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Summary
- If $A=\begin{bmatrix}a_{11}& a_{12}\\ a_{21}& a_{22} \end{bmatrix}$, we define the determinant of $A$ to be the scalar $$\det A= \begin{vmatrix}a_{11}& a_{12}\\ a_{21}& a_{22} \end{vmatrix} = a_{11}a_{22} - a_{12}a_{21}$$
- Minor
Let $M_{ij}(A)$ (or $M_{ij}$ if there is no ambiguity) denote the determinant of the $(n-1) \times (n-1)$ submatrix of $A$ formed by deleting the $i$-th row and $j$-th column of $A$. $M_{ij}(A)$ is called the $(i, j)$ minor of $A$. - Laplace expansion
Assume that the determinant function has been defined for matrices of size $(n-1) \times (n-1)$. Then $\det A$ is defined by the so-called first row Laplace expansion: $$\det A=a_{11}M_{11} -a_{12}M_{12} + \cdots +(-1)^{n+1}a_{1n}M_{1n} = \sum_{j=1}^{n}(-1)^{1+j}a_{1j}M_{1j}$$ For example, the Laplace expansion of a $3\times3$ matrix is $$\det A = a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13}$$ $$=a_{11}\begin{vmatrix}a_{22}& a_{23}\\ a_{32}& a_{33} \end{vmatrix} - a_{12}\begin{vmatrix}a_{21}& a_{23}\\ a_{31}& a_{33} \end{vmatrix} + a_{13}\begin{vmatrix}a_{21}& a_{22}\\ a_{31}& a_{32} \end{vmatrix}$$ $$=a_{11}(a_{22}a_{33}-a_{32}a_{23}) - a_{12}(a_{21}a_{33} - a_{31}a_{23}) + a_{13}(a_{21}a_{32}- a_{31}a_{22})$$ $$=a_{11}a_{22}a_{33} - a_{11}a_{32}a_{23} -a_{12}a_{21}a_{33}+a_{12}a_{31}a_{23} + a_{13}a_{21}a_{32} - a_{13}a_{31}a_{22}$$ - If a row (column) of a matrix is zero, then the value of the determinant is zero. For example, $$\begin{vmatrix}1& 2& 3\\ 4& 5& 6\\ 0& 0& 0 \end{vmatrix} = 0$$
- Let $A = [a_{ij}]$, where $a_{ij} = 0$ if $i \neq j$. Then $$\det A = a_{11}a_{22}\cdots a_{nn}$$ That is, if matrix $A$ is upper triangular, lower triangular, or diagonal, then its determinant is the product of the diagonal elements. For example, $$\begin{vmatrix}2& 0& 0\\ 4& 5& 0\\ 0& 1& 6 \end{vmatrix} = 2\times5\times6=60$$
- We can expand the determinant along any row or column: $$\det A = \sum_{j=1}^{n} (-1)^{i+j}a_{ij}M_{ij}$$ for $i = 1, \cdots, n$ (the so-called $i$-th row expansion) and $$\det A = \sum_{i=1}^{n}(-1)^{i+j} a_{ij}M_{ij}$$ for $j=1, \cdots, n$ (the so-called $j$-th column expansion).
- A matrix and its transpose have equal determinant: $$\det A = \det A^{t}$$ For example, $$\begin{vmatrix}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9 \end{vmatrix} = \begin{vmatrix}1&4& 7\\ 2& 5&8\\3& 6& 9 \end{vmatrix}$$
- If two rows of a matrix are interchanged, the determinant changes sign. For example, $$\begin{vmatrix}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9 \end{vmatrix} = -\begin{vmatrix}4& 5& 6\\ 1& 2& 3\\ 7& 8& 9 \end{vmatrix}$$
- Cofactor
The $(i, j)$ cofactor of $A$, denoted by $C_{ij}(A)$ (or $C_{ij}$ if there is no ambiguity) is defined by $$C_{ij}(A) = (-1)^{i+j}M_{ij}(A)$$ - In terms of the cofactor notation, the determinant can be rewritten as $$\det A = \sum_{j=1}^{n} a_{ij}C_{ij}$$ for $i = 1, \cdots, n$ and $$\det A = \sum_{i=1}^{n} a_{ij}C_{ij}$$ for $j=1, \cdots, n$.
- An important result is $$\sum_{j=1}^{n}a_{ij}C_{kj} = 0$$ if $i\neq k$; and $$\sum_{i=1}^{n}a_{ij}C{ik} = 0$$ if $j\neq k$.
This is intuitive, for instance, $a_{ij}C_{kj}$ for $i\neq k$ can be considered as there is another matrix $B$ which is obtained from $A$ by replacing row $k$ by row $i$. Note that $\det B=0$ since it has two identical rows. - Adjoint
If $A = [a_{ij}]$ is an $n\times n$ matrix, the adjoint of $A$, denoted by $\text{adj} A$, is the transpose of the matrix of cofactors. That is, $$\text{adj} A = \begin{bmatrix}C_{11}& C_{21}& \cdots& C_{n1}\\ C_{12}& C_{22}& \cdots& C_{n2}\\ \vdots& & &\vdots \\C_{1n}& C_{2n}& \cdots& C_{nn} \end{bmatrix}$$ - In terms of the definitions of cofactor and adjoint, we have $$A(\text{adj} A) = (\det A)I_n = (\text{adj} A)A$$
- Formula for the inverse
If $\det A \neq 0$, then $A$ is non-singular and $$A^{-1} = {1\over\det A} \text{adj} A$$ This is obtained from: $$A(\text{adj} A) = \det A \Rightarrow I (\text{adj} A) = A^{-1}\det A \Rightarrow A^{-1} = {1\over\det A} \text{adj} A$$ - The determinant is a linear function of each row and column. For example, $$\begin{vmatrix}a_{11}+a'_{11}& a_{12}+a'_{12} & a_{13} + a'_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix} =\begin{vmatrix}a_{11}& a_{12}& a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix} + \begin{vmatrix}a'_{11}& a'_{12} & a'_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix}$$ $$\begin{vmatrix}ta_{11}& ta_{12}& ta_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix} = t\begin{vmatrix}a_{11}& a_{12}& a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix}$$ We can deduce the following results:
- Interchange two rows $$\det(E_{ij}A) = -\det A$$
- Multiply a row by a scalar $$\det(E_{i}(t)A) = t\det A$$
- Multiplying a row by a scalar and adding to another row $$\det(E_{ij}(t)A) = \det A$$
- Let $A$ be an $n\times n$ matrix. Then
- $A$ is non-singular if and only if $\det A\neq0$;
- $A$ is singular if and only if $\det A=0$;
- the homogeneous system $AX=0$ has a non-trivial solution if and only if $\det A=0$.
