Determinants & Inverses

Where $ad - bc$ comes from: $A$ sends the unit square to the parallelogram spanned by its columns $(a, c)$ and $(b, d)$, and that parallelogram's signed area is $ad - bc$ (the proof below cuts it out of a rectangle). For $A = \begin{pmatrix}3&1\\1&2\end{pmatrix}$ the columns are $(3, 1)$ and $(1, 2)$ and $\det A = 6 - 1 = 5$: the square of area $1$ becomes a parallelogram of area $5$. Since every region is filled by tiny squares, every area is multiplied by the same $5$.

The geometry explains the test. If $\det A = 0$, the square is flattened onto a line (or a point), and different inputs land on the same output — no map can undo that, so there is no inverse. If $\det A \ne 0$, the columns span the plane and every $\mathbf b$ is hit exactly once. The product rule is the area story twice: applying $B$ scales areas by $\det B$, then $A$ scales them by $\det A$.

Watch it happen through the columns $(2, 3)$ and $(4, k)$. At $k = 8$ the determinant is $16 - 12 = 4$, and the columns span a thin parallelogram of area $4$. As $k$ falls toward $6$ the second column swings onto the line through $(2, 3)$, the parallelogram's area $2k - 12$ shrinks to $0$, and at $k = 6$ the second column is exactly twice the first.

The third rule is the important one, and it is a shear. Since $\det A = \det A^{\mathsf T}$, rows and columns obey the same rules, so picture columns: replacing $\mathbf v$ by $\mathbf v + \mathbf u$ slides the top of the parallelogram along a line parallel to $\mathbf u$. Base and height are unchanged, so the area is too. Scaling one side by $c$ scales the area by $c$, and a swap reverses the order of the two sides, which reverses orientation. A triangular matrix's determinant is the product of its diagonal because expanding down the first column meets only one nonzero entry at each stage.

Now replacements only, which change nothing: $R_2 \gets R_2 - R_1$ gives $(0, 1, 3)$ and $R_3 \gets R_3 - 3R_1$ gives $(0, 1, 6)$; then $R_3 \gets R_3 - R_2$ gives $(0, 0, 3)$. The triangular matrix has diagonal $1, 1, 3$, so

Check by cofactors along row 1: $2(27 - 28) - 4(9 - 12) + 2(7 - 9) = -2 + 12 - 4 = 6$ ✓. Elimination took three row replacements; for an $n\times n$ matrix it needs about $\tfrac{n^3}{3}$ operations, while cofactor expansion needs about $n!$.

Clear column 1: $R_3 \gets R_3 - 4R_1$ gives $(0, -3, -4)$. Clear column 2: $R_3 \gets R_3 + 3R_2$ gives $(0, 0, 2)$. The pivots are $1, 1, 2$, with product $2$, so $\det = -2$. Check by cofactors along the first row of the original: $0\cdot(\ldots) - 1(8 - 12) + 2(-3 - 0) = 4 - 6 = -2$ ✓.

Each row scaling contributes a factor $3$, so $\det(3A) = 3^3\det A = 27\cdot(-2) = -54$. From $AA^{-1} = I$ and the product rule, $\det A\cdot\det(A^{-1}) = 1$, so $\det(A^{-1}) = -\tfrac12$. And $\det(A^{\mathsf T}A) = \det A^{\mathsf T}\det A = (-2)^2 = 4$ — always nonnegative, a fact that returns with least squares and singular values.

Why Cramer's rule holds, in the plane: $\mathbf b = x_1\mathbf a_1 + x_2\mathbf a_2$, so the parallelogram on $\mathbf b$ and $\mathbf a_2$ is the one on $x_1\mathbf a_1$ and $\mathbf a_2$ sheared along $\mathbf a_2$ — same area. Hence $\det(\mathbf b, \mathbf a_2) = x_1\det(\mathbf a_1, \mathbf a_2)$, which is $\det A_1 = x_1\det A$. The rule is elegant and good for one unknown or for proofs; for solving whole systems, elimination is far cheaper.

So $x = \tfrac{10}{5} = 2$. Replace column 2 instead: $\det A_2 = \det\begin{pmatrix}2&7\\1&11\end{pmatrix} = 22 - 7 = 15$, so $y = \tfrac{15}{5} = 3$. Check: $4 + 3 = 7$ and $2 + 9 = 11$ ✓. In the figure, the parallelogram on $\mathbf b$ and $\mathbf a_2$ (dashed) has area $10$, exactly twice the area $5$ of the one on $\mathbf a_1$ and $\mathbf a_2$ — that factor $2$ is $x$.

Replace column 3 by $\mathbf b = (6, -4, 27)$: $\det A_3 = \det\begin{pmatrix}1&1&6\\0&2&-4\\2&5&27\end{pmatrix} = 1(54 + 20) - 1(0 + 8) + 6(0 - 4) = 74 - 8 - 24 = 42$. So $z = \tfrac{42}{-21} = -2$. This is where Cramer's rule earns its keep: one unknown from two determinants, without solving for the others. (The full solution is $(5, 3, -2)$; check the second equation: $6 - 10 = -4$ ✓.)

They span a parallelogram of area $\left|\det\begin{pmatrix}3&1\\1&4\end{pmatrix}\right| = |12 - 1| = 11$, and the triangle is half of it: area $\tfrac{11}{2}$. The determinant is positive, which says the order $P \to Q \to R$ runs counterclockwise; listing the vertices the other way round would give $-11$. The formula needs no heights, no angles, and no square roots.

Embed in 3D and take the cross product: $(a, c, 0)\times(b, d, 0) = (0, 0, ad - bc)$, so the parallelogram's area is $|ad - bc|$ — the factor by which every area scales, since any region is a limit of little squares. The sign records whether orientation flips. $\blacksquare$ A cut-and-paste proof avoids the cross product. With columns $(a, c) = (4, 1)$ and $(b, d) = (1, 3)$, the parallelogram sits in a $5\times4$ rectangle of area $(a + b)(c + d) = 20$. Around it are two triangles of area $\tfrac12ac = 2$, two of area $\tfrac12bd = \tfrac32$, and two $b\times c$ rectangles of area $1$. What is left is $20 - 4 - 3 - 2 = 11 = 4\cdot3 - 1\cdot1$. In general $(a + b)(c + d) - ac - bd - 2bc = ad - bc$.

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

Geometry: $|\det A|$ is the factor by which $A$ 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 $0$ (swap them: the sign flips yet nothing changed); triangular matrices have determinant $=$ product of the diagonal; and $\det(A^{\mathsf T}) = \det A$.

Assemble with the $+, -, +$ signs: $2(-2) - 1(-1) + 3(-4) = -4 + 1 - 12 = -15$. Now by elimination: swap $R_1 \leftrightarrow R_3$ to get the pivot $1$ on top (the swap flips the sign), then $R_3 \gets R_3 - 2R_1$ gives $(0, -3, 3)$, and $R_3 \gets R_3 + \tfrac34R_2$ gives $(0, 0, \tfrac{15}{4})$. The pivots are $1, 4, \tfrac{15}{4}$ with product $15$, and the one swap makes it $-15$ ✓. Same number, two engines — cofactors for small or sparse matrices, elimination for everything else.

Step 1 — determinant: $3\cdot2 - 1\cdot5 = 1$.

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

Step 3 — multiply: $\mathbf x = \begin{pmatrix}2&-1\\-5&3\end{pmatrix}\begin{pmatrix}7\\12\end{pmatrix} = \begin{pmatrix}14 - 12\\-35 + 36\end{pmatrix} = \begin{pmatrix}2\\1\end{pmatrix}$.

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