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Matrix Determinants

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Core Definition

The determinant is a scalar value calculated from the elements of a square matrix. For a matrix AA, it is denoted as det⁡(A)\det(A) or ∣A∣|A|. It acts as a scaling factor for linear transformations, indicating how area (in 2D) or volume (in 3D) is stretched or squished under the transformation.

Concept Explanation

Determinants are defined only for square matrices (matrices with the same number of rows and columns). To compute them, we use distinct algebraic methods depending on the dimensions of the matrix.

Calculations

1. 2x2 Matrices

The determinant of a 2x2 matrix is the difference of the cross-products of its diagonals:

det⁡(abcd)=∣abcd∣=ad−bc\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc

2. 3x3 Matrices (Cofactor Expansion)

To calculate the determinant of a 3x3 matrix, we expand it along the first row. This breaks the 3x3 matrix into three smaller 2x2 determinants (minors).

Note: The signs of the terms alternate: positive (++), negative (−-), positive (++).

∣abcdefghi∣=a∣efhi∣−b∣dfgi∣+c∣degh∣\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a \begin{vmatrix} e & f \\ h & i \end{vmatrix} - b \begin{vmatrix} d & f \\ g & i \end{vmatrix} + c \begin{vmatrix} d & e \\ g & h \end{vmatrix}
=a(ei−fh)−b(di−fg)+c(dh−eg)= a(ei - fh) - b(di - fg) + c(dh - eg)

Why We Use It in Multivariable Calculus (Calculus III)

Determinants are used extensively in Calculus III to formulate geometry in 3D space and scale integrations:

1. Vector Cross Product

The cross product u⃗×v⃗\vec{u} \times \vec{v} of two 3D vectors results in a vector perpendicular to both. We write this operation symbolically as a 3x3 determinant where the first row contains the unit basis vectors i^\hat{i}, j^\hat{j}, and k^\hat{k}:

u⃗×v⃗=∣i^j^k^u1u2u3v1v2v3∣\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}

2. Triple Scalar Product (Volume of Parallelepiped)

The scalar value u⃗⋅(v⃗×w⃗)\vec{u} \cdot (\vec{v} \times \vec{w}) represents the volume of the 3D parallel-sided box (parallelepiped) spanned by the three vectors. It is calculated by taking the determinant of the 3x3 matrix whose rows are the components of the three vectors. If the determinant is 00, the vectors lie in the same plane (coplanar).

3. The Jacobian Determinant (Change of Variables)

When converting a double or triple integral from Cartesian coordinates to another system (like polar, cylindrical, spherical, or a general u,vu,v-substitution), we must scale the differential area/volume elements. The scale factor is the determinant of the Jacobian Matrix (a matrix of partial derivatives). For a 2D transformation x(u,v),y(u,v)x(u,v), y(u,v), the Jacobian determinant is:

J(u,v)=∂(x,y)∂(u,v)=∣∂x∂u∂x∂v∂y∂u∂y∂v∣  ⟹  dA=∣J(u,v)∣ du dvJ(u,v) = \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \implies dA = |J(u,v)| \, du \, dv

Example Problems

Example Problem 1: 2x2 Determinant (Jacobian of Polar Coordinates)

Verify the scaling factor rr for polar integration by calculating the Jacobian determinant of the polar coordinate transformation:

x=rcos⁡θ,y=rsin⁡θx = r\cos\theta, \quad y = r\sin\theta

Solution

Step 1: Calculate the partial derivatives

  • ∂x∂r=cos⁡θ\frac{\partial x}{\partial r} = \cos\theta
  • ∂x∂θ=−rsin⁡θ\frac{\partial x}{\partial \theta} = -r\sin\theta
  • ∂y∂r=sin⁡θ\frac{\partial y}{\partial r} = \sin\theta
  • ∂y∂θ=rcos⁡θ\frac{\partial y}{\partial \theta} = r\cos\theta

Step 2: Set up the Jacobian matrix and its determinant

J=∣∂x∂r∂x∂θ∂y∂r∂y∂θ∣=∣cos⁡θ−rsin⁡θsin⁡θrcos⁡θ∣J = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix}

Step 3: Calculate the determinant (ad−bcad - bc)

det⁡(J)=(cos⁡θ)(rcos⁡θ)−(−rsin⁡θ)(sin⁡θ)\det(J) = (\cos\theta)(r\cos\theta) - (-r\sin\theta)(\sin\theta)
=rcos⁡2θ+rsin⁡2θ= r\cos^2\theta + r\sin^2\theta

Step 4: Simplify using trig identities Factor out rr:

det⁡(J)=r(cos⁡2θ+sin⁡2θ)=r(1)=r\det(J) = r(\cos^2\theta + \sin^2\theta) = r(1) = r

This shows that the polar area element is dA=r dr dθdA = r \, dr \, d\theta.

Example Problem 2: 3x3 Determinant (Cross Product)

Find the cross product vector c⃗=a⃗×b⃗\vec{c} = \vec{a} \times \vec{b} for the vectors:

a⃗=⟨1,3,−2⟩andb⃗=⟨2,−1,4⟩\vec{a} = \langle 1, 3, -2 \rangle \quad \text{and} \quad \vec{b} = \langle 2, -1, 4 \rangle

Solution

Step 1: Set up the symbolic 3x3 determinant Place the basis vectors in the top row:

a⃗×b⃗=∣i^j^k^13−22−14∣\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -2 \\ 2 & -1 & 4 \end{vmatrix}

Step 2: Expand along the first row (alternating signs)

a⃗×b⃗=i^∣3−2−14∣−j^∣1−224∣+k^∣132−1∣\vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 3 & -2 \\ -1 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -2 \\ 2 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 3 \\ 2 & -1 \end{vmatrix}

Step 3: Calculate the 2x2 determinants

  • For i^\hat{i}: (3)(4)−(−2)(−1)=12−2=10(3)(4) - (-2)(-1) = 12 - 2 = 10
  • For j^\hat{j}: (1)(4)−(−2)(2)=4−(−4)=8(1)(4) - (-2)(2) = 4 - (-4) = 8
  • For k^\hat{k}: (1)(−1)−(3)(2)=−1−6=−7(1)(-1) - (3)(2) = -1 - 6 = -7

Step 4: Combine the results

a⃗×b⃗=10i^−8j^−7k^\vec{a} \times \vec{b} = 10\hat{i} - 8\hat{j} - 7\hat{k}

In bracket notation, the cross product is ⟨10,−8,−7⟩\langle 10, -8, -7 \rangle.

Connections & References