- Vandermonde determinant
$$\begin{vmatrix}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2 \end{vmatrix} = (b-a)(c-a)(c-b)$$ Prove: $$\begin{vmatrix}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2 \end{vmatrix} = \begin{vmatrix}1& 0& 0\\ a& b-a& c-a\\ a^2& b^2-a^2& c^2-a^2 \end{vmatrix}$$ $$= \begin{vmatrix} b-a& c-a\\ b^2-a^2& c^2-a^2 \end{vmatrix} = (b-a)(c-a) \begin{vmatrix} 1& 1\\ b+a& c+a \end{vmatrix}$$ $$=(b-a)(c-a)(c+a-b-a) = (b-a)(c-a)(c-b)$$ - Cramer's rule
The system of $n$ linear equations in $n$ unknowns $x_1, \cdots, x_n$ $$\begin{cases}a_{11}x_1+a_{12}x_2+\cdots + a_{1n}x_n = b_1\\ a_{21}x_1 +a_{22}x_2 +\cdots +a_{2n}x_n =b_2\\ \vdots \\ a_{n1}x_1+ a_{n2}x_2+\cdots + a_{nn}x_n=b_n \end{cases}$$ has a unique solution if $\Delta = \det[a_{ij}]\neq0$, namely $$x_1={\Delta_1\over\Delta}, x_2={\Delta_2\over\Delta}, \cdots, x_n={\Delta_n\over\Delta}$$ where $\Delta_i$ is the determinant of the matrix formed by replacing the $i$-th column of the coefficient matrix $A$ by the entries $b_1, b_2, \cdots, b_n$.
Proof:
Since $\Delta\neq0$, so $A^{-1}$ exists and is given by $A^{-1} = {1\over\det A} \text{adj} A$. From $AX=b$ we have $$\begin{bmatrix}x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix} = A^{-1}\begin{bmatrix}b_1\\ b_2\\ \vdots\\ b_n \end{bmatrix}$$ $$= {1\over\Delta} \text{adj} A\begin{bmatrix}b_1\\ b_2\\ \vdots\\ b_n \end{bmatrix}$$ $$= {1\over\Delta} \begin{bmatrix}C_{11}& C_{21}& \cdots& C_{n1}\\ C_{12}& C_{22}& \cdots& C_{n2}\\ \vdots& & &\vdots \\C_{1n}& C_{2n}& \cdots& C_{nn} \end{bmatrix} \begin{bmatrix}b_1\\ b_2\\ \vdots\\ b_n \end{bmatrix}$$ $$= {1\over\Delta} \begin{bmatrix}b_1C_{11} + b_2C_{21}+\cdots +b_nC_{n1}\\ b_1C_{12}+b_2C_{22}+\cdots + b_nC_{n2}\\ \vdots\\ b_1C_{1n} +b_2C_{2n}+\cdots +b_nC_{nn} \end{bmatrix}$$ However the $i$-th component of the last vector is the expansion of $\Delta_i$ along column $i$. Hence $$\begin{bmatrix}x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix} = {1\over\Delta} \begin{bmatrix}\Delta_1\\ \Delta_2\\ \vdots\\ \Delta_n \end{bmatrix} = \begin{bmatrix}\Delta_1\big/ \Delta\\ \Delta_2\big/\Delta\\ \vdots\\ \Delta_n\big/\Delta \end{bmatrix}.$$
Problems 4.1
1. If the points $P_{i}=(x_i, y_i)$, $i=1, 2, 3, 4$ form a quadrilateral with vertices in anti-clockwise orientation, prove that the area of the quadrilateral equals $${1\over2}\left(\begin{vmatrix}x_1 & x_2\\ y_1 & y_2\end{vmatrix} + \begin{vmatrix}x_2 & x_3\\ y_2 & y_3\end{vmatrix} + \begin{vmatrix}x_3 & x_4\\ y_3 & y_4\end{vmatrix} + \begin{vmatrix}x_4 & x_1\\ y_4 & y_1\end{vmatrix}\right)$$
Solution:
Firstly, we calculate the area of $\bigtriangleup OPQ$ where $P=(x_1, y_1)$, $Q=(x_2, y_2)$ and $O=(0, 0)$. See the following figure.
