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Topos Uranos
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Determinants: properties and Cramer’s rule

Summary

This lesson extends to square matrices of any order the number that, in the lesson on linear systems of the course Algebra and Geometry I, decided whether a system of two equations in two unknowns has a unique solution. It interprets the determinant of order two as the signed area of the parallelogram spanned by the columns of the matrix; it defines the determinant of order three by expansion along the first row and shows that it agrees with the rule of Sarrus; and it defines the determinant of order nn recursively. It proves by induction on the order, starting from that definition, the properties of the determinant (linearity in each row, the interchange of rows, zero or equal rows, adding a multiple of another row, the transpose and triangular matrices), and applies them to computation by reduction and to the relation between the determinant and the rank. It proves the product theorem for determinants in order two, whose general proof is given in the lesson on the inverse matrix, and Cramer’s rule in every order; the problems apply determinants to the area of a triangle, to the Vandermonde determinant, to Cassini’s identity for the Fibonacci numbers and to the circle through three points.

Learning objectives

  1. Compute determinants of order two and three by expansion along a row and by the rule of Sarrus, and interpret the determinant of order two as a signed area.
  2. Prove by induction on the order the properties of the determinant and use them to compute determinants of any order by reduction to a triangular matrix.
  3. Relate the determinant to the rank and to the matrix product, and use it to decide whether a matrix is singular.
  4. Solve systems of nn equations in nn unknowns by Cramer’s rule, and apply determinants to problems of analytic geometry and of sequences.

The determinant of order two and the signed area

In the Course I lesson on first-degree equations and linear systems, the number D=a1b2−a2b1D = a_1 b_2 - a_2 b_1 was called the determinant of a system of two equations, and Cramer’s rule was proved: if DD is not zero, the system has exactly one solution. In the lesson on vectors in the plane, the same number, formed with the components of two vectors, decided whether they are parallel. Now, in the lesson on matrices and matrix operations square matrices were introduced, and in the lesson on the Gauss-Jordan method every linear system was written in the form AX=BAX = B. If this is so, it is natural to regard that number as a function of the coefficient matrix, and to extend it to square matrices of any order.

The determinant of a square matrix AA of order two is the product of the entries on its main diagonal minus the product of the entries on the other diagonal. It is written det⁡A\det A, or with the entries of the matrix between two vertical bars.
A=(abcd),det⁡A=∣abcd∣=ad−bcA = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \qquad \det A = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc

The determinant of a system of two equations is that of its coefficient matrix, which has in the first column the coefficients of xx and in the second those of yy: its determinant is a1b2−a2b1=Da_1 b_2 - a_2 b_1 = D. The bars do not denote an absolute value here, because the determinant can be negative; for example:

∣3112∣=6−1=5,∣1321∣=1−6=−5\begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} = 6 - 1 = 5, \qquad \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = 1 - 6 = -5

The two columns of a matrix of order two are vectors of the plane, and the signed area theorem, stated below, gives the determinant a geometric meaning that also explains the parallelism criterion of the Course I lesson on vectors.

TheoremSigned area

Let u⃗=(u1,u2)\vec{u} = \left( u_1, u_2 \right) and v⃗=(v1,v2)\vec{v} = \left( v_1, v_2 \right) be vectors of the plane, and let AA be the matrix whose columns are u⃗\vec{u} and v⃗\vec{v}. The area of the parallelogram spanned by the arrows of u⃗\vec{u} and of v⃗\vec{v} drawn from a common origin is the absolute value of det⁡A\det A. Moreover, if the vectors are neither zero nor parallel, det⁡A\det A is positive when the rotation of less than a half turn that carries the direction of u⃗\vec{u} to that of v⃗\vec{v} is counterclockwise, and negative when it is clockwise.

det⁡A=∣u1v1u2v2∣=u1v2−u2v1\det A = \begin{vmatrix} u_1 & v_1 \\ u_2 & v_2 \end{vmatrix} = u_1 v_2 - u_2 v_1

If u⃗\vec{u} is zero, the parallelogram degenerates into a line segment and the determinant is zero; we therefore assume that u⃗\vec{u} is not. The proof uses the distance from a point to a line, obtained in the Course I lesson on vectors, the polar coordinates of the lesson on polar coordinates and the formula for the sine of a difference from the lesson on the sum and difference formulas. The figure follows it with u⃗=(3,1)\vec{u} = \left( 3, 1 \right) and v⃗=(1,2)\vec{v} = \left( 1, 2 \right).

