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Summary
- Non-singular matrix
A matrix $A\in M_{n\times n}(F)$ is called non-singular or invertible if there exists a matrix $B\in M_{n\times n}(F)$ such that $$AB=I_n=BA.$$ Any matrix $B$ with the above property is called an inverse of $A$. If $A$ does not have an inverse, $A$ is called singular. - Inverses are unique
If $A$ has inverse $B$ and $C$, then $B=C$.
(Since $B=BI_n=B(AC)=(BA)C=I_nC=C$) - If $A$ and $B$ are non-singular matrices of the same size, then so is $AB$. Moreover $$(AB)^{-1}=B^{-1}A^{-1}.$$
(Since $(AB)(B^{-1}A^{-1})=ABB^{-1}A^{-1}=AI_nA^{-1}=I_n$)
The above result generalizes to a product of $m$ non-singular matrices: If $A_1,\cdots, A_m$ are non-singular $n\times n$ matrices, then the product $A_1\cdots A_n$ is also non-singular. Moreover $$(A_1\cdots A_m)^{-1}=A_m^{-1}\cdots A_{1}^{-1}.$$ - Let $A=\begin{bmatrix}a& b \\ c&d \end{bmatrix}$ and $\Delta=ad-bc\neq0$. Then $A$ is non-singular. Also $$A^{-1}=\Delta^{-1}\begin{bmatrix}d & -b\\ -c &a \end{bmatrix}$$
- If the coefficient matrix $A$ of a system of $n$ equations in $n$ unknown is non-singular, then the system $AX=B$ has the unique solution $X=A^{-1}B$.
- Cramer's rule for 2 equations in 2 unknowns
The system $$\begin{cases}ax+by=e \\ cx+dy=f \end{cases}$$ has a unique solution if $\Delta=\begin{vmatrix}a & b\\ c&d \end{vmatrix} \neq0$, namely $$x={\Delta_1\over\Delta},\ y={\Delta_2\over\Delta}$$ where $$\Delta_1=\begin{vmatrix}e & b \\ f & d \end{vmatrix},\ \Delta_2=\begin{vmatrix}a & e \\ c & f \end{vmatrix}. $$ Based on the above theorem, we have the following corollary:
The homogeneous system $$\begin{cases}ax+by=0 \\ cx+dy=0 \end{cases}$$ has only the trivial solution (i.e. zero) if $\Delta=\begin{vmatrix}a & b\\ c & d \end{vmatrix}\neq0.$ - Let $A$ be a square matrix. If $A$ is non-singular, the homogeneous system $AX=0$ has only the trivial solution. Equivalently, if the homogeneous system $AX=0$ has a non-trivial solution, then $A$ is singular.
(Since if $A$ is non-singular then it has inverse matrix and hence $AX=0\Rightarrow X=A^{-1}0=0$) - Elementary row matrices
To each of the three types of elementary row operation, there corresponds an elementary row matrix, denoted by $E_{ij},\ E_{i}(t),\ E_{ij}(t)$:
- $E_{ij}, (i\neq j)$ is obtained from the identity matrix $I_n$ by interchanging rows $i$ and $j$.
- $E_{i}(t)$ is obtained by multiplying the $i-$th row of $I_n$ by $t$.
- $E_{ij}(t),\ (i\neq j)$ is obtained from $I_n$ by adding $t$ times the $j-$th row of $I_n$ to the $i-$th row.
- If a matrix $A$ is pre-multiplied by an elementary row matrix, the resulting matrix is the one obtained b performing the corresponding elementary row-operation on $A$.
- Elementary row-matrices are non-singular. Indeed,
- $E_{ij}^{-1}=E_{ij}$
- $E_{i}^{-1}(t)=E_{i}(t^{-1})$
- $E_{ij}^{-1}(t)=E_{ij}(-t)$
- Let $A$ be non-singular $n\times n$ matrix. Then
- $A$ is row-equivalent to $I_n$,
- $A$ is a product of elementary row matrices.
- Let $A$ be $n\times n$ and suppose that $A$ is row-equivalent to $I_n$. Then $A$ is non-singular and $A^{-1}$ can be found by performing the same sequence of elementary row operation on $I_n$ as were used to convert $A$ to $I_n$.
(Since $I_n=E_r\cdots E_1A=BA\Rightarrow A^{-1}=B=(E_r\cdots E_1)I_n$, which means $A^{-1}$ is obtained from $I_n$ by performing the same sequence of elementary row operations as were used to convert $A$ to $I_n$.)