Therefore, the area of the quadrilateral $P_1P_2P_3P_4$ is (the following figure a) $$S_{P_1P_2P_3P_4} = S_{\bigtriangleup P_1P_2P_3} + S_{\bigtriangleup P_1P_3P_4}$$ $$= {1\over2} \begin{vmatrix}x_1 & x_2\\ y_1& y_2 \end{vmatrix} + {1\over2} \begin{vmatrix}x_2 & x_3\\ y_2& y_3 \end{vmatrix} + {1\over2} \begin{vmatrix}x_3 & x_1\\ y_3& y_1 \end{vmatrix}$$ $$+{1\over2} \begin{vmatrix}x_1 & x_3\\ y_1& y_3 \end{vmatrix} + {1\over2} \begin{vmatrix}x_3 & x_4\\ y_3& y_4 \end{vmatrix} + {1\over2} \begin{vmatrix}x_4 & x_1\\ y_4& y_1 \end{vmatrix}$$ $$={1\over2} \begin{vmatrix}x_1 & x_2\\ y_1& y_2 \end{vmatrix} + {1\over2} \begin{vmatrix}x_2 & x_3\\ y_2& y_3 \end{vmatrix} - {1\over2} \begin{vmatrix}x_1 & x_3\\ y_1& y_3 \end{vmatrix}$$ $$+{1\over2} \begin{vmatrix}x_1 & x_3\\ y_1& y_3 \end{vmatrix} + {1\over2} \begin{vmatrix}x_3 & x_4\\ y_3& y_4 \end{vmatrix} + {1\over2} \begin{vmatrix}x_4 & x_1\\ y_4& y_1 \end{vmatrix}$$ $$= {1\over2} \left(\begin{vmatrix}x_1 & x_2\\ y_1& y_2 \end{vmatrix} + \begin{vmatrix}x_2 & x_3\\ y_2& y_3 \end{vmatrix} + \begin{vmatrix}x_3 & x_4\\ y_3& y_4 \end{vmatrix} + {1\over2} \begin{vmatrix}x_4 & x_1\\ y_4& y_1 \end{vmatrix}\right)$$
Another possible configuration for the quadrilateral occurs when it is not convex, as in the above figure b. We can obtain the similar result as before.
2. Prove that the following identity holds by expressing the left-hand side as the sum of 8 determinants: $$\begin{vmatrix}a+x & b+y & c+z\\ x+u & y+v & z+w\\ u+a & v+b & w+c \end{vmatrix} = 2\begin{vmatrix}a & b & c\\ x & y& z\\ u & v &w \end{vmatrix}$$
Solution:
$$\begin{vmatrix}a+x & b+y & c+z\\ x+u & y+v & z+w\\ u+a & v+b & w+c \end{vmatrix} = \begin{vmatrix}a & b & c\\ x+u & y+v & z+w\\ u+a & v+b & w+c \end{vmatrix} + \begin{vmatrix}x & y & z\\ x+u & y+v & z+w\\ u+a & v+b & w+c \end{vmatrix}$$ The first determinant on the right hand is $$\begin{vmatrix}a & b & c\\ x+u & y+v & z+w\\ u+a & v+b & w+c \end{vmatrix} = \begin{vmatrix}a & b & c\\ x+u & y+v & z+w\\ u & v & w\end{vmatrix} + \begin{vmatrix}a & b & c\\ x+u & y+v & z+w\\ a & b & c \end{vmatrix}$$ $$= \begin{vmatrix}a & b & c\\ x+u & y+v & z+w\\ u & v & w\end{vmatrix} = \begin{vmatrix}a & b & c\\ x & y & z\\ u & v & w\end{vmatrix} + \begin{vmatrix}a & b & c\\ u & v & w\\ u & v & w\end{vmatrix} = \begin{vmatrix}a & b & c\\ x & y & z\\ u & v & w\end{vmatrix}$$ Similarly, the second determinant on the right hand is $$\begin{vmatrix}x & y & z\\ x+u & y+v & z+w\\ u+a & v+b & w+c \end{vmatrix} = \begin{vmatrix}x & y & z\\ x & y & z\\ u+a & v+b & w+c \end{vmatrix} + \begin{vmatrix}x & y & z\\ u & v & w\\ u+a & v+b & w+c \end{vmatrix}$$ $$\begin{vmatrix}x & y & z\\ u & v & w\\ u+a & v+b & w+c \end{vmatrix} = \begin{vmatrix}x & y & z\\ u & v & w\\ u & v & w \end{vmatrix} + \begin{vmatrix}x & y & z\\ u & v & w\\ a & b & c \end{vmatrix} = \begin{vmatrix}x & y & z\\ u & v & w\\ a & b & c \end{vmatrix}$$ $$= -\begin{vmatrix}u & v & w\\ x & y & z\\ a & b & c \end{vmatrix} = \begin{vmatrix}a & b & c\\ x & y & z\\ u & v & w\end{vmatrix}$$ Hence we have $$\begin{vmatrix}a+x & b+y & c+z\\ x+u & y+v & z+w\\ u+a & v+b & w+c \end{vmatrix} = 2\begin{vmatrix}a & b & c\\ x & y& z\\ u & v &w \end{vmatrix}$$
3. Prove that $$\begin{vmatrix}n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2\\ (n+2)^2 & (n+3)^2 & (n+4)^2 \end{vmatrix} = -8$$
Solution:
$$\begin{vmatrix}n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2\\ (n+2)^2 & (n+3)^2 & (n+4)^2 \end{vmatrix} = \begin{cases}R_2-R_1\\ R_3-R_2\end{cases} \begin{vmatrix}n^2 & (n+1)^2 & (n+2)^2 \\ 2n+1 & 2n+3 & 2n+5\\ 2n+3 & 2n+5 & 2n+7 \end{vmatrix}$$ $$= [R_3-R_2]\ \begin{vmatrix}n^2 & (n+1)^2 & (n+2)^2 \\ 2n+1 & 2n+3 & 2n+5\\ 2 & 2 & 2 \end{vmatrix}$$ $$=\begin{cases}C_2-C_1\\ C_3-C_2\end{cases} \begin{vmatrix}n^2 & 2n+1 & 2n+3 \\ 2n+1 & 2 & 2\\ 2 & 0& 0 \end{vmatrix}$$ $$=2\times[(2n+1)\times2-2\times(2n+3)] = -8$$
4. Evaluate the following determinants: $$(a)\ \begin{vmatrix}246 & 427 & 327\\ 1014 & 543 & 443\\ -342 & 721 & 621\end{vmatrix};\ (b)\ \begin{vmatrix}1& 2& 3& 4\\ -2& 1& -4& 3\\ 3& -4& -1& 2\\ 4& 3& -2& -1 \end{vmatrix}. $$
Solution:
$$(a)\ \begin{vmatrix}246 & 427 & 327\\ 1014 & 543 & 443\\ -342 & 721 & 621\end{vmatrix} = \begin{vmatrix}246 & 100 & 327\\ 1014 & 100 & 443\\ -342 & 100 & 621\end{vmatrix} = 100\begin{vmatrix}246 & 1 & 327\\ 1014 & 1 & 443\\ -342 & 1 & 621\end{vmatrix}$$ $$ = 100\begin{vmatrix}-768 & 0 & -116\\ 1356 & 0 & -178\\ -342 & 1 & 621\end{vmatrix} = 100\times(-1)^{5}\times[(-768)\times(-178) - 1356\times(-116)]= -29400000$$ $$(b)\ \begin{vmatrix}1& 2& 3& 4\\ -2& 1& -4& 3\\ 3& -4& -1& 2\\ 4& 3& -2& -1 \end{vmatrix} = \begin{vmatrix}1& 2& 3& 4\\ 0& 5& 2& 11\\ 0& -10& -10& -10\\ 0& -5& -14& -17 \end{vmatrix} = \begin{vmatrix}5& 2& 11\\ -10& -10& -10\\ -5& -14& -17 \end{vmatrix}$$ $$ = \begin{vmatrix}5& 2& 11\\ 0& -6& 12\\ 0& -12& -6 \end{vmatrix} = 5\times [(-6)\times(-6)- (-12)\times12] = 900$$
5. Compute the inverse of the matrix $$A = \begin{bmatrix}1& 0& -2\\ 3& 1& 4\\ 5& 2& -3 \end{bmatrix}$$ by first computing the adjoint matrix.
Solution:
Recall that $$A^{-1} = {1\over\det A} \text{adj} A$$ where $$\text{adj} A = \begin{bmatrix}C_{11}& C_{21}& \cdots& C_{n1}\\ C_{12}& C_{22}& \cdots& C_{n2}\\ \vdots& & &\vdots \\C_{1n}& C_{2n}& \cdots& C_{nn} \end{bmatrix}$$ where $C_{ij}$ is the cofactor and given as $C_{ij} = (-1)^{i+j}M_{ij}$. Note that both of $C_{ij}$ and $M_{ij}$ do not depend on $a_{ij}$. For instance, $$C_{11} = \begin{vmatrix}1& 4\\2& -3 \end{vmatrix} = -11$$ $$C_{21} = -\begin{vmatrix}0& -2\\2& -3 \end{vmatrix} = -4$$
$$C_{12} = -\begin{vmatrix}3& 4\\5& -3 \end{vmatrix} =29$$ and so forth. Hence we have $$\det A=1\times(-3-8) + (-2)\times(6-5) = -13$$ $$\Rightarrow A^{-1} = -{1\over13} \begin{bmatrix}-11& -4& 2\\ 29& 7& -10\\ 1& -2& 1 \end{bmatrix}$$
6. Prove that the following identities hold: $$\begin{vmatrix}2a& 2b& b-c \\ 2b& 2a& a+c\\ a+b& a+b& b \end{vmatrix} = -2(a-b)^2(a+b)$$ $$\begin{vmatrix}b+c& b& c \\ c& c+a& a\\ b& a& a+b \end{vmatrix} = 2a(b^2+c^2)$$
Solution:
$$\begin{vmatrix}2a& 2b& b-c \\ 2b& 2a& a+c\\ a+b& a+b& b \end{vmatrix} =[R_1+R_2]\ \begin{vmatrix}2a+2b& 2b+2a& a+b \\ 2b& 2a& a+c\\ a+b& a+b& b \end{vmatrix}$$ $$= (a+b)\begin{vmatrix}2& 2& 1 \\ 2b& 2a& a+c\\ a+b& a+b& b \end{vmatrix} =[C_2-C_1]\ (a+b)\begin{vmatrix}2& 0& 1 \\ 2b& 2a-2b& a+c\\ a+b& 0& b \end{vmatrix}$$ $$=(a+b)(2a-2b)[2b-(a+b)] = -2(a+b)(a-b)^{2}$$ $$\begin{vmatrix}b+c& b& c \\ c& c+a& a\\ b& a& a+b \end{vmatrix} = [C_1-C_2]\ \begin{vmatrix}c& b& c \\ -a& c+a& a\\ b-a& a& a+b \end{vmatrix}$$ $$= [C_3-C_1]\ \begin{vmatrix}c& b& 0 \\ -a& c+a& 2a\\ b-a& a& 2a \end{vmatrix} = 2a\begin{vmatrix}c& b& 0 \\ -a& c+a& 1\\ b-a& a& 1 \end{vmatrix} $$ $$=[R_3-R_2]\ 2a\begin{vmatrix}c& b& 0 \\ -a& c+a& 1\\ b& -c& 0 \end{vmatrix}$$ $$=-2a(-c^2-b^2) = 2a(b^2+c^2)$$
7. Let $P_i = (x_i, y_i)$, $i=1, 2, 3$. If $x_1, x_2, x_3$ are distinct, prove that there is precisely one curve of the form $y=ax^2 + bx +c$ passing through $P_1$, $P_2$ and $P_3$.