Proof

  1. n⃗=(−u2,u1),n⃗⋅u⃗=−u2u1+u1u2=0,∥n⃗∥=∥u⃗∥\vec{n} = \left( -u_2, u_1 \right), \qquad \vec{n} \cdot \vec{u} = \resaltar{-u_2 u_1 + u_1 u_2} = \resaltar{0}, \qquad \dato{n}{\norma{\vec{n}} = \norma{\vec{u}}}

    We admit, as in elementary geometry, that the area of a parallelogram is the product of a base and the corresponding height. We take as base the side representing u⃗\vec{u}, on the line through the origin with that direction; its equation is −u2x+u1y=0-u_2 x + u_1 y = 0, which the origin and the tip of u⃗\vec{u} satisfy. By the normal vector theorem, the vector n⃗\vec{n} is perpendicular to that line, and its norm is that of u⃗\vec{u}.

  2. h=∣v⃗⋅n⃗∣∥n⃗∥=∣u1v2−u2v1∣∥n⃗∥h = \frac{\abs{\resaltar{\vec{v} \cdot \vec{n}}}}{\norma{\vec{n}}} = \dato{h}{\frac{\abs{\resaltar{u_1 v_2 - u_2 v_1}}}{\norma{\vec{n}}}}

    The height is the distance from the tip of v⃗\vec{v} to that line. In the distance formula we take the origin as the point of the line, so that the vector from the line to the point is v⃗\vec{v} itself.

  3. ∥u⃗∥⋅h=∣u1v2−u2v1∣=∣det⁡A∣,∣3⋅2−1⋅1∣=5\resaltar{\norma{\vec{u}}} \cdot h = \resaltar{\abs{u_1 v_2 - u_2 v_1}} = \abs{\det A}, \qquad \abs{3 \cdot 2 - 1 \cdot 1} = 5

    The area is the base times the height, and the norm of u⃗\vec{u} cancels with that of n⃗\vec{n}. In the figure, the base measures 10\sqrt{10}, the height 510\frac{5}{\sqrt{10}}, and the area is 55.

  4. det⁡A=(rcos⁡α)(ssin⁡β)−(rsin⁡α)(scos⁡β)=rs(sin⁡βcos⁡α−cos⁡βsin⁡α)=rssin⁡(β−α)\det A = (r \cos \alpha)(s \sin \beta) - (r \sin \alpha)(s \cos \beta) = rs \resaltar{(\sin \beta \cos \alpha - \cos \beta \sin \alpha)} = \dato{s}{rs \resaltar{\sin(\beta - \alpha)}}

    For the sign, we write the vectors in polar coordinates: rr and ss are their norms, and α\alpha and β\beta the angles they form with the positive half of the horizontal axis, so that u⃗=(rcos⁡α,rsin⁡α)\vec{u} = \left( r \cos \alpha, r \sin \alpha \right) and v⃗=(scos⁡β,ssin⁡β)\vec{v} = \left( s \cos \beta, s \sin \beta \right). By the formula for the sine of a difference, the determinant is rsrs times the sine of the angle from u⃗\vec{u} to v⃗\vec{v}.

  5. ∣3112∣=5,∣1321∣=−5\begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} = \resaltar{5}, \qquad \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = \resaltar{-5}

    Since rs>0rs > 0, the determinant has the sign of sin⁡(β−α)\sin(\beta - \alpha), which is positive if β−α\beta - \alpha lies between 0∘0^\circ and 180∘180^\circ, up to a multiple of 360∘360^\circ, that is, if v⃗\vec{v} is reached from u⃗\vec{u} by turning counterclockwise through less than a half turn, and negative otherwise. It is zero if and only if β−α\beta - \alpha is a multiple of 180∘180^\circ, that is, if the vectors are parallel. In the figure, β−α=45∘\beta - \alpha = 45^\circ; when the columns are interchanged, the rotation carrying v⃗\vec{v} to u⃗\vec{u} is clockwise, and the sign changes.

The theorem justifies calling the determinant of order two the signed area: its absolute value measures the area, and its sign, the direction of the rotation carrying the first column to the second. The parallelism criterion of the Course I lesson on vectors is the case of zero area: two vectors are parallel exactly when the parallelogram they span collapses onto a line. The relation between the determinant and the areas of arbitrary figures, when the matrix acts as a transformation of the plane, will be studied in the lesson on matrices as transformations of the plane.