Properties of Determinant

THEOREM 15.3.1 (Properties of Determinant)   Let $A=[a_{ij}]$ be an $n \times n$ matrix. Then

1. if $B$ is obtained from $A$ by interchanging two rows, then
   $\det (B) = - \det (A).$
2. if $B$ is obtained from $A$ by multiplying a row by $c$ then
   $\det (B) = c \det (A).$
3. if all the elements of one row is 0 then $\det (A) = 0.$
4. if $A$ is a square matrix having two rows equal then $\det (A) = 0.$
5. Let $B= [b_{ij}]$ and $C = [c_{ij}]$ be two matrices which differ from the matrix $A=[a_{ij}]$ only in the $m^{\mbox{th}}$ row for some $m$ . If $c_{mj} = a_{mj} + b_{mj}$ for $1 \le j \le n$ then $\det(C) = \det(A) + \det(B)$ .
6. if $B$ is obtained from $A$ by replacing the $\ell$ th row by itself plus $k$ times the $m$ th row, for $\ell \neq m$ then $\det (B) = \det (A).$
7. if $A$ is a triangular matrix then $\det (A) = a_{11} a_{22} \cdots a_{nn},$ the product of the diagonal elements.
8. If $E$ is an elementary matrix of order $n$ then $\det(EA) = \det(E)\det(A)$ .
9. $A$ is invertible if and only if $\det(A) \ne 0$ .
10. If $B$ is an $n \times n$ matrix then $\det(A B) = \det(A) \det(B)$ .
11. $\det(A) = \det(A^t)$ , where recall that $A^t$ is the transpose of the matrix $A$ .

Proof. Proof of Part 1. Suppose $B= [b_{ij}]$ is obtained from $A=[a_{ij}]$ by the interchange of the $\ell^{\mbox{th}}$ and $m^{\mbox{th}}$ row. Then $b_{\ell j} = a_{m j}, \; b_{m j} = a_{\ell j}$ for $1 \le j \le n$ and $b_{ij} = a_{ij}$ for $1 \le i \ne \ell, m \le n, \; 1 \le j \le n$ .

Let $\tau = (\ell \; m)$ be a transposition. Then by Proposition 14.2.4, ${\mathcal S}_n = \{ \sigma \circ \tau: \; \sigma \in {\mathcal S}_n \}$ . Hence by the definition of determinant and Example 14.2.14.2, we have

$$\det(B) = \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\sigma) \pro... ...} {\mbox{sgn}}(\sigma\circ\tau) \prod\limits_{i=1}^n b_{i (\sigma\circ\tau)(i)}$$

$$= \sum\limits_{\sigma\circ\tau \in {\mathcal S}_n} {\mbox{sgn}}(\ta... ...tau)(\ell)} \cdots b_{m (\sigma\circ\tau)(m)} \cdots b_{n (\sigma\circ\tau)(n)}$$

$$= {\mbox{sgn}}(\tau ) \sum\limits_{\sigma \in {\mathcal S}_n} {\mbo... ...(2)} \cdots b_{\ell \sigma(m)} \cdots b_{m \sigma(\ell)} \cdots b_{n \sigma(n)}$$

$$= - \left(\sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\sig... ... \sigma(\ell)} \cdots a_{n \sigma(n)} \right) \;\;\; {\mbox{ as sgn}}(\tau)= -1$$

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

Proof of Part 2. Suppose that $B= [b_{ij}]$ is obtained by multiplying the $m^{\mbox{th}}$ row of $A$ by $c \ne 0$ . Then $b_{mj} = c \;a_{mj}$ and $b_{ij} = a_{ij}$ for $1 \le i \ne m \le n, \; 1 \le j \le n$ . Then

$$\det(B) = \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\sigma) b_{1 \sigma(1)} b_{2 \sigma(2)} \cdots b_{m \sigma(m)} \cdots b_{n \sigma(n)}$$

$$= \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\sigma) a_{1 \sigma(1)} a_{2 \sigma(2)} \cdots c a_{m \sigma(m)} \cdots a_{n \sigma(n)}$$

$$= c \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\sigma) a_{1 \sigma(1)} a_{2 \sigma(2)} \cdots a_{m \sigma(m)} \cdots a_{n \sigma(n)}$$

$$= c \det(A).$$

Proof of Part 3. Note that $\det(A) = \sum\limits_{\sigma\in {\mathcal S}_n} {\mbox{sgn}}(\sigma) a_{1 \sigma(1)} a_{2\sigma(2)} \ldots a_{n\sigma(n)}$ . So, each term in the expression for determinant, contains one entry from each row. Hence, from the condition that $A$ has a row consisting of all zeros, the value of each term is 0 . Thus, $\det (A) = 0$ .

Proof of Part 4. Suppose that the $\ell^{\mbox{th}}$ and $m^{\mbox{th}}$ row of $A$ are equal. Let $B$ be the matrix obtained from $A$ by interchanging the $\ell^{\mbox{th}}$ and $m^{\mbox{th}}$ rows. Then by the first part, $\det (B) = - \det (A).$ But the assumption implies that $B = A$ . Hence, $\det (B) = \det (A)$ . So, we have $\det(B) = - \det(A) = \det(A).$ Hence, $\det (A) = 0$ .

