11. 행렬식의 정의Definition of Determinants

Solution. Apply the $2\times 2$ determinant formula $\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad-bc$.

(1) $$\det\begin{pmatrix} 6 & -3 \\ 2 & 4 \end{pmatrix} = 6\cdot 4-(-3)\cdot 2 = 24+6=30.$$

(2) Applying the same formula to the complex entries, compute the two products first: $$(5-2i)(7i) = 35i-14i^2 = 14+35i, \qquad (6+4i)(-3+i) = -18+6i-12i+4i^2 = -22-6i,$$ so $$\det\begin{pmatrix} 5-2i & 6+4i \\ -3+i & 7i \end{pmatrix} = (14+35i)-(-22-6i) = 36+41i.$$ (Check: recomputing each product with $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$ again gives $14+35i$ and $-22-6i$, confirming the result.)

Hence the answers are $30$ for (1) and $36+41i$ for (2).

Solution. The area of the parallelogram equals the absolute value of the determinant of the matrix whose rows are the two side vectors. With $u=(3,-2)$ and $v=(2,5)$ as rows, $$\det\begin{pmatrix} 3 & -2 \\ 2 & 5 \end{pmatrix} = 3\cdot 5-(-2)\cdot 2 = 15+4=19.$$ (Check: swapping the row order gives $\det\begin{pmatrix} 2 & 5 \\ 3 & -2 \end{pmatrix} = 2\cdot(-2)-5\cdot 3=-4-15=-19$, the negative of the above, so the absolute value agrees.)

Hence the area is $|19|=19$.

Proof. Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in M_{2\times 2}(F)$, so $\det(A) = ad-bc$.

(1) Since $A^t = \begin{pmatrix} a & c \\ b & d \end{pmatrix}$, $$\det(A^t) = a\cdot d-c\cdot b = ad-bc = \det(A).$$

(2) If $B$ is obtained by interchanging the rows of $A$, then $B = \begin{pmatrix} c & d \\ a & b \end{pmatrix}$, so $$\det(B) = c\cdot b-d\cdot a = -(ad-bc) = -\det(A).$$

Proof. Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $B = \begin{pmatrix} e & f \\ g & h \end{pmatrix}$, so $$AB = \begin{pmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{pmatrix}.$$

By definition, $$\det(AB) = (ae+bg)(cf+dh)-(af+bh)(ce+dg),$$ and expanding the right-hand side, $$(ae+bg)(cf+dh) = acef+adeh+bcfg+bdgh, \qquad (af+bh)(ce+dg) = acef+adfg+bceh+bdgh,$$ so $$\det(AB) = adeh+bcfg-adfg-bceh = ad(eh-fg)-bc(eh-fg) = (ad-bc)(eh-fg).$$

Since $ad-bc = \det(A)$ and $eh-fg = \det(B)$, $$\det(AB) = \det(A)\det(B).$$

Solution.

1. Expanding along the first row, $$\det\begin{pmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ 2 & 3 & 0 \end{pmatrix} = 0\cdot\begin{vmatrix} 0 & -3 \\ 3 & 0 \end{vmatrix} - 1\cdot\begin{vmatrix} -1 & -3 \\ 2 & 0 \end{vmatrix} + 2\cdot\begin{vmatrix} -1 & 0 \\ 2 & 3 \end{vmatrix} = 0 - 1\cdot(0+6) + 2\cdot(-3-0) = -6-6=-12.$$ (Check: expanding along the first column instead gives $0-(-1)\cdot\begin{vmatrix} 1 & 2 \\ 3 & 0 \end{vmatrix}+2\cdot\begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix} = 0-(-1)(-6)+2(-3)=-6-6=-12$, the same value.)
2. Expanding along the first row, $$\det\begin{pmatrix} 0 & 1+i & 2 \\ -2i & 0 & 1-i \\ 3 & 4i & 0 \end{pmatrix} = 0 - (1+i)\begin{vmatrix} -2i & 1-i \\ 3 & 0 \end{vmatrix} + 2\begin{vmatrix} -2i & 0 \\ 3 & 4i \end{vmatrix},$$ and $$\begin{vmatrix} -2i & 1-i \\ 3 & 0 \end{vmatrix} = 0-3(1-i)=-3+3i, \qquad \begin{vmatrix} -2i & 0 \\ 3 & 4i \end{vmatrix} = -8i^2=8,$$ so $$\det = -(1+i)(-3+3i)+2\cdot 8 = 6+16=22.$$ (Check: expanding along the first column instead also gives $22$.)
3. Use row operations. Since $R_2 \to R_2+3R_1$, $R_3 \to R_3-2R_1$, $R_4 \to R_4+2R_1$ leave the determinant unchanged, $$\begin{pmatrix} 1 & -1 & 2 & -1 \\ -3 & 4 & 1 & -1 \\ 2 & -5 & -3 & 8 \\ -2 & 6 & -4 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & -1 & 2 & -1 \\ 0 & 1 & 7 & -4 \\ 0 & -3 & -7 & 10 \\ 0 & 4 & 0 & -1 \end{pmatrix}.$$ Expanding along the first column, $$\det = 1\cdot\det\begin{pmatrix} 1 & 7 & -4 \\ -3 & -7 & 10 \\ 4 & 0 & -1 \end{pmatrix},$$ and applying $R_2 \to R_2+3R_1$, $R_3 \to R_3-4R_1$ again, $$\begin{pmatrix} 1 & 7 & -4 \\ -3 & -7 & 10 \\ 4 & 0 & -1 \end{pmatrix} \to \begin{pmatrix} 1 & 7 & -4 \\ 0 & 14 & -2 \\ 0 & -28 & 15 \end{pmatrix},$$ so $$\det = 1\cdot(14\cdot 15-(-2)\cdot(-28)) = 210-56=154.$$ (Check: expanding the original matrix directly along the first column also gives $154$.)

