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## Properties of Determinants

Theorem. For an $$n\times n$$ matrix $$A$$ and $$n\times n$$ elementary matrices $$E_{ij}$$, $$E_i(c)$$, $$E_{ij}(c)$$, we have $$\det E_{ij}=-1$$, $$\det E_i(c)=c$$, $$\det E_{ij}(c)=1$$, and \begin{align*} \det (E_{ij}A)& =-\det A=(\det E_{ij}) (\det A),\\ \det (E_i(c)A)& =c\det A=(\det E_i(c)) (\det A),\\ \det(E_{ij}(c)A)& =\det A=(\det E_{ij}(c)) (\det A). \end{align*}

Use cofactor expansion and induction on $$n$$.

Theorem. Let $$A$$ be an $$n\times n$$ matrix. Then $$A$$ is invertible if and only if $$\det(A)\neq 0$$.

Suppose $$A$$ is invertible. Then $$A^{-1}$$ is invertible and there are elementary matrices $$E_1,E_2,\ldots,E_k$$ such that $$E_kE_{k-1}\cdots E_1A^{-1}=I_n$$. Postmultiplying by $$A$$, we get $E_kE_{k-1}\cdots E_1 =A \implies \det(E_kE_{k-1}\cdots E_1)=\det(A).$ By successively applying the preceding theorem, we get $\det(A)=\det(E_kE_{k-1}\cdots E_1)=\det(E_k)\det(E_{k-1})\cdots\det(E_1)\neq 0.$ For the converse, suppose that $$A$$ is not invertible. Then the RREF $$R$$ of $$A$$ is not $$I_n$$. So $$R$$ is an upper-triangular matrix with the last row being a zero row and consequently $$\det(R)=0$$. Suppose $$E_1',E_2',\ldots,E_t'$$ are elementary matrices for which $$E_t'E_{t-1}'\cdots E_1'A=R$$. Then $\det(E_t'E_{t-1}'\cdots E_1'A)=\det(R)=0 \implies \det(E_t')\det(E_{t-1}')\cdots\det(E_1')\det(A)=0,$ by the preceding theorem. Since $$\det(E_i')\neq 0$$ for $$i=1,2,\ldots,t$$, $$\det(A)= 0$$.

Remark. We extend the IMT by adding one more equivalent condition:

• (a) $$A$$ is invertible.

• (m) $$\det(A)\neq 0$$.

Theorem. Let $$A$$ and $$B$$ be two $$n\times n$$ matrices. Then $$\det(AB)=\det(A)\det(B)$$.

Case 1. $$A$$ is not invertible.
By the IMT, $$\operatorname{rank}(A) < n$$. Since $$\operatorname{CS}\left(AB\right)\subseteq \operatorname{CS}\left(A\right)$$, $$\operatorname{rank}(AB)\leq \operatorname{rank}(A) < n$$ and consequently $$AB$$ is also not invertible. By the IMT, $$\det(A)=0$$ and $$\det(AB)=0$$. Thus $\det(AB)=0=\det(A)\det(B).$
Case 2. $$A$$ is invertible.
There are elementary matrices $$E_1,E_2,\ldots,E_k$$ such that $$E_kE_{k-1}\cdots E_1=A$$. Postmultiplying by $$B$$, we get $$AB=E_kE_{k-1}\cdots E_1B$$. By successively applying the first theorem of this section, we get \begin{align*} \det(AB) &=\det(E_kE_{k-1}\cdots E_1B)\\ &=\det(E_k)\det(E_{k-1})\cdots\det(E_1)\det(B)\\ &=\det(E_kE_{k-1}\cdots E_1)\det(B)\\ &=\det(A)\det(B). \end{align*}

Corollary. Let $$A$$ be an $$n\times n$$ matrix.

1. For all scalars $$c$$, $$\det(cA)=\det(cI_nA)=\det(cI_n)\det(A)=c^n\det(A)$$.

2. If $$A$$ is invertible, then $$\det(A) \det(A^{-1})=\det(AA^{-1})=\det(I_n)=1$$.

Example. $$A=\left[\begin{array}{rrr} 1&2&3\\ 3&5&1\\ 0&0&2 \end{array} \right].$$ Is $$A$$ invertible? Compute $$\det(A^T)$$, $$\det(4A^5)$$, and $$\det(A(^{-1})$$.

