Properties derived from cofactor expansion

Section 4.4 Properties derived from cofactor expansion

The Laplace expansion theorem turns out to be a powerful tool, both for computation and for the derivation of theoretical results. In this section we derive several of these results.

All matrices under discussion in the section will be square of order \(n\text{.}\)

Subsection 4.4.1 All zero rows

Theorem 4.4.1. All zero row or column implies \(\det A=0\).

If \(A\) has an all zero row or all zero column, then \(\det(A)=0\text{.}\)

Proof.

Suppose row \(R_i\) is all zero. Then the expansion on \(R_i\) is

\begin{equation*} \sum_{j=1}^n a_{i,j}C_{i,j} = \sum_{j=1}^n 0C_{i,j} =0. \end{equation*}

Subsection 4.4.2 Triangular matrices

Theorem 4.4.2. The determinant of a triangular matrix.

If \(A\) is triangular, then \(\det(A)=a_{1,1}a_{2,2}\cdots a_{n,n}\text{.}\)

Proof.

Suppose \(A\) is lower triangular (the upper triangular case uses essentially the same argument). We repeatedly expand on the first row.

\begin{align*} \det A \amp = \det \begin{bmatrix} a_{1,1} \amp 0 \amp 0 \amp \cdots \amp 0\\ * \amp a_{2,2} \amp 0 \amp \cdots \amp 0\\ * \amp * \amp a_{3,3} \amp \cdots \amp 0\\ \amp \amp \amp \vdots\\ * \amp * \amp * \amp \cdots \amp a_{n,n}\\ \end{bmatrix}\\ \amp = a_{1,1} \det \begin{bmatrix} a_{2,2} \amp 0 \amp \cdots \amp 0\\ * \amp a_{3,3} \amp \cdots \amp 0\\ \amp \amp \vdots\\ * \amp * \amp \cdots \amp a_{n,n}\\ \end{bmatrix}\\ \amp = a_{1,1} a_{2,2} \det \begin{bmatrix} a_{3,3} \amp \cdots \amp 0\\ \amp \vdots\\ * \amp \cdots \amp a_{n,n}\\ \end{bmatrix}\\ \amp \phantom{x}\vdots\\ \amp = a_{1,1}a_{2,2}\cdots a_{n,n} \end{align*}

Corollary 4.4.3. The determinant of a diagonal matrix the product of its diagonal entries.

If \(A\) is a diagonal matrix, then \(\det(A)=a_{1,1}a_{2,2}\cdots a_{n,n}\text{.}\)

Proof.

A diagonal matrix is certainly triangular.

Example 4.4.4. \(\det I=1\).

Since the identity matrix \(I\) is a diagonal matrix,

\begin{equation*} \det I=1 \end{equation*}

Subsection 4.4.3 Interchanging rows

The purpose of this section is to show that if a matrix \(B\) is derived from \(A\) by interchanging two rows, then \(\det B = -\det A\text{.}\) We do this in three steps:

Lemma 4.4.5. Interchanging rows \(R_1\) and \(R_2\) changes the sign of the determinant.

If \(B\) is derived from \(A\) by interchanging the first and second rows (that is, \(R_1\leftrightarrow R_2\)), then \(\det B = -\det A\text{.}\)

Proof.

We compute the determinant of \(A\) by cofactor expansion along the first row and the determinant of \(B\) by cofactor expansion along the second row. This means that

\begin{equation*} \det A =\sum_{j=1}^n (-1)^{1+j} a_{1,j}M_{1,j}\\ \det B =\sum_{j=1}^n (-1)^{2+j} b_{2,j}M'_{2,j} \end{equation*}

Since the first row of \(A\) is the second row of \(B\text{,}\) we have \(a_{1,j}=b_{2,j}\) for \(j=1,2,\ldots,n\text{.}\) In addition, deleting \(R_1\) and \(C_j\) from \(A\) yields exactly the same matrix as deleting \(R_2\) and \(C_j\) from \(B\text{,}\) that is to say \(M_{1,j}=M'_{2,j}\text{.}\) Hence we have

\begin{align*} \det B \amp=\sum_{j=1}^n (-1)^{2+j} b_{2,j}M'_{2,j}\\ \amp= \sum_{j=1}^n (-1)^{2+j} a_{1,j}M_{1,j}\\ \amp= -\sum_{j=1}^n (-1)^{1+j} a_{1,j}M_{1,j}\\ \amp= -\det A \end{align*}

Lemma 4.4.6. Interchanging rows \(R_i\) and \(R_{i+1}\) changes the sign of the determinant.