An important corollary is that if $A$ is singular, then $A$ is row-equivalent to a matrix whose last row is zero. - The transpose of a matrix
Let $A$ be an $m\times n$ matrix. Then $A^{t}$, the transpose of $A$, is the matrix obtained by interchanging the rows and columns of $A$. In other words if $A=[a_{ij}]$, then $(A^{t})_{ji}=a_{ij}$. Consequently $A^{t}$ is $n\times m$.
- $(A^{t})^{t}=A$;
- $(A\pm B)^{t}=A^{t}\pm B^{t}$ if $A$ and $B$ are $m\times n$;
- $(sA)^{t}=sA^{t}$ if $s$ is a scalar;
- $(AB)^{t}=B^{t}A^{t}$ if $A$ is $m\times n$ and $B$ is $n\times p$;
(Suppose $A=[a_{ij}],\ B=[b_{jk}]$, we have $((AB)^{t})_{ki}=(AB)_{ik}=\displaystyle\sum_{j=1}^{n}a_{ij}b_{jk} = \sum_{j=1}^{n}(B^{t})_{kj}(A^{t})_{ji}=(B^{t}A^{t})_{ki}$) - If $A$ is non-singular, then $A^{t}$ is also non-singular and $$(A^{t})^{-1}(A^{-1})^{t};$$
- $X^{t}X=x_1^2+\cdots+x_n^2$ if $X=[x_1,\cdots,x_n]^{t}$ is a column vector.
- Symmetric matrix
A matrix $A$ is symmetric if $A^{t}=A$. In other words $A$ is square (say $n\times n$) and $a_{ji}=a_{ij}$ for all $1\leq i \leq n,\ 1 \leq j \leq n$. Hence $$A=\begin{bmatrix}a&b\\b&c \end{bmatrix}$$ is a general $2\times2$ symmetric matrix. - Skew-symmetric matrix
A matrix $A$ is called skew-symmetric if $A^{t}=-A$. In other words $A$ is square (say $n\times n$) and $a_{ji}=-a_{ij}$ for all $1\leq i \leq n,\ 1 \leq j \leq n$ (note that $a_{ii}=0$). Hence $$A=\begin{bmatrix}0&b\\-b&0 \end{bmatrix}$$ is a general $2\times2$ skew-symmetric matrix. - Normal equations
Suppose $AX=B$ represents a system of linear equations with real coefficients. The associated system $A^{t}AX=A^{t}B$ is always consistent and that the any solution of this system minimize the sum $r_1^2+\cdots+r_m^2$, where $r_1,\cdots,r_m$ (the residuals) are defined by $$r_i=a_{i1}x_1+\cdots+a_{in}x_n-b_i$$ for $i=1,\cdots, m$. The equations represented by $A^{t}AX=A^{t}B$ are called normal equations corresponding to the system $AX=B$ and any solution of the system of normal equations is called a least squares solution of the original system. More details can be found at HERE.
Problems 2.7
1. Let $A=\begin{bmatrix}1 & 4\\ -3 &1 \end{bmatrix}$. Prove that $A$ is non-singular, find $A^{-1}$ and express $A$ as a product of elementary row matrices.
Solution:
$$\begin{bmatrix}1& 4& 1& 0\\ -3 &1 &0& 1 \end{bmatrix}\Rightarrow R_2+3R_1 \begin{bmatrix}1& 4& 1& 0\\ 0 &13 &3& 1 \end{bmatrix}$$ $$\Rightarrow {1\over13}R_2 \begin{bmatrix}1& 4& 1& 0\\ 0 &1 &{3\over13}& {1\over13} \end{bmatrix} \Rightarrow R_1-4R_2 \begin{bmatrix}1& 0& {1\over13}& -{4\over13}\\ 0 &1 &{3\over13}& {1\over13} \end{bmatrix}$$ Thus $A$ is non-singular and $A^{-1}= \begin{bmatrix}{1\over13}& -{4\over13}\\ {3\over13}& {1\over13} \end{bmatrix}$.
Next, we have $$I_2=E_{12}(-4)E_{2}({1\over13})E_{21}(3)A$$ $$\Rightarrow A^{-1}=E_{12}(-4)E_{2}({1\over13})E_{21}(3)$$ $$\Rightarrow A=(A^{-1})^{-1}=(E_{12}(-4)E_{2}({1\over13})E_{21}(3))^{-1}$$ $$=E_{21}(3)^{-1}E_{2}({1\over13})^{-1}E_{12}(-4)^{-1} = E_{21}(-3)E_{2}(13)E_{12}(4)$$
2. A square matrix $D=[d_{ij}]$ is called diagonal if $d_{ij}=0$ for $i\neq j$. (That is the off-diagonal elements are zero.) Prove that pre-multiplication of a matrix $A$ by a diagonal matrix $D$ results in matrix $DA$ whose rows are the rows of $A$ multiplied by the respective diagonal elements of $D$. State and prove a similar result for post-multiplication by a diagonal matrix.
