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## Determinant of a Matrix

In this section we study the determinant of an $$n\times n$$ matrix $$A=[a_{ij}]$$, denoted by $$\det(A)$$ or $$\det A$$ or $$|A|$$ or $\left| \begin{array}{cccc} a_{11}&a_{12}&\cdots &a_{1n}\\ a_{21}&a_{22}&\cdots &a_{2n}\\ \vdots&\vdots& \ddots &\vdots\\ a_{m1}&a_{m2}&\cdots &a_{mn} \end{array} \right|.$

To define $$\det(A)$$ recursively, we denote $$A(i,j)$$ for the the matrix obtained from $$A$$ by deleting row $$i$$ and column $$j$$ of $$A$$.

Definition. If $$A=[a_{11}]$$, then $$\det(A)=a_{11}$$. If $$A=\left[\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right]$$, then $$\det(A)=a_{11}a_{22}-a_{12}a_{21}$$. For an $$n\times n$$ matrix $$A=[a_{ij}]$$ where $$n\geq 3$$, $\det(A)=\sum_{i=1}^n (-1)^{1+i} a_{1i} \det A(1,i)=a_{11} \det A(1,1)-a_{12}\det A(1,2)+\cdots+(-1)^{n+1} a_{1n} \det A(1,n).$

Example. We find $$\det(A)$$ for $$A=\left[\begin{array}{rrr} 1&2&3\\ 1&3&5\\ 1&4&2\end{array} \right]$$. \begin{align*} \det(A) &=a_{11} \det A(1,1)-a_{12}\det A(1,2)+a_{13} \det A(1,3)\\ &=1 \left| \begin{array}{rr}3&5\\4&2\end{array} \right| -2 \left| \begin{array}{rrr}1&5\\1&2\end{array} \right| +3 \left| \begin{array}{rr}1&3\\1&4\end{array} \right| \\ &=1(3\cdot 2-5\cdot 4)-2(1\cdot 2-5\cdot 1) +3(1\cdot 4-3\cdot 1)\\ &=-5 \end{align*}

Definition. For an $$n\times n$$ matrix $$A=[a_{ij}]$$ where $$n\geq 2$$, the $$(i,j)$$ minor, denoted by $$m_{ij}$$, is $$m_{ij}=\det A(i,j)$$ and the $$(i,j)$$ cofactor, denoted by $$c_{ij}$$, is $c_{ij}=(-1)^{i+j} m_{ij} =(-1)^{i+j}\det A(i,j).$

Remark. We defined $$\det(A)$$ as the cofactor expansion along the first row of $$A$$: $\det(A)=\sum_{i=1}^n (-1)^{1+i}a_{1i} \det A(1,i)= \sum_{i=1}^n a_{1i} c_{1i}.$ But it can be proved that $$\det(A)$$ is the cofactor expansion along any row or column of $$A$$.

Theorem. Let $$A$$ be an $$n\times n$$ matrix. Then for each $$i,j=1,2,\ldots,n$$, $\det(A)= \sum_{j=1}^n a_{ij} c_{ij} = \sum_{i=1}^n a_{ij} c_{ij} .$

The preceding theorem can be proved using the following equivalent definition of determinant: $\det(A)=\sum_{\sigma \in S_n} \left( \operatorname{sign}(\sigma) \prod_{i=1}^n a_{i\sigma(i)} \right),$ where $$\sigma$$ runs over all $$n!$$ permutations $$\sigma$$ of $$\{1,2,\ldots,n\}$$. (This requires study of permutations)

Corollary. Let $$A=[a_{ij}]$$ be an $$n\times n$$ matrix.

1. $$\det(A^T)=\det(A)$$.

2. If $$A$$ is a triangular matrix, then $$\det(A)=a_{11}a_{22}\cdots a_{nn}$$.

(a) Note that the $$(i,j)$$ cofactor of $$A$$ is the $$(j,i)$$ cofactor of $$A^T$$. The cofactor expansions along the first rows to get $$\det(A)$$ would be same as cofactor expansions along the first columns to get $$\det(A^T)$$.