If \(B\) is derived from \(A\) by interchanging \(R_i\) and \(R_{i+1}\) (that is, \(R_i\leftrightarrow R_{i+1}\)), then \(\det B = -\det A\text{.}\)

Proof.

We compute the determinants by cofactor expansion along the \(i\)-th row of \(A\) and along the the \(i+1\)-st row of \(B\text{.}\) This means that

\begin{equation*} \det A =\sum_{j=1}^n (-1)^{i+j} a_{i,j}M_{i,j}\\ \det B =\sum_{j=1}^n (-1)^{i+1+j} b_{i+1,j}M'_{i+1,j} \end{equation*}

We have \(a_{i,j}=b_{{i+1},j}\) for \(j=1,2,\ldots,n\text{.}\) In addition, deleting \(R_i\) and \(C_j\) from \(A\) yields exactly the same matrix as deleting \(R_{i+1}\) and \(C_j\) from \(B\) and so \(M_{i,j}=M'_{i+1,j}\text{.}\) Hence we have

\begin{align*} \det B \amp=\sum_{j=1}^n (-1)^{i+1+j} b_{i+1,j}M'_{i+1,j}\\ \amp= \sum_{j=1}^n (-1)^{i+1+j} a_{i,j}M_{i,j}\\ \amp= -\sum_{j=1}^n (-1)^{i+j} a_{i,j}M_{i,j}\\ \amp= -\det A \end{align*}

Theorem 4.4.7. Interchanging rows \(R_i\) and \(R_j\) changes the sign of the determinant.

If \(B\) is derived from \(A\) by interchanging the \(i\)-th and \(j\)-th rows (that is, \(R_i\leftrightarrow R_j\)), then \(\det B = -\det A\text{.}\)

Proof.

With no loss of generality, we assume that \(i\lt j\text{.}\) Interchange \(R_i\) with the one below it so that it has moved one row lower. Repeat the process until it is just below \(R_j\text{.}\) This take \(j-i\) interchanges. Now interchange \(R_j\) with the one above it repeatedly until it is in the \(i\)-th row. This takes \(j-i-1\) interchanges. The net effect is to interchange \(R_i\) and \(R_j\text{.}\) Each interchange multiplies the determinant by \(-1\text{.}\) Since there are \(2(j-i)-1\) (an odd number) interchanges in total, we have

\begin{equation*} \det B = (-1)^{2(j-i)-1} \det A = -\det A. \end{equation*}

Here is an example to see how the proof actually works. The second row (in red) and sixth row (in green) will be interchanged. The red row is interchanged with the one below it until it is just below the green row. Then the green row is interchanged with the one above it until it is in the position originally occupied by the red row. It takes four interchanges to get the red row below the green row and three interchanges to get the green row in the original position of the red row.

Subsection 4.4.4 Multiplying a row by a constant \(\lambda\)

Theorem 4.4.9. Multiplying a row by \(\lambda\) multiplies the determinant by \(\lambda\).

If \(B\) is derived from \(A\) by multiplying the \(i\)-th row by \(\lambda\) (that is, \(R_i\gets \lambda R_i\)), then \(\det B=\lambda \det A\text{.}\)

Proof.

Expanding on the \(i\)-th row:

\begin{align*} \det B \amp = \sum_{j=1}^n b_{i,j}M_{i,j} \\ \amp = \sum_{j=1}^n \lambda a_{i,j}M_{i,j} \\ \amp = \lambda \sum_{j=1}^n a_{i,j}M_{i,j} \\ \amp = \lambda\det A \end{align*}

We use this theorem to evaluate the determinant of \(\lambda A\text{.}\)

Corollary 4.4.10. The determinant of \(\lambda A\).