Proof of Part 5. By definition and the given assumption, we have

$$\det(C) = \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\sigma) c_{1 \sigma(1)} c_{2 \sigma(2)} \cdots c_{m \sigma(m)} \cdots c_{n \sigma(n)}$$

$$= \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\sigma) c_{1... ...{2 \sigma(2)} \cdots (b_{m \sigma(m)} + a_{m \sigma(m)}) \cdots c_{n \sigma(n)}$$

$$= \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\sigma) b_{1 \sigma(1)} b_{2 \sigma(2)} \cdots b_{m \sigma(m)} \cdots b_{n \sigma(n)}$$

$$\hspace{1in} + \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn... ...) a_{1 \sigma(1)} a_{2 \sigma(2)} \cdots a_{m \sigma(m)} \cdots a_{n \sigma(n)}$$

$$= \det(B) + \det(A).$$

Proof of Part 6. Suppose that $B= [b_{ij}]$ is obtained from $A$ by replacing the $\ell$ th row by itself plus $k$ times the $m$ th row, for $\ell \neq m$ . Then $b_{\ell j} = a_{\ell j} + k \;a_{mj}$ and $b_{ij} = a_{ij}$ for $1 \le i \ne m \le n, \; 1 \le j \le n$ . Then

$$\det(B) = \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\sigma) b_{1... ...(2)} \cdots b_{\ell \sigma(\ell)} \cdots b_{m \sigma(m)} \cdots b_{n \sigma(n)}$$

$$= \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\sigma) a_{1... ...sigma(\ell)} + k a_{m \sigma(m)}) \cdots a_{m \sigma(m)} \cdots a_{n \sigma(n)}$$

$$= \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\sigma) a_{1... ...(2)} \cdots a_{\ell \sigma(\ell)} \cdots a_{m \sigma(m)} \cdots a_{n \sigma(n)}$$

$$\hspace{1in} + k \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{s... ...\sigma(2)} \cdots a_{m \sigma(m)} \cdots a_{m \sigma(m)} \cdots a_{n \sigma(n)}$$

$$= \sum\limits_{\sigma \in {\mathcal S... ...s a_{n \sigma(n)} \hspace{.25in} {\mbox{ use Part }}~\ref{app:lem:tworows:same}$$

$$= \det(A).$$

Proof of Part 7. First let us assume that $A$ is an upper triangular matrix. Observe that if $\sigma \in {\mathcal S}_n$ is different from the identity permutation then $n(\sigma) \ge 1$ . So, for every $\sigma \ne Id_n \in {\mathcal S}_n$ , there exists a positive integer $m, \; 1 \le m \le n-1$ (depending on $\sigma$ ) such that $m > \sigma(m)$ . As $A$ is an upper triangular matrix, $a_{m \sigma(m) } = 0$ for each $\sigma (\ne Id_n) \in {\mathcal S}_n$ . Hence the result follows.

A similar reasoning holds true, in case $A$ is a lower triangular matrix.

Proof of Part 8. Let $I_n$ be the identity matrix of order $n$ . Then using Part 7, $\det(I_n) = 1$ . Also, recalling the notations for the elementary matrices given in Remark 2.3.14, we have $\det(E_{ij}) = -1,$ (using Part 1) $\; \det(E_i(c)) = c$ (using Part 2) and $\det(E_{ij}(k) = 1$ (using Part 6). Again using Parts 1, 2 and 6, we get $\det(EA) = \det(E)\det(A)$ .

Proof of Part 9. Suppose $A$ is invertible. Then by Theorem 2.5.8, $A$ is a product of elementary matrices. That is, there exist elementary matrices $E_1, E_2, \ldots, E_k$ such that $A = E_1 E_2 \cdots E_k$ . Now a repeated application of Part 8 implies that $\det(A) = \det(E_1) \det(E_2) \cdots \det(E_k)$ . But $\det(E_i) \ne 0$ for $1 \le i \le k$ . Hence, $\det(A) \ne 0$ .

Now assume that $\det(A) \ne 0$ . We show that $A$ is invertible. On the contrary, assume that $A$ is not invertible. Then by Theorem 2.5.8, the matrix $A$ is not of full rank. That is there exists a positive integer $r < n$ such that ${\mbox{rank }}(A) = r$ . So, there exist elementary matrices $E_1, E_2, \ldots, E_k$ such that $E_1 E_2 \cdots E_k A = \left[\begin{array}{c} B \\ {\mathbf 0}\end{array}\right]$ . Therefore, by Part 3 and a repeated application of Part 8,

$$\det(E_1) \det(E_2) \cdots \det(E_k) \det(A) = \det(E_1 E_2 \cdot... ...et \left( \left[\begin{array}{c} B \\ {\mathbf 0}\end{array}\right] \right)= 0.$$

But $\det(E_i) \ne 0$ for $1 \le i \le k$ . Hence, $\det (A) = 0$ . This contradicts our assumption that $\det(A) \ne 0$ . Hence our assumption is false and therefore $A$ is invertible.