(b) If $$A$$ is an upper-triangular matrix, then by cofactor expansions along the first rows we get $$\det(A)=a_{11}a_{22}\cdots a_{nn}$$. Similarly if $$A$$ is a lower-triangular matrix, then by cofactor expansions along the first columns we get $$\det(A)=a_{11}a_{22}\cdots a_{nn}$$.

Example. $$A=\left[\begin{array}{rrrrr} 1&2&3&4&5\\ 3&0&1&3&2\\ 0&0&4&3&0\\ 0&0&0&2&1\\ 2&0&0&0&3 \end{array} \right].$$ We compute $$\det(A)$$ using rows or columns with maximum number of zeros at a step. So first we choose column 2 and do cofactor expansion along it: $\det(A)=-2 \left|\begin{array}{rrrr} 3&1&3&2\\ 0&4&3&0\\ 0&0&2&1\\ 2&0&0&3 \end{array} \right|$ Now we have 5 choices: row 2,3,4 and column 1,2. We do cofactor expansion along row 4: $\det(A)=-2 \left( -2 \left|\begin{array}{rrr} 1&3&2\\ 4&3&0\\ 0&2&1 \end{array} \right| +3 \left|\begin{array}{rrr} 3&1&3\\ 0&4&3\\ 0&0&2 \end{array} \right| \right)$ Since the second determinant is a determinant of an upper-triangular matrix, its determinant is $$3\cdot 4 \cdot 2=24$$. We do cofactor expansion along column 3 for the first determinant. \begin{align*} \det(A) &=-2 \left( -2 \left( 2 \left|\begin{array}{rr} 4&3\\ 0&2\end{array} \right| +1 \left|\begin{array}{rrrr} 1&3\\ 4&3 \end{array} \right| \right)+3\cdot 24 \right)\\ &= -2 \left( -2 \left(2(4\cdot 2-0) +1 (1\cdot 3-3\cdot 4)\right)+72 \right)\\ &=-116 \end{align*}

Some applications of determinants:

1. Determinant as volume: Suppose a hypersolid $$S$$ in $$\mathbb R^n$$ is given by $$n$$ concurrent edges that are represented by column vectors of an $$n\times n$$ matrix $$A$$. Then the volume of $$S$$ is $$|\det(A)|$$.

Example. Let $$\overrightarrow{r_1}=[a_1,b_1,c_1]^T$$, $$\overrightarrow{r_2}=[a_2,b_2,c_2]^T$$, $$\overrightarrow{r_3}=[a_3,b_3,c_3]^T$$. $$A=[\overrightarrow{r_1}\;\overrightarrow{r_2}\;\overrightarrow{r_3}]=\left[\begin{array}{ccc}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{array}\right]$$ and the volume of the parallelepiped with concurrent edges given by $$\overrightarrow{r_1},\overrightarrow{r_2},\overrightarrow{r_3}$$ is $|\det(A)|=|a_1(b_2c_3-b_3c_2)-a_2(b_1c_3-b_3c_1)+a_3(b_1c_2-b_2c_1)|.$

2. Equation of a plane: Consider the plane passing through three distinct points $$P_1(x_1,y_1,z_1)$$, $$P_2(x_2,y_2,z_2)$$, and $$P_3(x_3,y_3,z_3)$$. Let $$P(x,y,z)$$ be a point on the plane. So the volume of the parallelepiped with concurrent edges $$\overrightarrow{P_1P}$$, $$\overrightarrow{P_2P}$$, and $$\overrightarrow{P_3P}$$ is zero. $\left|\begin{array}{ccc}x-x_1&x-x_2&x-x_3\\y-y_1&y-y_2&y-y_3\\z-z_1&z-z_2&z-z_3\end{array}\right|=0.$

3. Volume after transformation: Let $$T:\mathbb R^n\to \mathbb R^n$$ be a linear transformation with the standard matrix $$A$$. Let $$S$$ be a bounded hypersolid in $$\mathbb R^n$$. Then the volume of $$T(S)$$ is $$|\det(A)|$$ times the volume of $$S$$.