If \(A\) is a square matrix of order \(n\) and \(\lambda\) is any real number, then

\begin{equation*} \det \lambda A=\lambda^n A. \end{equation*}

Proof.

The matrix \(\lambda A\) is derived from \(A\) by applying \(R_i\gets \lambda R_i\) for \(i=1,2,\ldots,n\text{.}\) Each application multiplies the determinant by \(\lambda\text{,}\) and so after the \(n\) applications we have

\begin{equation*} \det \lambda A=\lambda^n A. \end{equation*}

Subsection 4.4.5 Row additivity

We wish to consider two matrices \(A\) and \(B\) that are identical except for the \(i\)-th row. We may visualize this at

\begin{equation*} A= \begin{bmatrix} R_1\\ R_2\\ \vdots\\ R_i\\ \vdots\\ R_n \end{bmatrix} \textrm{ and } B= \begin{bmatrix} R_1\\ R_2\\ \vdots\\ R_i'\\ \vdots\\ R_n \end{bmatrix}. \end{equation*}

We then define the matrix \(C\) by

\begin{equation*} C= \begin{bmatrix} R_1\\ R_2\\ \vdots\\ R_i+R_i'\\ \vdots\\ R_n \end{bmatrix}. \end{equation*}

Theorem 4.4.11. Row additivity theorem.

If \(A\text{,}\) \(B\) and \(C\) are as above, then \(\det C=\det A + \det B\text{.}\)

Proof.

We expand by cofactors on the \(i\)-th row.

\begin{align*} \det C \amp = \sum_{j=1}^n (-1)^{i+j}c_{i,j} M_{i,j} \\ \amp = \sum_{j=1}^n (-1)^{i+j}(a_{i,j}+b_{i,j}) M_{i,j} \\ \amp = \sum_{j=1}^n (-1)^{i+j}a_{i,j} M_{i,j} + \sum_{j=1}^n (-1)^{i+j}b_{i,j} M_{i,j} \\ \amp =\det A + \det B \end{align*}

Subsection 4.4.6 Identical and proportional rows

Theorem 4.4.12. A matrix \(A\) with two identical rows has \(\det A=0\).

Suppose a matrix \(A\) has two equal rows: \(R_i=R_j\) with \(i\not=j\text{.}\) Then \(\det A=0\text{.}\)

Proof.

Let \(B\) be the matrix obtained by the elementary row operation \(R_i\leftrightarrow R_j\text{.}\) The equality of the two rows implies \(B=A\text{,}\) and so \(\det B = \det A\text{.}\) On the other hand, Theorem 4.4.7 implies \(\det B = -\det A\text{.}\) Hence \(\det A=0\text{.}\)

Definition 4.4.13. Proportional rows of a matrix.

Two rows, \(R_i\) and \(R_j\) are proportional if \(R_i=\lambda R_j\) for some \(\lambda\not=0\text{.}\)

Theorem 4.4.14. A matrix \(A\) with two proportional rows has \(\det A=0\).

Suppose a matrix \(A\) has two proportional rows: \(R_i=\lambda R_j\) with \(i\not=j\) and \(\lambda\not=0\text{.}\) Then \(\det A=0\text{.}\)

Proof.

Let \(B\) be the matrix obtained by the elementary row operation \(R_i\gets \lambda R_i\text{.}\) By Theorem 4.4.9, \(\det B = \lambda \det A\text{.}\) However, \(B\) has two identical rows and so \(\det B=0\text{.}\) Hence \(\lambda\not=0\) implies \(\det A=0\text{.}\)

Subsection 4.4.7 Adding a multiple of one row to another

We can now use Theorem 4.4.11 and Theorem 4.4.14 to find the effect of the third elementary row operation on the determinant of a matrix.

Theorem 4.4.15. Adding a multiple of one row to another leaves the determinant unchanged.

If \(B\) is derived from \(A\) by adding a multiple of one row to another (that is, \(R_i\gets R_i+\lambda R_j\)) then \(\det B=\det A\text{.}\)

Proof.