Let $\text{diag}(a_1,\cdots,a_n)$ denote the diagonal matrix whose diagonal elements $d_{ii}$ are $a_1, \cdots, a_n$, respectively. Show that $$\text{diag}(a_1,\cdots,a_n)\text{diag}(b_1,\cdots,b_n)=\text{diag}(a_1b_1,\cdots,a_nb_n)$$ and deduce that if $a_1\cdots a_n\neq0$, then $\text{diag}(a_1,\cdots,a_n)$ is non-singular and $$(\text{diag}(a_1,\cdots,a_n))^{-1}=\text{diag}(a_1^{-1},\cdots,a_n^{-1}).$$ Also prove that $\text{diag}(a_1,\cdots,a_n)$ is singular if $a_i=0$ for some $i$.
Solution:
Let $D=[d_{ij}]$ be an $m\times m$ diagonal matrix and let $A=[a_{jk}]$ be an $m\times n$ matrix. Thus $$(DA)_{ik}=\sum_{j=1}^{m}d_{ij}a_{jk} = d_{ii}a_{ik}$$ for $i=1,\cdots,m$ as $d_{ij}=0$ while $i\neq j$. It follows that the $i-$th row of $DA$ is obtained by multiplying the $i$th row of $A$ by $d_{ii}$.
Similarly, post-multiplication of a matrix $A$ by a diagonal matrix $D$ results in a matrix whose columns are those of $A$, multiplied by the respective diagonal elements of $D$.
Next, $$\text{diag}(a_1,\cdots,a_n)\text{diag}(b_1,\cdots,b_n) = \text{diag}(a_1b_1,\cdots,a_nb_n)$$ can be seen as the pre-multiplication of the matrix $\text{diag}(b_1,\cdots,b_n)$ by the diagonal matrix $\text{diag}(a_1,\cdots,a_n)$.
Finally, if $a_i\neq0$ for $i=1,\cdots,n$, then $a_i^{-1}$ exists and hence $$\text{diag}(a_1,\cdots,a_n)\text{diag}(a_1^{-1},\cdots,a_n^{-1})=\text{diag}(1,\cdots,1)=I_n$$ which means $\text{diag}(a_1,\cdots,a_n)$ is non-singular and its inverse is $\text{diag}(a_1^{-1},\cdots,a_n^{-1})$.
3. Let $A=\begin{bmatrix}0& 0& 2\\1& 2& 6\\ 3&7&9 \end{bmatrix}$. Prove that $A$ is non-singular, find $A^{-1}$ and express $A$ as a product of elementary row matrices.
Solution:
$$\begin{bmatrix}0& 0& 2& 1 & 0 & 0\\ 1& 2&6 & 0 & 1 & 0\\ 3& 7& 9 & 0& 0& 1 \end{bmatrix}$$ $$\Rightarrow R_1\leftrightarrow R_2 \begin{bmatrix} 1& 2&6 & 0 & 1 & 0\\ 0& 0& 2& 1 & 0 & 0\\3& 7& 9 & 0& 0& 1 \end{bmatrix}$$ $$\Rightarrow R_3-3R_1 \begin{bmatrix} 1& 2&6 & 0 & 1 & 0\\ 0& 0& 2& 1 & 0 & 0\\0& 1&- 9 & 0& -3& 1 \end{bmatrix}$$ $$\Rightarrow R_2\leftrightarrow R_3 \begin{bmatrix} 1& 2&6 & 0 & 1 & 0\\ 0& 1&- 9 & 0& -3& 1\\0& 0& 2& 1 & 0 & 0 \end{bmatrix}$$ $$\Rightarrow {1\over2}R_3 \begin{bmatrix} 1& 2&6 & 0 & 1 & 0\\ 0& 1&- 9 & 0& -3& 1\\0& 0& 1& {1\over2} & 0 & 0 \end{bmatrix}$$ $$\Rightarrow \begin{cases}R_1-2R_2\\ R_2+9R_3 \end{cases} \begin{bmatrix} 1& 0&24 & 0 & 7 & -2\\ 0& 1&0 & {9\over2}& -3& 1\\0& 0& 1& {1\over2} & 0 & 0 \end{bmatrix}$$ $$\Rightarrow R_1-24R_3 \begin{bmatrix} 1& 0&0& -12 & 7 & -2\\ 0& 1&0 & {9\over2}& -3& 1\\0& 0& 1& {1\over2} & 0 & 0 \end{bmatrix}$$ Thus $A$ is non-singular and $A^{-1}= \begin{bmatrix} -12 & 7 & -2\\ {9\over2}& -3& 1\\{1\over2} & 0 & 0 \end{bmatrix}$.