Solution. Since $$\det(A)=-2\neq 0$$, $$A$$ is invertible and we have $$\det(A^T)=\det(A)=-2$$, $$\det(4A^5)=4^3(\det A)^5=-2048$$, and $$\det(A^{-1})=(\det A)^{-1}=(-2)^{-1}=-1/2$$.

Theorem.(Cramer's Rule) Let $$A$$ be an $$n\times n$$ invertible matrix and $$\overrightarrow{b}\in \mathbb R^n$$. The unique solution $$\overrightarrow{x}=[x_1,x_2,\ldots,x_n]^T$$ of $$A\overrightarrow{x}=\overrightarrow{b}$$ is given by $x_i=\frac{\det(A_i(\overrightarrow{b}))}{\det(A)}, \;i=1,2,\ldots,n,$ where $$A_i(\overrightarrow{b})$$ is the matrix obtained from $$A$$ by replacing its $$i$$th column by $$\overrightarrow{b}$$.

Let $$i\in \{1,2,\ldots,n\}$$. Note that \begin{align*} A[\overrightarrow{e_1}\cdots \overrightarrow{e_{i-1}} \overrightarrow{x} \overrightarrow{e_{i+1}}\cdots \overrightarrow{e_n}] &= [A\overrightarrow{e_1}\cdots A\overrightarrow{e_{i-1}} A\overrightarrow{x} A\overrightarrow{e_{i+1}}\cdots A\overrightarrow{e_n}]\\ &=[\overrightarrow{A_1}\cdots \overrightarrow{A_{i-1}}\; A\overrightarrow{x}\; \overrightarrow{A_{i+1}}\cdots \overrightarrow{A_n}]\\ &=[\overrightarrow{A_1}\cdots \overrightarrow{A_{i-1}}\; \overrightarrow{b}\; \overrightarrow{A_{i+1}}\cdots \overrightarrow{A_n}]\\ &=A_i(\overrightarrow{b}). \end{align*} Since $$\det([\overrightarrow{e_1}\cdots \overrightarrow{e_{i-1}} \overrightarrow{x} \overrightarrow{e_{i+1}}\cdots \overrightarrow{e_n}])=x_i$$, \begin{align*} \det(A_i(\overrightarrow{b})) &=\det(A[\overrightarrow{e_1}\cdots \overrightarrow{e_{i-1}} \overrightarrow{x} \overrightarrow{e_{i+1}}\cdots \overrightarrow{e_n}])\\ &=\det(A)\det([\overrightarrow{e_1}\cdots \overrightarrow{e_{i-1}} \overrightarrow{x} \overrightarrow{e_{i+1}}\cdots \overrightarrow{e_n}])\\ &=\det(A)x_i. \end{align*} Thus $$x_i=\frac{\det(A_i(\overrightarrow{b}))}{\det(A)}$$.

Example. We solve $$A\overrightarrow{x}=\overrightarrow{b}$$ by Cramer's Rule, where $A=\left[\begin{array}{rrr} 1&0&2\\3&2&5\\1&1&4 \end{array}\right], \overrightarrow{x}=\left[\begin{array}{c}x_1\\x_2\\ x_3 \end{array} \right], \mbox{ and } \overrightarrow{b}=\left[\begin{array}{c} 1\\8\\1 \end{array} \right].$ Since $$\det(A)=5\neq 0$$, there is a unique solution $$[x_1,x_2,x_3]^T$$ and by Cramer's Rule, \begin{align*} x_1 &=\frac{\det(A_1(\overrightarrow{b}))}{\det(A)}=\frac{ \left|\begin{array}{rrr} 1&0&2\\8&2&5\\1&1&4 \end{array}\right| }{5}=\frac{15}{5}=3\\ x_2 &=\frac{\det(A_2(\overrightarrow{b}))}{\det(A)}=\frac{ \left|\begin{array}{rrr} 1&1&2\\3&8&5\\1&1&4 \end{array}\right| }{5}=\frac{10}{5}=2\\ x_3 &=\frac{\det(A_3(\overrightarrow{b}))}{\det(A)}=\frac{ \left|\begin{array}{rrr} 1&0&1\\3&2&8\\1&1&1 \end{array}\right| }{5}=\frac{-5}{5}=-1. \end{align*} Thus the unique solution is $$[x_1,x_2,x_3]^T=[3,2,-1]^T$$.

