Why an Inverse Has a Reciprocal Determinant
The determinant of a square matrix records a scalar quantity that behaves predictably under matrix multiplication. In particular, the determinant of a product is the product of the determinants. An inverse matrix undoes multiplication by the original matrix, so the product of a matrix and its inverse is the identity. Combining these facts forces the determinants of a matrix and its inverse to multiply to \(1\).
Throughout this tutorial, matrices are square \(n\times n\) real matrices, where \(n\in\mathbb{N}\). A matrix \(\mathbf{A}\) is invertible if there is a matrix \(\mathbf{A}^{-1}\) such that \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}_n\). The determinant \(\det(\mathbf{A})\) is a real number. We use the multiplicative determinant law: for any two \(n\times n\) matrices \(\mathbf{A}\) and \(\mathbf{B}\), \(\det(\mathbf{A}\mathbf{B})=\det(\mathbf{A})\det(\mathbf{B})\). We also use \(\det(\mathbf{I}_n)=1\).
Proof. Since \(\mathbf{A}\) is invertible, \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}_n\). Apply the multiplicative determinant law to the left-hand side and use \(\det(\mathbf{I}_n)=1\): $$ \det(\mathbf{A})\det(\mathbf{A}^{-1})=\det(\mathbf{A}\mathbf{A}^{-1})=\det(\mathbf{I}_n)=1. $$ A product of real numbers equal to \(1\) cannot have either factor equal to \(0\). Thus \(\det(\mathbf{A})\ne0\), and division by \(\det(\mathbf{A})\) gives \(\det(\mathbf{A}^{-1})=1/\det(\mathbf{A})\). \(\square\)
The order of reasoning matters: invertibility is the hypothesis, and the determinant identity is a consequence. The proof does not assume a determinant-based test for whether a matrix is invertible. In fact, it directly shows that an invertible matrix has nonzero determinant. Conversely, the theorem as stated does not claim that every matrix with nonzero determinant is invertible; that direction belongs to the determinant test for invertibility.
Computing and Checking the Identity
In an example, there are two useful checks. First, verify that the proposed inverse really multiplies with the original matrix to give the identity. Then compute the determinant of the inverse and compare it with the reciprocal of the original determinant. The second check illustrates the theorem; the first confirms that the matrix being used is indeed an inverse.
Worked Example: A Positive Determinant
Let $$ \mathbf{A}=\begin{bmatrix}2&1\\3&4\end{bmatrix}. $$ Its determinant is \(2\cdot4-1\cdot3=5\). Consider $$ \mathbf{B}=\begin{bmatrix}\frac45&-\frac15\\-\frac35&\frac25\end{bmatrix}. $$ We verify that \(\mathbf{B}=\mathbf{A}^{-1}\) by multiplying in both orders:
Thus \(\mathbf{B}=\mathbf{A}^{-1}\). Its determinant is $$ \det(\mathbf{B})=\frac45\cdot\frac25-\left(-\frac15\right)\left(-\frac35\right) =\frac{8}{25}-\frac{3}{25}=\frac15. $$ The theorem predicts \(1/\det(\mathbf{A})=1/5\), which agrees with the direct calculation.
Worked Example: A Negative Determinant
Take $$ \mathbf{A}=\begin{bmatrix}1&2\\3&4\end{bmatrix}, \qquad \mathbf{B}=\begin{bmatrix}-2&1\\\frac32&-\frac12\end{bmatrix}. $$ The determinant of \(\mathbf{A}\) is \(4-6=-2\). Check the candidate inverse by multiplication:
Therefore \(\mathbf{B}=\mathbf{A}^{-1}\). Directly, $$ \det(\mathbf{B})=(-2)\left(-\frac12\right)-1\cdot\frac32 =1-\frac32=-\frac12. $$ The reciprocal of \(\det(\mathbf{A})=-2\) is \(-1/2\), not \(+1/2\). The result has the same sign as the original determinant.
For any nonzero real number \(q\), \(q\) and \(1/q\) have the same sign: if \(q>0\), then \(1/q>0\), and if \(q<0\), then \(1/q<0\). Thus when \(\det(\mathbf{A})\) is negative, the determinant of the inverse is negative as well. Its magnitude is the reciprocal of the original magnitude. This sign observation uses the fact that the scalar field here is \(\mathbb{R}\); sign is not defined for a general complex number.