Solution:
Since the curve passes through the points, so we have $$\begin{cases}ax_1^2 + bx_1 + c=0\\ ax_2^2 + bx_2 + c=0\\ ax_3^2 + bx_3 + c=0 \end{cases}$$ We hope to show that there is unique solution (i.e. $(a, b, c)$) of the above set of equations. The coefficient matrix is $$\begin{vmatrix}x_1^2 & x_1 & 1\\ x_2^2 & x_2 & 1\\ x_3^2 & x_3 & 1\end{vmatrix} = \begin{vmatrix}x_1^2 & x_2^2 & x_3^2\\ x_1 & x_2 & x_3\\ 1 & 1 & 1\end{vmatrix} = -\begin{vmatrix}1 & 1 & 1\\ x_1 & x_2 & x_3\\ x_1^2 & x_2^2 & x_3^2\end{vmatrix}$$ which is the Vandermonde matrix and hence we have $$\begin{vmatrix}x_1^2 & x_1 & 1\\ x_2^2 & x_2 & 1\\ x_3^2 & x_3 & 1\end{vmatrix} = -(x_2-x_1)(x_3-x_1)(x_3-x_2) \neq 0$$ for $x_i$ are distinct. By Cramer's rule, there is an unique solution for $a$, $b$ and $c$.
8. Let $$A=\begin{bmatrix}1& 1& -1\\ 2& 3& k\\ 1& k& 3 \end{bmatrix}$$ Find the vector of $k$ for which $\det A=0$ and hence, or otherwise, determine the value of $k$ for which the following system has more than one solution $$\begin{cases}x+y-z =1\\ 2x+3y+kz =3\\ x+ky+3z =2 \end{cases}$$ Solve the system for this value of $k$ and determinant the solution for which $x^2+y^2+z^2$ has least value.
Solution:
$$\Delta =\begin{vmatrix}1& 1& -1\\ 2& 3& k\\ 1& k& 3 \end{vmatrix} = \begin{vmatrix}1& 1& -1\\ 0& 1& k+2\\ 0& k-1& 4 \end{vmatrix} = 4-(k-1)(k+2)$$ $$= -k^2 - k +6 =0\Rightarrow (k+3)(k-2) =0\Rightarrow k_1=-3, k_2=2$$ Hence $\det A=0$ if and only if $k=-3, 2$. That is, the system has unique solution if $k\neq -3, 2$.
When $k=-3$, the augmented matrix is $$\begin{bmatrix}1& 1& -1 & 1\\ 2& 3& -3 & 3\\ 1& -3& 3 &2 \end{bmatrix}\Rightarrow \begin{bmatrix}1& 1& -1 &1\\ 0& 1& -1 & 1\\ 0& -4& 4 &1\end{bmatrix} \Rightarrow \begin{bmatrix}1& 0& 0 &0\\ 0& 1& -1 &1\\ 0& 0& 0 & 5\end{bmatrix}$$ from which we read off inconsistency.
When $k=2$, the augmented matrix is $$\begin{bmatrix}1& 1& -1 & 1\\ 2& 3& 2 & 3\\ 1& 2& 3 &2 \end{bmatrix}\Rightarrow \begin{bmatrix}1& 1& -1 &1\\ 0& 1& 4 & 1\\ 0& 1& 4 &1\end{bmatrix} \Rightarrow \begin{bmatrix}1& 0& -5 &0\\ 0& 1& 4 &1\\ 0& 0& 0 & 0\end{bmatrix}$$ from which we read off the solution $x=5z, y=1-4z$ where $z$ is arbitrary. Finally, we will determine the solution for which $x^2+y^2+z^2$ is least: $$f(z) = 25z^2 + (1-4z)^2 +z^2 = 42z^2 - 8z+1 \Rightarrow f'(z) = 84z-8=0\Rightarrow z={2\over21}$$ Since $f''(z) = 84 > 0$ which means $f'(z)=0$ is the minimum solution in terms of $z$ for $f(z)$. Thus, $k=2$ we can obtain the solution $$\begin{cases}x={10\over21}\\ y={13\over21}\\ z={2\over21} \end{cases}$$ for which $x^2+y^2+z^2$ is least.
9. By considering the coefficient determinant, find all rational numbers $a$ and $b$ for which the following system has (1) solution, (2) exactly one solution, (3) infinitely many solutions: $$\begin{cases}x-2y+bz=3 \\ ax+2z=2\\ 5x+2y=1\end{cases}$$ Solve the system in case(3).