Hence (1) $-12$, (2) $22$, (3) $154$.

Solution. Write $a=(a_1,a_2,a_3)$, $b=(b_1,b_2,b_3)$, $c=(c_1,c_2,c_3)$ for the row vectors, and treat the determinant as a trilinear function of its rows.

1. The scalars $2$ and $7$ factor out of the first row $2a$ and the third row $7c$, respectively. The remaining second row $3b+5c$ is a sum of two rows, so by linearity $$\det\begin{pmatrix} a \\ 3b+5c \\ c \end{pmatrix} = 3\det\begin{pmatrix} a \\ b \\ c \end{pmatrix} + 5\det\begin{pmatrix} a \\ c \\ c \end{pmatrix},$$ and the second term is $0$ since its second and third rows coincide. Hence $$\det\begin{pmatrix} 2a_1 & 2a_2 & 2a_3 \\ 3b_1+5c_1 & 3b_2+5c_2 & 3b_3+5c_3 \\ 7c_1 & 7c_2 & 7c_3 \end{pmatrix} = 2\cdot 7\cdot 3\det\begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix} = 42\det\begin{pmatrix} a \\ b \\ c \end{pmatrix},$$ so $k=42$. (Check: taking $a=(1,0,0)$, $b=(0,1,0)$, $c=(0,0,1)$ makes the right side $\det I = 1$, while the left-side matrix is upper triangular with diagonal entries $2,3,7$, so its determinant is $2\cdot 3\cdot 7=42$, matching.)
2. Applying linearity in each of the rows $b+c$, $a+c$, $a+b$ produces $2^3=8$ terms: $$\det\begin{pmatrix} b+c \\ a+c \\ a+b \end{pmatrix} = \det\begin{pmatrix} b \\ a \\ a \end{pmatrix}+\det\begin{pmatrix} b \\ a \\ b \end{pmatrix}+\det\begin{pmatrix} b \\ c \\ a \end{pmatrix}+\det\begin{pmatrix} b \\ c \\ b \end{pmatrix}+\det\begin{pmatrix} c \\ a \\ a \end{pmatrix}+\det\begin{pmatrix} c \\ a \\ b \end{pmatrix}+\det\begin{pmatrix} c \\ c \\ a \end{pmatrix}+\det\begin{pmatrix} c \\ c \\ b \end{pmatrix}.$$ Six of these terms have a repeated row and vanish, leaving $$\det\begin{pmatrix} b \\ c \\ a \end{pmatrix} + \det\begin{pmatrix} c \\ a \\ b \end{pmatrix}.$$ Both $(a,b,c)\to(b,c,a)$ and $(a,b,c)\to(c,a,b)$ are obtained by two row interchanges, so neither changes the sign of the determinant: $$\det\begin{pmatrix} b \\ c \\ a \end{pmatrix} = \det\begin{pmatrix} a \\ b \\ c \end{pmatrix}, \qquad \det\begin{pmatrix} c \\ a \\ b \end{pmatrix} = \det\begin{pmatrix} a \\ b \\ c \end{pmatrix}.$$ Therefore $$\det\begin{pmatrix} b_1+c_1 & b_2+c_2 & b_3+c_3 \\ a_1+c_1 & a_2+c_2 & a_3+c_3 \\ a_1+b_1 & a_2+b_2 & a_3+b_3 \end{pmatrix} = 2\det\begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix},$$ so $k=2$. (Check: taking $a=(1,0,0)$, $b=(0,1,0)$, $c=(0,0,1)$ gives left side $\det\begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}=2$ and right side $2\det I=2$, matching.)

Proof. Let the rows of $A \in M_{n\times n}(F)$ be $a_1, \cdots, a_n$. The determinant is linear in each row (it is $n$-linear), so scaling one row by $k$ scales the determinant by $k$: $$\det\begin{pmatrix} ka_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} = k\det\begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix}.$$

Since $kA$ is the matrix with rows $ka_1, \cdots, ka_n$, applying this property to each row in turn gives $$\det(kA) = \det\begin{pmatrix} ka_1 \\ ka_2 \\ \vdots \\ ka_n \end{pmatrix} = k\det\begin{pmatrix} a_1 \\ ka_2 \\ \vdots \\ ka_n \end{pmatrix} = k^2\det\begin{pmatrix} a_1 \\ a_2 \\ ka_3 \\ \vdots \\ ka_n \end{pmatrix} = \cdots = k^n\det\begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} = k^n\det(A).$$

Solution. Reversing the rows, i.e. $a_1, \cdots, a_n \to a_n, \cdots, a_1$, can be realized by interchanging the $i$-th row with the $(n+1-i)$-th row from the outside in: swap row $1$ with row $n$, then row $2$ with row $n-1$, and so on. This requires exactly $\lfloor n/2 \rfloor$ interchanges (if $n$ is odd, the middle row is left in place and needs no swap).

Each row interchange negates the determinant, so after $\lfloor n/2 \rfloor$ interchanges, $$\det(B) = (-1)^{\lfloor n/2 \rfloor}\det(A).$$