 2k=122k = 12: at k=6k = 6 the rows become parallel and the matrix flattens the plane onto a line.

Side note
Significance: one number, many jobs

Jacobians in calculus are determinants (area scaling); Cramer's rule solves small systems by determinants; the characteristic polynomial that finds eigenvalues IS a determinant; and orientation (does a transformation flip space?) is its sign. It is the only scalar that summarizes an entire matrix's invertibility and volume behavior.

Try it
Try it: zero without computing

What is det⁡(123456579)\det\begin{pmatrix}1&2&3\\4&5&6\\5&7&9\end{pmatrix}? Work: row 3 == row 1 ++ row 2 — a dependence, so the determinant is 00: no expansion needed. Spotting structure beats grinding cofactors; check by expansion if unconvinced.

Proofs & Why It Matters

Tip
Proof: ad − bc is the area factor

The matrix sends the unit square (spanned by e1,e2\mathbf e_1, \mathbf e_2) to the parallelogram spanned by the columns (a,c)(a, c) and (b,d)(b, d).

Embed in 3D and take the cross product: (a,c,0)×(b,d,0)=(0,0,ad−bc)(a, c, 0)\times(b, d, 0) = (0, 0, ad - bc), so the parallelogram's area is ∣ad−bc∣|ad - bc| — the factor by which EVERY area scales, since any region is a limit of little squares. The sign records whether orientation flips. ■\blacksquare

Tip
Proof: the 2×2 inverse formula

Multiply directly: (abcd)(d−b−ca)=(ad−bc00ad−bc)=(det⁡A) I\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}d&-b\\-c&a\end{pmatrix} = \begin{pmatrix}ad-bc & 0\\ 0 & ad-bc\end{pmatrix} = (\det A)\,I.

When det⁡A≠0\det A \ne 0, dividing by it gives A−1A^{-1}; when det⁡A=0\det A = 0 the product is the zero matrix, so no inverse can exist (an invertible product would force det⁡A⋅det⁡A−1=1\det A \cdot \det A^{-1} = 1). ■\blacksquare

Tip
Proof: det(AB) = det A · det B (2×2, by areas)

The map x↦ABx\mathbf x \mapsto AB\mathbf x first scales areas by det⁡B\det B (applying BB), then by det⁡A\det A (applying AA) — so the composite scales areas by the product. Since the determinant IS the (signed) area factor, det⁡(AB)=det⁡A det⁡B\det(AB) = \det A\,\det B. Setting B=An−1B = A^{n-1} repeatedly gives det⁡(An)=(det⁡A)n\det(A^n) = (\det A)^n. ■\blacksquare

Going Deeper: Explanations & Worked Problems

Concept
Three faces of one number

The determinant answers three seemingly different questions with one number. ALGEBRA: Ax=bA\mathbf x = \mathbf b has a unique solution iff det⁡A≠0\det A \ne 0 — the determinant is the product of the pivots (up to sign from row swaps), so "no zero pivot" and "nonzero determinant" are the same statement.

GEOMETRY: ∣det⁡A∣|\det A| is the factor by which AA scales all areas (volumes in 3D), and its sign records orientation — negative means the plane got flipped. COMPUTATION: expand along any row or column (each entry times its minor, alternating signs +,−,+,…+,-,+,\ldots in a checkerboard), or better, row reduce and multiply pivots — for large matrices, elimination is exponentially faster than cofactors. Useful consequences worth having at your fingertips: a matrix with two equal rows has determinant 00 (swap them: the sign flips yet nothing changed); triangular matrices have determinant == product of the diagonal; and det⁡(AT)=det⁡A\det(A^{\mathsf T}) = \det A.

Example
Worked: a 3×3 determinant by cofactors, then again by elimination

Let M=(213041120)M = \begin{pmatrix}2&1&3\\0&4&1\\1&2&0\end{pmatrix}. COFACTORS along row 1: det⁡M=2det⁡(4120)−1det⁡(0110)+3det⁡(0412)\det M = 2\det\begin{pmatrix}4&1\\2&0\end{pmatrix} - 1\det\begin{pmatrix}0&1\\1&0\end{pmatrix} + 3\det\begin{pmatrix}0&4\\1&2\end{pmatrix}. Evaluate each 2×22\times2: (0−2)=−2(0 - 2) = -2; (0−1)=−1(0 - 1) = -1; (0−4)=−4(0 - 4) = -4.

Assemble with the +,−,++,-,+ signs: 2(−2)−1(−1)+3(−4)=−4+1−12=−152(-2) - 1(-1) + 3(-4) = -4 + 1 - 12 = -15. ELIMINATION: swap R1↔R3R_1 \leftrightarrow R_3 (determinant flips sign), then clear below: the pivots come out 1,4,1541, 4, \tfrac{15}{4}... product 1515, times the swap sign gives −15-15 ✓. Same number, two engines — cofactors for small or sparse matrices, elimination for everything else.

Example
Worked: using the inverse formula end to end

Solve (3152)x=(712)\begin{pmatrix}3&1\\5&2\end{pmatrix}\mathbf x = \begin{pmatrix}7\\12\end{pmatrix} by the inverse.

Step 1 — determinant: 3⋅2−1⋅5=13\cdot2 - 1\cdot5 = 1.

Step 2 — the 2×22\times2 inverse recipe (swap diagonal, negate off-diagonal, divide): A−1=11(2−1−53)A^{-1} = \tfrac11\begin{pmatrix}2&-1\\-5&3\end{pmatrix}.

Step 3 — multiply: x=(2−1−53)(712)=(14−12−35+36)=(21)\mathbf x = \begin{pmatrix}2&-1\\-5&3\end{pmatrix}\begin{pmatrix}7\\12\end{pmatrix} = \begin{pmatrix}14 - 12\\-35 + 36\end{pmatrix} = \begin{pmatrix}2\\1\end{pmatrix}.

Step 4 — check in the original system: 3⋅2+1=73\cdot2 + 1 = 7 ✓, 5⋅2+2=125\cdot2 + 2 = 12 ✓. Because det⁡=1\det = 1, this matrix even preserves areas exactly — an example of the geometry and the algebra agreeing to the digit.