Let $A\in {\text{M}}_{n\times n}(F).$ If $n=1,$ so that $A=(A_{11}),$ we define $\det (A)=A_{11}.$ For $n\ge 2,$ we define det(A) recursively as

The scalar det(A) is called the determinant of A and is also denoted by $\left|A\right|.$ The scalar

is called the cofactor of the entry of A in row i, column j.

Letting

denote the cofactor of the row i, column j entry of A, we can express the formula for the determinant of A as

Thus the determinant of A equals the sum of the products of each entry in row 1 of A multiplied by its cofactor. This formula is called cofactor expansion along the first row of A. Note that, for $2\times 2$ matrices, this definition of the determinant of A agrees with the one given in Section 4.1 because

The proof is by mathematical induction on n. The result is immediate if $n=1.$ Assume that for some integer $n\ge 2$ the determinant of any $(n-1)\times (n-1)$ matrix is a linear function of each row when the remaining rows are held fixed. Let A be an $n\times n$ matrix with rows $a_1,a_2,\dots ,a_n,$ respectively, and suppose that for some $r(1\le r\le n),$ we have $a_r=u+kv$ for some u, $v\in {\text{F}}^n$ and some scalar k. Let $u=(b_1,b_2,\dots ,b_n)$ and $v=(c_1,c_2,\dots ,c_n),$ and let B and C be the matrices obtained from A by replacing row r of A by u and v, respectively. We must prove that $\det (A)=\det (B)+k\det (C).$ We leave the proof of this fact to the reader for the case $r=1.$ For $r>1$ and $1\le j\le n,$ the rows of ${\tilde{A}}_{1j},{\tilde{B}}_{1j},$ and ${\tilde{C}}_{1j}$ are the same except for row $r-1.$ Moreover, row $r-1$ of ${\tilde{A}}_{ij}$ is

which is the sum of row $r-1$ of ${\tilde{B}}_{1j}$ and k times row $r-1$ of ${\tilde{C}}_{1j}.$ Since ${\tilde{B}}_{1j}$ and ${\tilde{C}}_{1j}$ are $(n-1)\times (n-1)$ matrices, we have

by the induction hypothesis. Thus since $A_{1j}=B_{1j}=C_{1j},$ we have

This shows that the theorem is true for $n\times n$ matrices, and so the theorem is true for all square matrices by mathematical induction.

If $A\in {\text{M}}_{n\times n}(F)$ has a row consisting entirely of zeros, then $\det (A)=0$.

See Exercise 24.

The definition of a determinant requires that the determinant of a matrix be evaluated by cofactor expansion along the first row. Our next theorem shows that the determinant of a square matrix can be evaluated by cofactor expansion along any row. Its proof requires the following technical result.

Let $B\in {\text{M}}_{n\times n}(F),$ where $n\ge 2.$ If row i of B equals $e_k$ for some $k(1\le k\le n),$ then $\det (B)={(-1)}^{i+k}\det ({\tilde{B}}_{ik})$.

The proof is by mathematical induction on n. The lemma is easily proved for $n=2.$ Assume that for some integer $n\ge 3,$ the lemma is true for $(n-1)\times (n-1)$ matrices, and let B be an $n\times n$ matrix in which row i of B equals $e_k$ for some $k(1\le k\le n).$ The result follows immediately from the definition of the determinant if $i=1.$ Suppose therefore that $1<i\le n.$ For each $j\ne k(1\le j\le n),$ let $C_{ij}$ denote the $(n-2)\times (n-2)$ matrix obtained from B by deleting rows 1 and i and columns j and k. For each j, row $i-1$ of ${\tilde{B}}_{1j}$ is the following vector in ${\text{F}}^{n-1}$:

Hence by the induction hypothesis and the corollary to Theorem 4.3, we have

Therefore

Because the expression inside the preceding bracket is the cofactor expansion of ${\tilde{B}}_{ik}$ along the first row, it follows that

This shows that the lemma is true for $n\times n$ matrices, and so the lemma is true for all square matrices by mathematical induction.

We are now able to prove that cofactor expansion along any row can be used to evaluate the determinant of a square matrix.

Cofactor expansion along the first row of A gives the determinant of A by definition. So the result is true if $i=1.$ Fix $i>1.$ Row i of A can be written as ${\displaystyle {\sum }_{j=1}^n{A}_{ij}{e}_j}.$ For $1\le j\le n,$ let $B_j$ denote the matrix obtained from A by replacing row i of A by $e_j.$ Then by Theorem 4.3 and the lemma, we have

If $A\in {\text{M}}_{n\times n}(F)$ has two identical rows, then $\det (A)=0$.