Proof of Part 10. Suppose $A$ is not invertible. Then by Part 9, $\det (A) = 0$ . Also, the product matrix $A B$ is also not invertible. So, again by Part 9, $\det(A B) = 0$ . Thus, $\det(A B) = \det(A) \det(B)$ .

Now suppose that $A$ is invertible. Then by Theorem 2.5.8, $A$ is a product of elementary matrices. That is, there exist elementary matrices $E_1, E_2, \ldots, E_k$ such that $A = E_1 E_2 \cdots E_k$ . Now a repeated application of Part 8 implies that

$$\det(AB) = \det ( E_1 E_2 \cdots E_k B) = \det(E_1) \det(E_2) \cdots \det(E_k) \det(B)$$

$$= \det ( E_1 E_2 \cdots E_k) \det( B) = \det(A) \det(B).$$

Proof of Part 11. Let $B = [b_{ij}]= A^t$ . Then $b_{ij} = a_{ji}$ for $1 \le i, j \le n$ . By Proposition 14.2.4, we know that ${\mathcal S}_n = \{\sigma^{-1} : \; \sigma \in {\mathcal S}_n\}$ . Also ${\mbox{sgn}}(\sigma) = {\mbox{sgn}}(\sigma^{-1})$ . Hence,

$$\det(B) = \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\sigma) b_{1 \sigma(1)} b_{2 \sigma(2)} \cdots b_{n \sigma(n)}$$

$$= \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\sigma^{-1}) b_{\sigma^{-1}(1) \;1} \; b_{\sigma^{-1}(2) \;2} \cdots b_{\sigma^{-1}(n)\; n}$$

$$= \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\sigma^{-1}) a_{1 \sigma^{-1}(1)} b_{2 \sigma^{-1}(2)} \cdots b_{n \sigma^{-1}(n)}$$

$$= \det(A).$$

height6pt width 6pt depth 0pt

Remark 15.3.2  

1. The result that $\det(A) = \det(A^t)$ implies that in the statements made in Theorem 14.3.1, where ever the word ``row" appears it can be replaced by ``column".
2. Let $A=[a_{ij}]$ be a matrix satisfying $a_{11} = 1$ and $a_{1j} = 0$ for $2 \le j \le n$ . Let $B$ be the submatrix of $A$ obtained by removing the first row and the first column. Then it can be easily shown that $\det(A) = \det(B)$ . The reason being is as follows:
   for every $\sigma \in {\mathcal S}_n$ with $\sigma(1) = 1$ is equivalent to saying that $\sigma$ is a permutation of the elements $\{2, 3, \ldots, n\}$ . That is, $\sigma \in {\mathcal S}_{n-1}$ . Hence,
   

   $$\det(A) = \sum\limits_{\sigma \in {\mathcal S}_n} {\mbox{sgn}}(\sigma) a_{1... ...}_n, \sigma(1) = 1} {\mbox{sgn}}(\sigma) a_{2 \sigma(2)} \cdots a_{n \sigma(n)}$$

   $$= \sum\limits_{\sigma \in {\mathcal S}_{n-1}} {\mbox{sgn}}(\sigma) b_{1 \sigma(1)} \cdots b_{n \sigma(n)} = \det(B).$$

   
   

We are now ready to relate this definition of determinant with the one given in Definition 2.6.2.

THEOREM 15.3.3   Let $A$ be an $n \times n$ matrix. Then $\det(A) = \sum\limits_{j=1}^n (-1)^{1 + j} a_{1j} \det\bigl(A(1\vert j)\bigr),$ where recall that $A(1\vert j)$ is the submatrix of $A$ obtained by removing the $1^{\mbox{st}}$ row and the $j^{\mbox{th}}$ column.

Proof. For $1 \le j \le n$ , define two matrices

$$B_j = \begin{bmatrix}0 & 0 & \cdots & a_{1j} & \cdots & 0 \\ a_{2... ...dots \\ a_{nj} & a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}_{n \times n}.$$

Then by Theorem 14.3.1.5,

$$\det(A) = \sum\limits_{j=1}^n \det(B_j).$$ (15.3.6)

We now compute $\det(B_j)$ for $1 \le j \le n$ . Note that the matrix $B_j$ can be transformed into $C_j$ by $j-1$ interchanges of columns done in the following manner:
first interchange the $1^{\mbox{st}}$ and $2^{\mbox{nd}}$ column, then interchange the $2^{\mbox{nd}}$ and $3^{\mbox{rd}}$ column and so on (the last process consists of interchanging the $(j-1)^{\mbox{th}}$ column with the $j^{\mbox{th}}$ column. Then by Remark 14.3.2 and Parts 1 and 2 of Theorem 14.3.1, we have $\det(B_j) = a_{1j} (-1)^{j-1} \det(C_j)$ . Therefore by (14.3.6),

$$\det(A) = \sum\limits_{j=1}^n (-1)^{j-1} a_{1j} \det\bigl( A(1\vert j)\bigr)=\sum\limits_{j=1}^n (-1)^{j+1} a_{1j} \det\bigl( A(1\vert j)\bigr).$$

height6pt width 6pt depth 0pt