Solution:
The determinant of the coefficient matrix is $$\begin{vmatrix} 1& -2& b\\ a& 0& 2\\ 5& 2& 0\end{vmatrix} = \begin{vmatrix} 1& -2& b\\ 0& 2a& 2-ab\\ 0& 12& -5b\end{vmatrix}$$ $$= 2a(-5b) - 12(2-ab) = -10ab-24+12ab =2(ab-12)$$ If $ab\neq12$ then the system has unique solution.\\
If $ab=12$, the augmented matrix is $$\begin{bmatrix}1& -2& b& 3\\ a& 0& 2& 2\\ 5& 2& 0& 1 \end{bmatrix}\Rightarrow \begin{bmatrix}1& -2& b& 3\\ 0& 2a& 2-ab& 2-3a\\ 0& 12& -5b& -14 \end{bmatrix}$$ $$\Rightarrow \begin{bmatrix}1& 0& {b\over6}& {2\over3}\\ 0& 0& 2-{ab\over6}& 2-{2a\over3}\\ 0& 12& -5b& -14 \end{bmatrix} = \begin{bmatrix}1& 0& {b\over6}& {2\over3}\\ 0& 0& 0& 2-{2a\over3}\\ 0& 12& -5b& -14 \end{bmatrix} $$ Hence, if $a\neq3$ then the system is inconsistency; if $a=3$ then the system has infinite solutions. $$\begin{bmatrix}1& 0& {b\over6}& {2\over3}\\ 0& 0& 0& 2-{2a\over3}\\ 0& 12& -5b& -14 \end{bmatrix} = \begin{bmatrix}1& 0& {2\over3}& {2\over3}\\ 0& 0& 0& 0\\ 0& 12& -20& -14 \end{bmatrix}\Rightarrow \begin{cases}x= -{2\over3}z+{2\over3}\\ y={20z-14\over12}={5\over3}z-{7\over6}\end{cases}$$ In conclusion, $$\begin{cases}ab=12, a\neq3 & \mbox{no solution}\\ ab\neq12 & \mbox{unique solution}\\ a=3, b=4 & \mbox{infinite solutions:} x=-{2\over3}z+{2\over3}, y={5\over3}z-{7\over6} \end{cases}$$
10. Express the determinant of the matrix $$B=\begin{bmatrix}1& 1& 2& 1\\ 1& 2& 3& 4\\ 2& 4& 7& 2t+6\\ 2& 2& 6-t& t \end{bmatrix}$$ as a polynomial in $t$ and hence determine the rational values of $t$ for which $B^{-1}$ exists.
Solution:
$$\det B=\begin{vmatrix}1& 1& 2& 1\\ 1& 2& 3& 4\\ 2& 4& 7& 2t+6\\ 2& 2& 6-t& t \end{vmatrix} = \begin{vmatrix}1& 1& 2& 1\\ 0& 1& 1& 3\\ 0& 2& 3& 2t+4\\ 0& 0& 2-t& t-2 \end{vmatrix} $$ $$=\begin{vmatrix}1& 1& 3\\2& 3& 2t+4\\0& 2-t& t-2 \end{vmatrix} =\begin{vmatrix}1& 1& 3\\0& 1& 2t-2\\0& 2-t& t-2 \end{vmatrix} $$ $$= \begin{vmatrix}1& 2t-2\\2-t& t-2 \end{vmatrix} = t-2-(2-t)(2t-2)$$ $$= t-2-4t+4+2t^2-2t = 2t^2-5t+2 = (2t-1)(t-2)$$ Hence $t\neq2, {1\over2}$ for which $B^{-1}$ exists.
11. If $A$ is a $3\times3$ matrix over a field and $\det A\neq0$, prove that
(1) $\det(\text{adj}A) = (\det A)^2$,
(2) $(\text{adj}A)^{-1}={1\over\det A}A = \text{adj}(A^{-1})$.
Solution:
(1) Since $\det(AB) = \det A\det B$ for $A$ and $B$ are same size square matrices. We have $$A\cdot\text{adj}A = \det AI_3$$ $$\Rightarrow \det(A)\det(\text{adj}A) = \det(\det AI_3 = \begin{vmatrix}\det A &0 &0\\ 0& \det A& 0\\ 0& 0& \det A \end{vmatrix} = (\det A)^3$$ $$\Rightarrow \det(\text{adj}A) = (\det A)^{2}$$
(2) Similar to (1), we have $$A\cdot\text{adj}A = \det AI_3$$ $$\Rightarrow {1\over\det A}A = {I_3\over\text{adj}A} = (\text{adj}A)^{-1}$$ Finally, note that $\det A\det A^{-1} = \det(AA^{-1}) =1$, hence $$A^{-1}\text{adj}(A^{-1}) = \det A^{-1}I_3$$ $$\Rightarrow \text{adj}(A^{-1}) = A\det A^{-1}I_3 = {1\over\det A}A$$
12. Suppose that $A$ is a real $3\times3$ matrix such that $A^{t}A=I_3$.
(1) Prove that $A^{t}(A - I_3)=-(A-I_3)^{t}$.
(2) Prove that $\det A=\pm1$.
(3) Use (1) to prove that if $\det A=1$ then $\det(A-I_3)=0$.
Solution:
(1) $$A^{t}(A-I_3) = I_3-A^{t}=-(A^{t}-I_3) =-(A^{t}-I_3^{t}) = -(A-I_3)^{t}$$
(2) $$\det(AA^{t})=1=\det A\det A^{t} = (\det A)^{2}\Rightarrow \det A=\pm1$$
(3) By the result of (1), we have $$\det(A^{t}(A-I_3))=\det(-(A-I_3)^{t})$$ $$\Rightarrow \det A^{t}\det(A-I_3) = (-1)^3\det((A-I_3)^{t})$$ $$\Rightarrow \det(A-I_3)=-\det(A-I_3)$$ $$\Rightarrow \det(A-I_3)=0$$
13. If $A$ is a square matrix such that one column is a linear combination of the remaining columns, prove that $\det A=0$. Prove that the converse also holds.
Solution:
Without losing generality, assuming that the first column is the combination of the remaining columns: $$A_{*1}=x_2A_{*2}+\cdots+ x_nA_{*n}$$ $$\Rightarrow \det A=\begin{vmatrix}x_2a_{12}+\cdots+ x_na_{1n}& a_{12}&\cdots& a_{1n}\\ \vdots & \vdots & \vdots & \vdots \\x_2a_{n2}+\cdots+ x_na_{nn}& a_{n2} &\cdots &a_{nn} \end{vmatrix}$$ then $C_1\rightarrow C_1-x_2C_2-\cdots -x_nC_{n}$ $$\det A= \begin{vmatrix}0&\cdots& a_{1n}\\ \vdots & \vdots & \vdots \\0 &\cdots &a_{nn} \end{vmatrix} =0$$ Conversely, suppose $\det A=0$ which means $A$ is singular or non-invertible. Thus the homogeneous system $AX=0$ has non-trivial solution such that $X=\begin{bmatrix}x_1\\ \vdots \\ x_n \end{bmatrix}$ while assuming that $x_1\neq0$, and hence we have $$\begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\ \vdots & \vdots & \vdots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix} \begin{bmatrix}x_1\\ \vdots \\ x_n \end{bmatrix} =0$$ $$\Rightarrow x_1A_{*1} + x_2A_{*2} +\cdots +x_nA_{*n} =0$$ $$\Rightarrow A_{*1} = -{x_2\over x_1}A_{*2} - \cdots - {x_{n}\over x_1}A_{*n}$$ which indicates that the first column of $A$ is a linear combination of the remaining columns.