The proof is by mathematical induction on n. We leave the proof of the result to the reader in the case that $n=2.$ Assume that for some integer $n\ge 3,$ it is true for $(n-1)\times (n-1)$ matrices, and let rows r and s of $A\in {\text{M}}_{n\times n}(F)$ be identical for $r\ne s.$ Because $n\ge 3,$ we can choose an integer $i(1\le i\le n)$ other than r and s. Now

by Theorem 4.4. Since each ${\tilde{A}}_{ij}$ is an $(n-1)\times (n-1)$ matrix with two identical rows, the induction hypothesis implies that each $\det ({\tilde{A}}_{ij})=0,$ and hence $\det (A)=0.$ This completes the proof for $n\times n$ matrices, and so the lemma is true for all square matrices by mathematical induction.

It is possible to evaluate determinants more efficiently by combining cofactor expansion with the use of elementary row operations. Before such a process can be developed, we need to learn what happens to the determinant of a matrix if we perform an elementary row operation on that matrix. Theorem 4.3 provides this information for elementary row operations of type 2 (those in which a row is multiplied by a nonzero scalar). Next we turn our attention to elementary row operations of type 1 (those in which two rows are interchanged).

Let the rows of $A\in {\text{M}}_{n\times n}(F)$ be $a_1,a_2,\dots ,a_n,$ and let B be the matrix obtained from A by interchanging rows r and s, where $r<s.$ Thus

Consider the matrix obtained from A by replacing rows r and s by $a_r+a_s.$ By the corollary to Theorem 4.4 and Theorem 4.3, we have

Therefore $\det (B)=-\det (A)$.

We now complete our investigation of how an elementary row operation affects the determinant of a matrix by showing that elementary row operations of type 3 do not change the determinant of a matrix.

Suppose that B is the $n\times n$ matrix obtained from A by adding k times row r to row s, where $r\ne s$. Let the rows of A be $a_1,a_2,\dots ,a_n$ and the rows of B be $b_1,b_2,\dots ,b_n$ Then $b_i\ne a_i$ for $i\ne s$ and $b_s=a_s+k{a}_r$. Let C be the matrix obtained from A by replacing row s with $a_r$. Applying Theorem 4.3 to row s of B, we obtain

because $\text{det}(C)=0$ by the corollary to Theorem 4.4.

In Theorem 4.2 (p. 201), we proved that a $2\times 2$ matrix is invertible if and only if its determinant is nonzero. As a consequence of Theorem 4.6, we can prove half of the promised generalization of this result in the following corollary. The converse is proved in the corollary to Theorem 4.7.

If $A\in {\text{M}}_{n\times n}(F)$ has rank less than n, then $\text{det}(A)=0$.

If the rank of A is less than n, then the rows $a_1,a_2,\dots ,a_n$ of A are linearly dependent. By Exercise 14 of Section 1.5, some row of A, say, row r, is a linear combination of the other rows. So there exist scalars $c_i$ such that

Let B be the matrix obtained from A by adding $-c_i$ times row i to row r for each $i\ne r$. Then row r of B consists entirely of zeros, and so $\text{det}(B)=0$. But by Theorem 4.6, $\text{det}(B)=\text{det}(A)$. Hence $\text{det}(A)=0$.

The following rules summarize the effect of an elementary row operation on the determinant of a matrix $A\in {\text{M}}_{n\times n}(F)$.

1. (a) If B is a matrix obtained by interchanging any two rows of A, then $\text{det}(B)=-\text{det}(A)$.

2. (b) If B is a matrix obtained by multiplying a row of A by a nonzero scalar k, then $\text{det}(B)=k\text{det}(A)$.

3. (c) If B is a matrix obtained by adding a multiple of one row of A to another row of A, then $\text{det}(B)=\text{det}(A)$.

These facts can be used to simplify the evaluation of a determinant. Consider, for instance, the matrix in Example 1:

Adding 3 times row 1 of A to row 2 and 4 times row 1 to row 3, we obtain

Since M was obtained by performing two type 3 elementary row operations on A, we have $\text{det}(A)=\text{det}(M)$. The cofactor expansion of M along the first row gives

Both ${\tilde{M}}_{12}$ and ${\tilde{M}}_{13}$ have a column consisting entirely of zeros, and so $\text{det}({\tilde{M}}_{12})=\text{det}({\tilde{M}}_{13})=0$ by the corollary to Theorem 4.6. Hence

Thus with the use of two elementary row operations of type 3, we have reduced the computation of det(A) to the evaluation of one determinant of a $2\times 2$ matrix.

But we can do even better. If we add $-4$ times row 2 of M to row 3 (another elementary row operation of type 3), we obtain

Evaluating det(P) by cofactor expansion along the first row, we have

as described earlier. Since $\text{det}(A)=\text{det}(M)=\text{det}(P)$, it follows that $\text{det}(A)=40$.

The preceding calculation of det(P) illustrates an important general fact. The determinant of an upper triangular matrix is the product of its diagonal entries. (See Exercise 23.) By using elementary row operations of types 1 and 3 only, we can transform any square matrix into an upper triangular matrix, and so we can easily evaluate the determinant of any square matrix. The next two examples illustrate this technique.