The matrices \(A\) and \(B\) are identical except for the \(i\)-th row. In \(A\) the \(i\)-th row is \(R_i\text{,}\) and in \(B\) the \(i\)-th row is \(R_i+\lambda R_j\text{.}\) The row additivity theorem then says

\begin{align*} \det B \amp = \det \begin{bmatrix} R_1\\ \vdots\\ R_{i-1}\\ R_i+\lambda R_j\\ R_{i+1}\\ \vdots\\ R_n \end{bmatrix} \\ \amp = \det \begin{bmatrix} R_1\\ \vdots\\ R_{i-1}\\ R_i\\ R_{i+1}\\ \vdots\\ R_n \end{bmatrix} +\det \begin{bmatrix} R_1\\ \vdots\\R_{i-1}\\\lambda R_j\\ R_{i+1}\\ \vdots\\ R_n \end{bmatrix} \\ \amp \textrm{(The second matrix has two proportional rows)}\\ \amp =\det A + 0 \\ \amp =\det A \end{align*}

Subsection 4.4.8 The determinant of elementary matrices

As seen in Definition 3.10.1, there are three types of elementary row operations, and each one has an elementary matrix associated with it. We can now evaluate the determinant of these matrices \(E_1\text{,}\) \(E_2\) and \(E_3\text{.}\)

\(R_i\leftrightarrow R_j\text{:}\) If we interchange \(R_i\) and \(R_j\) of \(E_1\text{,}\) we get the matrix \(I\text{.}\) Hence \(\det E_1=-\det I=-1\text{.}\)
\(R_i\gets \lambda R_i\) with \(\lambda\not=0\text{:}\) \(\det E_2=\det \mathrm{diag} (1,1,\ldots, 1,\lambda,1,\ldots,1)=\lambda\text{.}\)
\(R_i\gets R_i+\lambda R_j\text{:}\) \(\det E_3=1\) since \(E_3\) is triangular.

We may combine these three results into one wonderful theorem:

Theorem 4.4.16. Determinants and elementary row operations.

If \(B\) is derived from \(A\) by one elementary row operation whose elementary matrix is \(E\text{,}\) then

\begin{equation*} B=EA \textrm{ and } \det B= \det (EA) = \det E \det A. \end{equation*}

Proof.

There are three possible elementary row operations, and the equation is valid in each one of them.

Table 4.4.17.

Row operation	matrix determinant	\(\det B\)
\(R_i\leftrightarrow R_j\)	\(\det E_1=-1\)	\(\det B=-\det A\) by Theorem 4.4.7.
\(R_i\gets \lambda R_i\)	\(\det E_2=\lambda\)	\(\det B=\lambda\det A\) by Theorem 4.4.9.
\(R_i\gets R_i+\lambda R_j\)	\(\det E_3=1\)	\(\det B=\det A\) by Theorem 4.4.15.

Example 4.4.18. Using elementary row operations to evaluate a determinant.

We recalculate the determinant from Example 4.3.3. Let

\begin{equation*} A= \begin{bmatrix} 1\amp0\amp-1\amp2\\ 1\amp-1\amp1\amp0\\ 0\amp1\amp-2\amp1\\ -1\amp1\amp0\amp1 \end{bmatrix} \end{equation*}

We apply the two elementary row operations to \(A\text{:}\) \(R_2\gets R_2-R_1\) and \(R_3 \gets R_3+R_1\) to get

\begin{equation*} B= \begin{bmatrix} 1\amp0\amp-1\amp2\\ 0\amp-1\amp2\amp-2\\ 0\amp1\amp-2\amp1\\ 0\amp1\amp-1\amp3 \end{bmatrix} \end{equation*}

From Theorem 4.4.15 we have \(\det A=\det B\text{.}\) Expanding on the first column gives,

\begin{equation*} \det A=\det B = \det \begin{bmatrix} -1\amp2\amp-2\\ 1\amp-2\amp1\\ 1\amp-1\amp3 \end{bmatrix}\text{.} \end{equation*}

Now we use \(R_2\gets R_2+R_1\) and \(R_3\gets R_3+R_1\) to get

\begin{equation*} \det A=\det B = \det \begin{bmatrix} -1\amp2\amp-2\\ 0\amp0\amp-1\\ 0\amp1\amp1 \end{bmatrix} \end{equation*}