Proof. The theorem gives \(\det(\mathbf{A}^{-1})=1/\det(\mathbf{A})\), with \(\det(\mathbf{A})\ne0\). If \(\det(\mathbf{A})>0\), its reciprocal is positive; if \(\det(\mathbf{A})<0\), its reciprocal is negative. These are all possibilities for a nonzero real number, so the signs agree. \(\square\)
Worked Example: A Triangular Matrix of Order Three
Consider the upper triangular matrix $$ \mathbf{A}=\begin{bmatrix}1&1&0\\0&2&1\\0&0&3\end{bmatrix}. $$ Its determinant, by expansion along the first column, is \(1\cdot(2\cdot3-1\cdot0)=6\). A candidate inverse is $$ \mathbf{B}=\begin{bmatrix}1&-\frac12&\frac16\\0&\frac12&-\frac16\\0&0&\frac13\end{bmatrix}. $$ Multiplication gives $$ \mathbf{A}\mathbf{B} =\begin{bmatrix} 1&-\frac12+\frac12&\frac16-\frac16\\ 0&1&-\frac13+\frac13\\ 0&0&1 \end{bmatrix} =\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}. $$ In the other order, $$ \mathbf{B}\mathbf{A} =\begin{bmatrix} 1&1-1&-\frac12+\frac12\\ 0&1&\frac12-\frac12\\ 0&0&1 \end{bmatrix} =\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}. $$ So \(\mathbf{B}=\mathbf{A}^{-1}\). Expanding \(\det(\mathbf{B})\) along its first column gives $$ \det(\mathbf{B})=1\cdot\left(\frac12\cdot\frac13-\left(-\frac16\right)\cdot0\right)=\frac16. $$ This equals \(1/\det(\mathbf{A})=1/6\), as required.
Using the Formula Reliably
The identity is especially useful when the inverse is known or easy to describe but its determinant is not convenient to compute directly. If \(\mathbf{A}\) is invertible and \(\det(\mathbf{A})=q\), the determinant of its inverse is obtained by taking the scalar reciprocal \(1/q\). This is a statement about the determinant as a whole; it does not say that each entry of \(\mathbf{A}^{-1}\) is the reciprocal of the corresponding entry of \(\mathbf{A}\).
The hypotheses and conclusion can be kept straight with a short sequence:
The matrix must be square, and it must be invertible so that \(\mathbf{A}^{-1}\) exists.
Write \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}_n\).
Apply \(\det(\mathbf{A}\mathbf{B})=\det(\mathbf{A})\det(\mathbf{B})\) and \(\det(\mathbf{I}_n)=1\).
Since the product of the two determinants is \(1\), \(\det(\mathbf{A})\ne0\), and division gives the reciprocal formula.
A rectangular matrix has no inverse in this two-sided square-matrix sense, so the theorem does not apply to rectangular matrices. Also, if an exercise only provides a matrix \(\mathbf{B}\) and calls it an inverse, the formula applies only after \(\mathbf{B}\) is established as \(\mathbf{A}^{-1}\), or when that fact is part of the stated assumptions.
Common Mistake: Losing the Negative Sign
A frequent error is to read “reciprocal” as “positive reciprocal.” For the matrix in the negative-determinant example, one incorrect calculation is $$ \frac{1}{\det(\mathbf{A})}=\frac{1}{-2}=\frac12. $$ The correction is $$ \frac{1}{-2}=-\frac12. $$ Indeed, the inverse matrix has determinant \(-1/2\), as its entries verify. For every nonzero real \(q\), \(1/q\) has the same sign as \(q\), not the opposite sign.
A second error is to apply the identity before confirming invertibility. The expression \(\det(\mathbf{A}^{-1})\) is defined here only if \(\mathbf{A}^{-1}\) exists. The theorem starts with invertibility; the calculation then proves that \(\det(\mathbf{A})\) cannot be zero. It is not valid to divide by \(\det(\mathbf{A})\) without first obtaining its nonzero value from the product equation.
Finally, the result is not a rule for taking entrywise reciprocals. For example, the inverse of \(\begin{bmatrix}2&1\\3&4\end{bmatrix}\) is \(\begin{bmatrix}4/5&-1/5\\-3/5&2/5\end{bmatrix}\), not a matrix made by replacing each entry by its reciprocal. The theorem concerns the single number \(\det(\mathbf{A}^{-1})\), and the examples verify it by computing that determinant from the entire inverse matrix.
Why the Identity Matters
The formula lets determinant calculations pass between a matrix and its inverse without recomputing the inverse determinant from scratch. It also gives an immediate consistency check: if an invertible matrix has determinant \(q\), then the inverse must have determinant \(1/q\), and the two determinant values must multiply to \(1\). When working with exact fractions, keeping this product check visible is often the simplest way to catch a sign or arithmetic error.
Check Your Understanding
Give a reason for each answer, and use matrix multiplication where an inverse is supplied.
- If an invertible matrix \(\mathbf{A}\) has determinant \(7\), what is \(\det(\mathbf{A}^{-1})\)?
- If \(\det(\mathbf{A})=-5\) for an invertible real matrix, what is the sign and value of \(\det(\mathbf{A}^{-1})\)?
- Explain why the determinant-of-an-inverse identity requires a square invertible matrix.
- For \(\mathbf{A}=\begin{bmatrix}1&1\\1&2\end{bmatrix}\), verify that \(\begin{bmatrix}2&-1\\-1&1\end{bmatrix}\) is its inverse and compare the two determinants.
- State the equation obtained by taking determinants of \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}_n\), and explain why it permits division by \(\det(\mathbf{A})\).