14. Use Cramer's rule to solve the system $$\begin{cases} -2x+3y-z=1\\ x+2y-z=4\\ -2x-y+z=-3\end{cases}$$
Solution:
The determinant of the coefficient matrix is $$\Delta = \begin{vmatrix}-2& 3& -1\\ 1& 2& -1\\ -2& -1& 1 \end{vmatrix} = \begin{vmatrix}0& 7& -3\\ 1& 2& -1\\ 0& 3& -1 \end{vmatrix} = -1\times \begin{vmatrix}7& -3\\ 3& -1 \end{vmatrix} = -2$$ And $$\Delta_1 = \begin{vmatrix}1& 3& -1\\ 4& 2& -1\\ -3& -1& 1 \end{vmatrix} = \begin{vmatrix}1& 3& -1\\ 0& -10& 3\\ 0& 8& -2 \end{vmatrix} =1\times \begin{vmatrix}-10& 3\\ 8& -2 \end{vmatrix} = -4$$ $$\Delta_2 = \begin{vmatrix}-2& 1& -1\\ 1& 4& -1\\ -2& -3& 1 \end{vmatrix} = \begin{vmatrix}0& 9& -3\\ 1& 4& -1\\ 0& 5& -1 \end{vmatrix}= -1\times \begin{vmatrix}9& -3\\ 5& -1 \end{vmatrix} = -6$$ $$\Delta_3 = \begin{vmatrix}-2& 3& 1\\ 1& 2& 4\\ -2& -1& -3 \end{vmatrix} = \begin{vmatrix}0& 7& 9\\ 1& 2& 4\\ 0& 3& 5 \end{vmatrix} = -1\times \begin{vmatrix}7& 9\\ 3& 5 \end{vmatrix} = -8$$ Hence the set of the solution is $$\begin{cases}x={\Delta_1\over\Delta} = 2\\ y={\Delta_2\over\Delta}=3\\ z={\Delta_3\over\Delta} = 4 \end{cases}$$
15. Prove that $$\det E_{ij}=-1,\ \det E_{i}(t)=t,\ \det E_{ij}(t)=1$$ and prove that $$\det(BA) = \det B\det A$$ if $B$ is non-singular. Also prove that the formula holds when $B$ is singular.
Solution:
$$\det E_{ij}=-1$$ since the interchanging of two rows will change the sign and $\det I_n=1$. $$\det E_{i}(t)=t$$ since the determinant of a diagonal matrix is the product of the diagonal elements, that is, $t$. $$\det E_{ij}(t)=1$$ since the matrix is still an upper triangular or lower triangular matrix after the operation, hence its determinant is still 1.
If $B$ is non-singular, that is, $B$ is a product of elementary row matrices: $$B=E_1E_2\cdots E_m$$ that is, we need to prove that $$\det(E_1E_2\cdots E_mA) = \det(E_1E_2\cdots E_m)\det A$$ By mathematical induction, when $m=1$ we have $$\det (E_1A) = \begin{cases}-\det A & (E_1=E_{ij})\\ t\det A & (E_1 = E_{i}(t))\\ \det A & (E_1=E_{ij}(t)) \end{cases}= \det E_1\det A$$ Assuming that the proposition holds for products of $m$ elementary row matrices, that is, $$\det(E_1E_2\cdots E_mA) = \det (E_1E_2\cdots E_m) \det A$$ Then $$\det(E_1E_2\cdots E_mE_{m+1}A) = \det[(E_1E_2\cdots E_m)(E_{m+1}A)]$$ $$= \det(E_1E_2\cdots E_m)\det(E_{m+1}A)$$ $$= \det(E_1E_2\cdots E_m)\det E_{m+1}\det A$$ $$= \det(E_1E_2\cdots E_mE_{m+1})\det A$$ and the induction goes through. Hence, $$\det(BA) = \det B\det A$$ if $B$ is non-singular.