Definition. Let $$A$$ be an $$n\times n$$ matrix. The cofactor matrix, denoted by $$C=[c_{ij}]$$, is an $$n\times n$$ matrix where $$c_{ij}$$ is the $$(i,j)$$ cofactor of $$A$$. The adjoint or adjugate of $$A$$, denoted by $$\operatorname{adj} A$$ or $$\operatorname{adj}(A)$$, is the transpose of the cofactor matrix of $$A$$, i.e., $$\operatorname{adj} A=C^T$$.

Theorem. Let $$A$$ be an $$n\times n$$ invertible matrix. Then $A^{-1}=\frac{1}{\det(A)}\operatorname{adj} A.$

Since $$AA^{-1}=I_n$$, $$A(\text{the column \(j$$ of } A^{-1})=\overrightarrow{e_j}\). By Cramer's Rule, the $$(i,j)$$-entry of $$A^{-1}$$, i.e., the $$i$$th entry of the column $$j$$ of $$A^{-1}$$, is $\frac{\det(A_i(\overrightarrow{e_j}))}{\det(A)}=\frac{(-1)^{i+j}\det(A(j,i))}{\det(A)}=\frac{c_{ji}}{\det(A)}=\frac{(C^T)_{ij}}{\det(A)}=\frac{(\operatorname{adj} A)_{ij}}{\det(A)}.$

Example.

1. For invertible $$A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$$, $A^{-1}=\frac{1}{\det(A)}\operatorname{adj} A =\frac{1}{ad-bc}\left[\begin{array}{cc} c_{11}&c_{12}\\c_{21}&c_{22}\end{array}\right]^T =\frac{1}{ad-bc} \left[\begin{array}{rr}d&-c\\-b&a\end{array}\right]^T =\frac{1}{ad-bc} \left[\begin{array}{rr}d&-b\\-c&a\end{array}\right].$

2. For invertible $$A=\left[\begin{array}{rrr} 1&0&2\\3&2&5\\1&1&4\end{array}\right]$$, $A^{-1}=\frac{1}{\det(A)}\operatorname{adj} A =\frac{1}{5} \left[\begin{array}{ccc} c_{11}&c_{12}&c_{13}\\c_{21}&c_{22}&c_{23}\\c_{31}&c_{32}&c_{33}\end{array}\right]^T =\frac{1}{5} \left[\begin{array}{rrr} 3&-7&1\\2&2&-1\\-4&1&2\end{array}\right]^T =\frac{1}{5} \left[\begin{array}{rrr} 3&2&-4\\-7&2&1\\1&-1&2\end{array}\right].$

We end by the following useful multilinear property of determinant:
Theorem. Let $$A=[\overrightarrow{a_1}\:\overrightarrow{a_2}\:\cdots\overrightarrow{a_n}]$$ be an $$n\times n$$ matrix. Then for all $$\overrightarrow{x},\overrightarrow{y}\in \mathbb R^n$$ and for all scalars $$c,d$$, \begin{align*} det [\overrightarrow{a_1}\:\cdots \overrightarrow{a_{i-1}}\: (c\overrightarrow{x}+d\overrightarrow{y}) \:\overrightarrow{a_{i+1}} \cdots \overrightarrow{a_n}] &=c\det [\overrightarrow{a_1}\:\cdots \overrightarrow{a_{i-1}}\: \overrightarrow{x} \:\overrightarrow{a_{i+1}} \cdots \overrightarrow{a_n}]\\ &+d\det [\overrightarrow{a_1}\:\cdots \overrightarrow{a_{i-1}}\: \overrightarrow{y} \:\overrightarrow{a_{i+1}} \cdots \overrightarrow{a_n}]. \end{align*}

Find determinants by the cofactor expansion along the $$i$$th column.

Example. \begin{align*} \left|\begin{array}{cc}3a+4s&3b+4t\\c&d\end{array}\right| &=\left|\begin{array}{cc}3a+4s&c\\3b+4t&d\end{array}\right| (\text{by transposing})\\ &=\left|\begin{array}{cc}3a&c\\3b&d\end{array}\right| +\left|\begin{array}{cc}4s&c\\4t&d\end{array}\right| (\text{by multilinearity of determinant})\\ &=3\left|\begin{array}{cc}a&c\\b&d\end{array}\right| +4\left|\begin{array}{cc}s&c\\t&d\end{array}\right| (\text{by multilinearity of determinant})\\ &=3(ad-cb)+4(sd-ct). \end{align*}

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