Worked Determinants Of Matrix Products

Determinants of Products

Multiplying two matrices can require many entry-by-entry calculations. If the goal is only to find the determinant of their product, that work is often unnecessary. The multiplicative determinant law converts the task into two determinant calculations, one for each factor. This is especially useful when each factor has a simple form, even if their product does not.

Throughout, the matrices in a product will be square matrices of the same order with real entries. The order matters: for \(\mathbf{A},\mathbf{B}\in\mathbb{R}^{n\times n}\), the product \(\mathbf{A}\mathbf{B}\) is also an \(n\times n\) matrix, so its determinant is defined. The multiplicative law is a standard result proved in the earlier material on the determinant of a product.

The matrices themselves need not commute: it may be that \(\mathbf{A}\mathbf{B}\ne\mathbf{B}\mathbf{A}\). The formula concerns their determinants, which are real numbers, and real-number multiplication is commutative. Thus \(\det(\mathbf{A})\det(\mathbf{B})=\det(\mathbf{B})\det(\mathbf{A})\), even when the matrix products have different entries.

Using the Law Instead of Multiplying First

For a determinant of a product, first check that both factors are square and have the same order. Then calculate the determinant of each factor and multiply the two scalar answers. This route can be much shorter than forming the product and then using cofactor expansion or row reduction to find its determinant.

Here the determinant calculations for the factors are simpler than a full expansion of the product. This advantage depends on the matrices chosen: the law gives a valid alternative, but one should still choose a convenient way to find each factor’s determinant.

Products with Several Factors

The product law also handles a product of three or more square matrices of the same order. The result follows by repeatedly applying the two-factor law; the matrices do not have to commute. The grouping of a matrix product can change by associativity, while the resulting determinant is the product of the individual determinants.

Proof. We use induction on \(k\). For \(k=1\), the statement is \(\det(\mathbf{A}_1)=\det(\mathbf{A}_1)\). Suppose it holds for a product of \(k\) factors. The product \(\mathbf{A}_1\cdots\mathbf{A}_k\) is an \(n\times n\) matrix, so the two-factor law applies to it and \(\mathbf{A}_{k+1}\). Hence \(\det(\mathbf{A}_1\cdots\mathbf{A}_k\mathbf{A}_{k+1}) =\det(\mathbf{A}_1\cdots\mathbf{A}_k)\det(\mathbf{A}_{k+1})\). Using the induction hypothesis on the first determinant gives \(\det(\mathbf{A}_1)\cdots\det(\mathbf{A}_k)\det(\mathbf{A}_{k+1})\), which is the claimed formula for \(k+1\). Thus the statement holds for every \(k\in\mathbb{N}\). \(\square\)

What a Zero-Determinant Factor Does

The product law has a useful consequence when one factor has determinant zero. A square matrix is singular if it is not invertible, meaning there is no matrix \(\mathbf{B}\) such that both \(\mathbf{A}\mathbf{B}=\mathbf{I}_n\) and \(\mathbf{B}\mathbf{A}=\mathbf{I}_n\). The examples below verify singularity directly by finding a nonzero vector that the factor sends to \(\mathbf{0}\); they also calculate the zero determinant from the entries. No general determinant test for invertibility is needed here.

Proof. By the multiplicative determinant law, \(\det(\mathbf{A}\mathbf{B})=\det(\mathbf{A})\det(\mathbf{B})\). If either factor on the right is zero, their product is zero, proving one direction. Conversely, suppose \(\det(\mathbf{A}\mathbf{B})=0\). Then \(\det(\mathbf{A})\det(\mathbf{B})=0\). A product of two real numbers is zero only if at least one of them is zero, so \(\det(\mathbf{A})=0\) or \(\det(\mathbf{B})=0\). This proves the other direction. \(\square\)

Common Pitfalls

Do not replace a product by a sum. The multiplicative determinant law says \(\det(\mathbf{A}\mathbf{B})=\det(\mathbf{A})\det(\mathbf{B})\); it does not say that determinants distribute over matrix addition. For a concrete counterexample, take \(\mathbf{A}=\mathbf{B}=\mathbf{I}_2\). The incorrect calculation \(\det(\mathbf{A}+\mathbf{B})=\det(\mathbf{A})+\det(\mathbf{B})\) would give \(4=2\). In fact, \(\mathbf{A}+\mathbf{B}=\begin{bmatrix}2&0\\0&2\end{bmatrix}\), whose determinant is \(2\cdot2=4\), while \(\det(\mathbf{A})+\det(\mathbf{B})=1+1=2\). The correction is to calculate the determinant of the sum directly; the product law applies only to a matrix product.

Check the dimensions before applying the law. Both factors must be square matrices of the same order. A rectangular matrix may multiply another rectangular matrix, but the determinant product law as stated here does not apply to such a pair.

A zero determinant in either factor is enough. There is no need to multiply the matrices first to discover that the product determinant is zero. The theorem above gives that conclusion immediately. If neither factor has zero determinant, the theorem also tells us that the product determinant is not zero, since a product of two nonzero real numbers is nonzero.

In practice, the product law is a way to choose an efficient calculation, not a rule that every determinant must be evaluated by factoring the matrix product. Use it when the factors have determinants that are easier to compute than the determinant of their product, and check that all factors meet the square, same-order condition.