On the other hand, if $B$ is singular, then $BA$ is also singular. Hence $$\det (BA) = 0 = \det B\det A$$
16. Prove that $$\begin{vmatrix}a+b+c & a+b & a & a\\ a+b & a+b+c& a& a \\ a& a& a+b+c& a+b \\ a& a& a+b& a+b+c \end{vmatrix} = c^2(2b+c)(4a+2b+c)$$
Solution:
$$\begin{vmatrix}a+b+c & a+b & a & a\\ a+b & a+b+c& a& a \\ a& a& a+b+c& a+b \\ a& a& a+b& a+b+c \end{vmatrix} = \begin{cases}R_1-R_2\\ R_2-R_3\\ R_3-R_4 \end{cases} \begin{vmatrix}c & -c & 0 & 0\\ b & b+c& -b-c & -b \\ 0& 0& c& -c \\ a& a& a+b& a+b+c \end{vmatrix}$$ $$=[C_4+C_3]\ \begin{vmatrix}c & -c & 0 & 0\\ b & b+c& -b-c & -2b-c \\ 0& 0& c& 0 \\ a& a& a+b& 2a+2b+c \end{vmatrix} = c \begin{vmatrix}c & -c & 0\\ b & b+c & -2b-c \\ a& a & 2a+2b+c \end{vmatrix} $$ $$= [C_2+C_1]\ c \begin{vmatrix}c & 0& 0\\ b & 2b+c & -2b-c \\ a& 2a & 2a+2b+c \end{vmatrix}= c^2 \begin{vmatrix}2b+c & -2b-c \\ 2a & 2a+2b+c \end{vmatrix}$$ $$= [C_2+C_1]\ c^2 \begin{vmatrix}2b+c & 0 \\ 2a & 4a+2b+c \end{vmatrix} = c^2(2b+c)(4a+2b+c)$$
17. Prove that $$\begin{vmatrix}1+u_1& u_1& u_1& u_1\\ u_2& 1+u_2& u_2& u_2\\ u_3& u_3& 1+u_3& u_3\\ u_4& u_4& u_4& 1+u_4 \end{vmatrix} = 1+u_1+u_2+u_3+u_4$$
Solution:
$$\begin{vmatrix}1+u_1& u_1& u_1& u_1\\ u_2& 1+u_2& u_2& u_2\\ u_3& u_3& 1+u_3& u_3\\ u_4& u_4& u_4& 1+u_4 \end{vmatrix} = [R_1+R_2+R_3+R_4]\ \begin{vmatrix}t & t & t &t \\ u_2& 1+u_2& u_2& u_2\\ u_3& u_3& 1+u_3& u_3\\ u_4& u_4& u_4& 1+u_4 \end{vmatrix}$$ where $t= 1+u_1+u_2+u_3+u_4$. Then $$\begin{vmatrix}1+u_1& u_1& u_1& u_1\\ u_2& 1+u_2& u_2& u_2\\ u_3& u_3& 1+u_3& u_3\\ u_4& u_4& u_4& 1+u_4 \end{vmatrix} = t \begin{vmatrix}1 & 1 & 1 &1 \\ u_2& 1+u_2& u_2& u_2\\ u_3& u_3& 1+u_3& u_3\\ u_4& u_4& u_4& 1+u_4 \end{vmatrix}$$ $$=\begin{cases}C_2-C_1\\ C_3-C_1 \\ C_4-C_1 \end{cases} t \begin{vmatrix}1 & 0 & 0 &0 \\ u_2& 1& 0& 0\\ u_3& 0& 1& 0\\ u_4& 0& 0& 1 \end{vmatrix} = t = 1+u_1+u_2+u_3+u_4$$
18. Let $A\in M_{n\times n}(F)$. If $A^{t} = -A$, prove that $\det A=0$ if $n$ is odd and $1+1\neq0$ in F.
Solution:
$$A^{t}=A\Rightarrow \det A=\det A^{t}=\det (-A)=(-1)^n\det A = -\det A$$ Since $1+1\neq0$ in $F$, hence $\det A=0$.
19. Prove that $$\begin{vmatrix}1& 1& 1& 1\\ r& 1& 1& 1\\ r& r& 1& 1\\ r& r& r& 1 \end{vmatrix} = (1-r)^3$$
Solution:
$$\begin{vmatrix}1& 1& 1& 1\\ r& 1& 1& 1\\ r& r& 1& 1\\ r& r& r& 1 \end{vmatrix} = [R_1-R_2]\ \begin{vmatrix}1-r& 0& 0& 0\\ r& 1& 1& 1\\ r& r& 1& 1\\ r& r& r& 1 \end{vmatrix} = (1-r) \begin{vmatrix}1& 1& 1\\ r& 1& 1\\ r& r& 1 \end{vmatrix}$$ $$= [R_1-R_2]\ (1-r) \begin{vmatrix}1-r& 0& 0\\ r& 1& 1\\ r& r& 1 \end{vmatrix} = (1-r)^2(1-r) = (1-r)^3$$
20. Express the determinant $$\begin{vmatrix}1& a^2-bc& a^4\\ 1& b^2-ac& b^4\\ 1& c^2-ab& c^4 \end{vmatrix} $$ as the product of one quadratic and four linear factors.
Solution:
$$\begin{vmatrix}1& a^2-bc& a^4\\ 1& b^2-ac& b^4\\ 1& c^2-ab& c^4 \end{vmatrix} = \begin{cases}R_2-R_1\\ R_3-R_1 \end{cases} \begin{vmatrix}1& a^2-bc& a^4\\ 0& b^2-ac-a^2+bc& b^4-a^4\\ 0& c^2-ab-a^2+bc& c^4-a^4 \end{vmatrix} $$ $$= \begin{vmatrix}b^2-ac-a^2+bc& b^4-a^4\\ c^2-ab-a^2+bc& c^4-a^4 \end{vmatrix}$$ $$= \begin{vmatrix}(a+b+c)(b-a)& (b^2+a^2)(b+a)(b-a)\\ (a+b+c)(c-a)& (c^2+a^2)(c+a)(c-a) \end{vmatrix} $$ $$=(a+b+c)\begin{vmatrix}b-a& (b^2+a^2)(b+a)(b-a)\\ c-a& (c^2+a^2)(c+a)(c-a) \end{vmatrix}$$ $$= (a+b+c)(b-a)(c-a) \begin{vmatrix}1 & (b^2+a^2)(b+a)\\ 1& (c^2+a^2)(c+a) \end{vmatrix}$$ $$=(a+b+c)(b-a)(c-a)(c^3+ac^2+a^2c+a^3-b^3-ab^2-a^2b-a^3)$$ $$= (a+b+c)(b-a)(c-a)(c-b)(a^2+b^2 + c^2 + +bc+ ac+ ab)$$
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