Expanding on the first column once again gives

\begin{equation*} \det A=\det B =(-1) \det \begin{bmatrix} 0\amp-1\\ 1\amp1 \end{bmatrix} =-1 \end{equation*}

Subsection 4.4.9 The determinant of invertible matrices

From Theorem 3.11.2 we have a test for matrix invertibility: a matrix \(A\) of order \(n\) is invertible if and only if its reduced row echelon form is \(I\text{,}\) a matrix whose determinant is one. If a matrix is not invertible, then the number of leading ones in the reduced row echelon form is less than \(n\text{,}\) and so the last row is all zero. From Theorem 4.4.1 the determinant of this matrix must be zero.

Lemma 4.4.19. Determinant of the reduced row echelon form.

Let \(B\) be the reduced row echelon form of \(A\text{.}\) Then

\begin{equation*} \det B = \begin{cases} 1 \amp \textrm{if } A \textrm{ is invertible}\\ 0 \amp A \textrm{ otherwise} \end{cases} \end{equation*}

Next, we relate the determinant of a matrix \(A\) to that of its reduced row echelon form \(B\text{.}\) From Theorem 3.10.2 we write

\begin{equation*} B=E_k E_{k-1}E_{k-2}\cdots E_2 E_1 A \end{equation*}

and then note from Theorem 4.4.16 that

\begin{align*} \det B \amp = \det (E_k E_{k-1}E_{k-2}\cdots E_2 E_1 A)\\ \amp = \det E_k \det( E_{k-1}E_{k-2}\cdots E_2 E_1 A)\\ \amp = \det E_k \det E_{k-1}\det(E_{k-2}\cdots E_2 E_1 A)\\ \amp \,\vdots\\ \amp = \det E_k \det E_{k-1}\det E_{k-2}\cdots \det E_2 \det E_1 \det A \end{align*}

We further note that for any elementary matrix \(E\text{,}\) we have

\begin{equation*} \det E = \begin{cases} -1 \amp \textrm{ for } R_i\leftrightarrow R_j\\ \lambda \amp \textrm{ for } R_i\gets \lambda R_i \textrm{ where } \lambda\not=0\\ 1 \amp \textrm{ for } R_i\gets R_i+\lambda R_j\\ \end{cases} \end{equation*}

In particular, this means that for any elementary matrix \(E\) we have \(\det E\not=0\text{,}\) and so we have:

\begin{equation*} \det B = \underbrace{\det E_k \det E_{k-1} \cdots \det E_2 \det E_1}_{\not=0} \det A \end{equation*}

and so \(\det B=0\) if and only if \(\det A=0\text{.}\) In summary:

Theorem 4.4.20. Invertible matrices have nonzero determinants.

\(A\) is invertible if and only if \(\det A \not= 0\text{.}\)

We can now add an extra condition to Theorem 3.11.2.

Theorem 4.4.21. Equivalent forms of invertibility.

Suppose that \(A\) is an \(n\times n\) square matrix. Then the following statements are equivalent:

\(A\) is invertible
\(A\vec x=0\) if and only if \(\vec x=0\)
The reduced row echelon form of \(A\) is \(I_n\)
\(A\) is a product of elementary matrices
\(A\vec x=\vec b\) is consistent for any \(\vec b\)
\(A\vec x=\vec b\) has exactly one solution for any \(\vec b\)
\(\displaystyle \det A\not=0\)

Subsection 4.4.10 The determinant of the product of two matrices

In Theorem 4.4.16 we proved that \(\det E \det A = \det(EA)\) for any elementary matrix \(E\text{.}\) In other words, in this case the determinant of the product is the product of the determinants. We can now show that this is true for any pair of matrices.

Theorem 4.4.22. The determinant of \(AB\).

For any square matrices \(A\) and \(B\) of the same size

\begin{equation*} \det(AB)=\det A\; \det B. \end{equation*}

Proof.

