Omni Calculator logo

Cofactor Expansion

Welcome to Omni's cofactor expansion article! Here, you can quickly and easily learn how to compute the determinant of any matrix by cofactor expansion. Not sure what cofactors are? What does the phrase compute the determinant by cofactor expansion actually mean? Struggling with 4×4 cofactor expansion?

Stuck with a homework assignment asking you to find the determinant of a 5×5 matrix with cofactor expansion? Keep calm and scroll on down!

🔎 If you want to to learn about cofactors on their own, not in the context of determinants, try our dedicated cofactor matrix calculator.

Cofactor expansion is a way of computing the determinant of a matrix. Recall that a determinant is a number related to various important properties of a matrix. In particular, you can invert a matrix if, and only if, its determinant is not equal to zero. Inverting matrices is covered in detail in our inverse matrix calculator.

It's vital you remember that cofactor expansion is a recursive formula: it converts the task of computing a determinant of size n into computing several determinants of size n - 1. More precisely, the cofactor expansion allows you to express the determinant of size n as a weighted sum of minors. Recall that a minor is the determinant of some (n − 1) × (n − 1) submatrix of the initial matrix.

Let A be an n × n matrix with real (or complex) coefficients aij. Fix i = 1, …, n . Then:

det(A) = ai1 × Ci1 + ai2 × Ci2 + ... + ain × Cin

where Cij is the (i,j)-th cofactor of A. That is:

Cij = (-1)i+j × det(Aij)

where Aij is the (n-1) × (n-1) submatrix of A obtained by removing the i-th row and j-th column of A.

Plugging in the formula for the cofactor Cij, we obtain the following formula for det(A):

det(A) = (-1)i+1 × ai1 × det(Ai1) + (-1)i+2 × ai2 × det(Ai2) + ... + (-1)i+n × ain × det(Ain)

The above formula for det(A) is the cofactor expansion of the determinant along row i.
We see that to compute the determinant of a matrix by cofactor expansion, we only need to multiply the coefficients from some row of the matrix by the respective cofactors and then add everything together.

We can also write down an analogous formula for expanding along column j:

det(A) = a1j × C1j + a2j × C2j + ... + anj × Cnj

where j = 1, ..., n is fixed. Using the formula for Cij, we have:

det(A) = (-1)1+j × a1j × det(A1j) + (-1)2+j × a2j × det(A2j) + ... + (-1)n+j × anj × det(Anj)

Let's consider the following 5×5 matrix and find its determinant by cofactor expansion.

[−1−21082000−14−60−1012321001−11]\begin{bmatrix} -1& -2& 1& 0& 8\\ 2& 0& 0& 0& -1 \\ 4& -6& 0& -1& 0\\ 1& 2& 3& 2& 1\\ 0& 0& 1& -1& 1\\ \end{bmatrix}

We see there are three zeros in the second row: let's perform cofactor expansion along this row.

det(A) = a21 × C21 + a22 × C22 + a23 × C23 + a24 × C24 + a25 × C25

We see that, in principle, we have to compute as many as FIVE determinants. Fortunately,

a22 = a23 = a24 = 0

so the formula reduces to:

det(A) = a21 × C21 + a25 × C25

which is:

det(A) = (-1)2+1 × a21 × det(A21) + (-1)2+5 × a25 × det(A25)

Simplifying the sign factors and plugging in the coefficients a21 = 2 and a25 = -1, we obtain:

det(A) = -2 × det(A21) + 1 × det(A25)

where A21 is the following matrix:

[−2108−60−10232101−11]\begin{bmatrix} -2 &1 &0 &8 \\ -6& 0 &-1& 0 \\ 2& 3& 2 &1 \\ 0& 1& -1 &1 \\ \end{bmatrix}

and A25 is:

[−1−2104−60−11232001−1]\begin{bmatrix} -1 &-2& 1 &0 \\ 4 &-6 &0 &-1 \\ 1& 2 &3& 2 \\ 0 &0 &1& -1 \\ \end{bmatrix}

It turns out that det(A21) = -212 and det(A25) = -84 (see below for some details), so:

det(A) = -2 × (-212) + 1 × (-84) = 424 − 84 = 340

4×4 cofactor expansion

To show you how to solve the problem of 4×4 cofactor expansion, we now compute det(A25) using cofactor expansion once again. To avoid confusion, let us denote the 4×4 matrix under consideration by B; that is, B reads:

[−1−2104−60−11232001−1]\begin{bmatrix} -1 &-2& 1 &0 \\ 4 &-6 &0 &-1 \\ 1& 2 &3& 2 \\ 0 &0 &1& -1 \\ \end{bmatrix}

We choose the expansion along the last (fourth) row, as it contains two zeros (b41 = b42 = 0). In consequence,

det(B) = (-1)4+3 × b43 × det(B43) + (-1)4+4 × b44 × det(B44)

det(B) = (-1) × 1 × det(B43) + 1 × (-1) × det(B44)

det(B) = -det(B43) − det(B44)

where B43 is:

[−1−204−6−1122]\begin{bmatrix} -1& -2 &0 \\ 4& -6& -1 \\ 1& 2 &2 \\ \end{bmatrix}

and B44 is:

[−1−214−60123]\begin{bmatrix} -1& -2& 1\\ 4& -6& 0 \\ 1& 2& 3 \\ \end{bmatrix}

It turns out that det(B43) = 28 and det(B44) = 56 (use the Sarrus rule or see below for the calculation via cofactor expansion), so:

det(B) = -28 − 56 = -84

3×3 cofactor expansion

Finally, we show you how to deal with 3×3 cofactor expansion. We take the second submatrix above and denote it by M. That is, M is:

[−1−214−60123]\begin{bmatrix} -1& -2& 1\\ 4& -6& 0 \\ 1& 2& 3 \\ \end{bmatrix}

We choose the expansion along the last (third) column as it contains a zero (m23 = 0), which simplifies calculations. In consequence:

det(M) = (-1)1+3 × m13 × det(M13) + (-1)3+3 × m33 × det(M33)

det(M) = 1 × 1 × det(M13) + 1 × 3 × det(M33)

det(M) = det(M13) + 3 × det(M33)

where M13 is:

[4−612]\begin{bmatrix} 4& -6 \\ 1& 2 \\ \end{bmatrix}

and M33 is:

[−1−24−6]\begin{bmatrix} -1& -2 \\ 4& -6 \\ \end{bmatrix}

We can easily determine that det(M13) = det(M33) = 14, so:

det(M) = 14 + 3 × 14 = 56

Now, it's your turn! Try computing the remaining determinants to confirm our results.

In cofactor expansion, go for the column or row containing the most zeros. This is because the successive coefficients are to be multiplied by the respective cofactors. If the coefficient is zero, you don't need to compute the corresponding cofactor because the product is going to be zero anyway.

To compute the (i, j)-th cofactor of a matrix, follow these steps:

  1. Remove the i-th row and j-th column of your matrix. You get a submatrix of size one less than the initial matrix.

  2. Compute the determinant of this submatrix.

  3. Compute the sign factor (-1)i+j.

  4. Multiply the determinant from step 2 by the sign factor from step 3.

  5. This product is precisely the (i, j)-th cofactor you've been looking for! Congratulations!

To compute the cofactor expansion of a 3×3 matrix, you have to:

  1. Choose a row/column of your matrix. Go for the one containing the most zeros.

  2. For each coefficient in your row/column, compute the respective 2×2 cofactor.

  3. Multiply the coefficient by its cofactor.

  4. Add the three numbers obtained in steps 2 & 3.

  5. This is your determinant! By the way, have you heard of the rule of Sarrus?

To compute the cofactor expansion of a 4×4 matrix, follow these steps:

  1. Choose a row/column of your matrix. Tip: go for the one containing the most zeros.
  2. For each coefficient in your row/column, compute the respective 3×3 cofactor.
  3. Multiply the coefficient by its cofactor.
  4. Add the four numbers obtained in steps 2 & 3.
  5. Congratulate yourself on finding the determinant!

To compute the cofactor expansion of a 5×5 matrix, you have to:

  1. Choose a row/column of your matrix. Go for the one containing the most zeros.
  2. For each coefficient in your row/column, compute the respective 4×4 cofactor.
  3. Multiply the coefficient by its cofactor.
  4. Add the five numbers obtained in steps 2 & 3.
  5. Wow, you have just found a 5×5 determinant!

This article was written by Anna Szczepanek and reviewed by Steven Wooding.