Example. Let $$A=\left[\begin{array}{cc}a&0\\0&b\end{array}\right]$$ and $$D=\{(x,y)\;|\;x^2+y^2\leq 1\}$$. Consider $$T:\mathbb R^2\to \mathbb R^2$$ defined by $$T([x, y]^T)=A[x, y]^T$$. Note $$T(D)=\{(x,y)\;|\;\frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1\}$$. So the area of ellipse = the area of $$T(D)=\det(A)\cdot A(D)=ab\cdot \pi 1^2=\pi ab$$.

4. Change of variables: Suppose variables $$x_1,\ldots,x_n$$ are changed to $$v_1,\ldots,v_n$$ by $$n$$ differentiable functions $$f_1,\ldots,f_n$$ so that $\begin{eqnarray*} v_1&=&f_1(x_1,\ldots,x_n)\\ v_2&=&f_2(x_1,\ldots,x_n)\\ &\vdots &\\ v_n&=&f_n(x_1,\ldots,x_n). \end{eqnarray*}$ So we have a function $$F:\mathbb R^n\to \mathbb R^n$$ defined by $F(x_1,\ldots,x_n)=(f_1(x_1,\ldots,x_n),\ldots,f_n(x_1,\ldots,x_n)).$ The Jacobian matrix of $$F:\mathbb R^n\to \mathbb R^n$$ is the following $\frac{\partial(f_1,\ldots,f_n)}{\partial(x_1,\ldots,x_n)}= \left[\begin{array}{ccc}\frac{\partial f_1}{\partial x_1}&\cdots&\frac{\partial f_1}{\partial x_n}\\ \vdots&\ddots&\vdots\\ \frac{\partial f_n}{\partial x_1}&\cdots&\frac{\partial f_n}{\partial x_n} \end{array} \right].$ The change of variables formula for integrals is $\int_{F(U)}G(\overrightarrow{v})d\overrightarrow{v}= \int_{U}G(\overrightarrow{x})\left|\frac{\partial(f_1,\ldots,f_n)}{\partial(x_1,\ldots,x_n)}\right| d\overrightarrow{x}.$

Example. So $$(x,y)=F(r,\theta)=(ar\cos\theta,br\sin\theta)$$ and $$F([0,1]\times[0,2\pi])$$ is the region inscribed by the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$. The Jacobian matrix is $\frac{\partial(x,y)}{\partial(r,\theta)}= \left[\begin{array}{cc} \frac{\partial x}{\partial r}&\frac{\partial x}{\partial \theta}\\ \frac{\partial y}{\partial r}&\frac{\partial y}{\partial \theta} \end{array} \right]= \left[\begin{array}{cc} a\cos\theta&-ar\sin\theta\\ b\sin\theta&br\cos\theta \end{array} \right]\text{ and } \left|\frac{\partial(x,y)}{\partial(r,\theta)}\right|=abr.$ By the change of variables formula, $\int_{F([0,1]\times[0,2\pi])}1\;d\overrightarrow{v}= \int_{\theta=0}^{2\pi}\int_{r=0}^11\; \left|\frac{\partial(x,y)}{\partial(r,\theta)}\right| drd\theta=ab\cdot \pi.$

5. Wronskian: The Wroskian of $$n$$ real-values differentiable functions $$f_1,\ldots,f_n$$ is $W(f_1,\ldots,f_n)(x)= \left|\begin{array}{ccc} f_1(x)&\cdots&f_n(x)\\ f^{'}_1(x)&\cdots&f^{'}_n(x)\\ \vdots&\ddots&\vdots\\ f^{(n-1)}_1(x)&\cdots&f^{(n-1)}_n(x)\\ \end{array} \right|.$ $$f_1,\ldots,f_n$$ are linearly independent functions iff $$W(f_1,\ldots,f_n)$$ is not identically zero.

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