We proceed by considering three cases:

\(\det B=0\text{:}\) In this case \(\det A\;\det B=0\text{.}\) In addition, from Theorem 4.4.21, The is an \(\vec x\not=0\) so that \(B\vec x=0\text{.}\) Then \(AB\vec x=0\) and so \(\det(AB)=0\text{.}\) This gives

\begin{equation*} \det(AB)=0=\det A\;\det B. \end{equation*}
\(\det B\not=0\) and \(\det A=0\text{:}\) Once again, \(\det A \det B=0\text{.}\) Again, using Theorem 4.4.21 (twice), there is \(\vec y\not=0\) so that \(A\vec y=0\) and there is an \(\vec x\) so that \(B\vec x=\vec y\text{.}\) Notice that \(\vec x\not=0\text{,}\) for otherwise \(\vec y=B\vec x=0\text{.}\) We then have \(AB\vec x=A\vec y=0\) with \(\vec y\not=0\text{,}\) and so \(\det(AB)=0\text{,}\) which, once again gives

\begin{equation*} \det(AB)=0=\det A\;\det B. \end{equation*}
\(\det B\not=0\) and \(\det A\not=0\text{:}\) Once again, using Theorem 4.4.21,

\begin{equation*} A=F_1 F_2\cdots F_k \textrm{, a product of elementary matrices.} \end{equation*}

Using Theorem 4.4.16 repeatedly,

\begin{equation*} \det A= \det F_1\;\det F_2\cdots\det F_k \end{equation*}

and so

\begin{equation*} \det A\; \det B=\det F_1\;\det F_2\cdots\det F_k \det B \end{equation*}

But also by using Theorem 4.4.16 repeatedly,

\begin{align*} \det(AB) \amp =\det(F_1F_2\cdots F_k B)\\ \amp =\det F_1\;\det F_2\cdots\det F_k\;\det B\\ \amp =\det A\;\det B \end{align*}

Corollary 4.4.23. The determinant of the inverse of \(A\).

If \(A\) is an invertible matrix, then

\begin{equation*} \det A^{-1}=\frac 1{\det A}. \end{equation*}

Proof.

Since \(A A^{-1}=I\text{,}\) we have \(\det A\;\det A^{-1}=\det I=1\text{.}\)

Notice how this reinforces the idea that an invertible matrix must have a nonzero determinant.

Subsection 4.4.11 The determinant of the transpose of \(A\)

As discussed in Subsection 4.4.8, the determinants of the three types of elementary matrices can be evaluated easily. In two cases the matrices are symmetric, and in the third case it is triangular. This leads to an easy result:

Lemma 4.4.24. The determinant of the transpose of an elementary matrix.

For any elementary matrix \(E\text{,}\)

\begin{equation*} \det(E^T)=\det E \end{equation*}

We now extend this result to all matrices.

Theorem 4.4.25. The determinant of the transpose of \(A\).

\begin{equation*} \det A = \det A^T \end{equation*}

Proof.

If \(B\) is the reduced row echelon form of \(A\text{,}\) then

\begin{equation*} B=E_k E_{k-1}\cdots E_2 E_1 A \end{equation*}

and

\begin{equation*} B^T=A^T E_1^T E_2^T\cdots E_{k-1}^T E_k^T. \end{equation*}

Hence

\begin{equation*} \det B=\det E_k \det E_{k-1}\cdots \det E_2 \det E_1 \det A \end{equation*}

and

\begin{equation*} \det B^T=\det A^T \det E_1^T \det E_2^T\cdots \det E_{k-1}^T \det E_k^T. \end{equation*}

There are two possibilities for \(B\text{:}\)

When \(A\) is invertible, \(B=B^T=I\) and so \(\det B= \det B^T=1\text{.}\)
When \(A\) is singular, \(B\) has an all zero last row and \(B^T\) has an all zero last column. This implies \(\det B=\det B^T=0\)

In either case we have \(\det B=\det B^T\text{,}\) and so we can equate the values given above to get

\begin{equation*} \det E_k \cdots \det E_1 \det A= \det A^T \det E_1^T \cdots \det E_k^T. \end{equation*}

Using Lemma 4.4.24, we have \(\det E_j=\det E_j^T\) for \(j=1,2,\ldots,k.\) Hence \(\det A=\det A^T.\)