Next, since we have $$I_3=E_{13}(-24)E_{23}(9)E_{12}(-2)E_{3}({1\over2})E_{23}E_{31}(-3)E_{12}A$$ $$\Rightarrow A^{-1}= E_{13}(-24)E_{23}(9)E_{12}(-2)E_{3}({1\over2})E_{23}E_{31}(-3)E_{12}$$ Thus $$A=(A^{-1})^{-1}=(E_{13}(-24)E_{23}(9)E_{12}(-2)E_{3}({1\over2})E_{23}E_{31}(-3)E_{12})^{-1}$$ $$=E_{12}E_{31}(3)E_{23}E_{3}(2)E_{12}(2)E_{23}(-9)E_{13}(24)$$
4. Find the rational number $k$ for which the matrix $A=\begin{bmatrix}1& 2& k\\ 3& -1& 1\\ 5& 3& -5 \end{bmatrix}$ is singular.
Solution:
$$A=\begin{bmatrix}1& 2& k\\ 3& -1& 1\\ 5& 3& -5 \end{bmatrix}$$ $$\Rightarrow \begin{cases}R_2-3R_1\\ R_3-5R_1\end{cases}\begin{bmatrix}1& 2& k\\ 0& -7& 1-3k\\ 0& -7& -5-5k \end{bmatrix}$$ $$\Rightarrow R_3-R_2\begin{bmatrix}1& 2& k\\ 0& -7& 1-3k\\ 0& 0& -6-2k \end{bmatrix}$$ Thus $A$ is singular if $-6-2k=0\Rightarrow k=-3$.
5. Prove that $A=\begin{bmatrix}1& 2\\ -2& -4 \end{bmatrix}$ is singular and find a non-singular matrix $P$ such that $PA$ has last row zero.
Solution:
$$A=\begin{bmatrix}1& 2\\ -2& -4 \end{bmatrix}$$ $$\Rightarrow R_2+2R_1 \begin{bmatrix}1& 2\\ 0& 0 \end{bmatrix}$$ That is, $A$ is singular and $P=E_{21}(2)=\begin{bmatrix}1& 0\\ 2& 1 \end{bmatrix}$.
6. If $A=\begin{bmatrix}1& 4\\-3& 1 \end{bmatrix}$, verify that $A^2-2A+13I_2=0$ and deduce that $A^{-1}=-{1\over13}(A-2I_2)$.
Solution:
$$A=\begin{bmatrix}1& 4\\-3& 1 \end{bmatrix}\Rightarrow A^2=AA=\begin{bmatrix}1& 4\\-3& 1 \end{bmatrix}\cdot\begin{bmatrix}1& 4\\-3& 1 \end{bmatrix} =\begin{bmatrix}-11& 8\\-6& -11 \end{bmatrix}$$ $$\Rightarrow A^2-2A+13I_2= \begin{bmatrix}-11& 8\\-6& -11 \end{bmatrix} - \begin{bmatrix}2& 8\\-6& 2 \end{bmatrix} + \begin{bmatrix}13& 0\\0& 13 \end{bmatrix} = 0$$ Since $A^2-2A+13I_2=0$, we have $$A(A-2I_2)=-13I_2=(A-2I_2)A \Rightarrow A^{-1}=-{1\over13}(A-2I_2)$$
7. Let $A = \begin{bmatrix}1& 1& -1\\0& 0& 1\\ 2& 1& 2 \end{bmatrix}$.
(1) Verify that $A^3=3A^2-3A+I_3$.
(2) Express $A^4$ in terms of $A^2$, $A$ and $I_3$ and hence calculate $A^4$ explicitly.
(3) Use (1) to prove that $A$ is non-singular and find $A^{-1}$ explicitly.
Solution:
(1) $$A = \begin{bmatrix}1& 1& -1\\0& 0& 1\\ 2& 1& 2 \end{bmatrix} \Rightarrow A^2=AA=\begin{bmatrix}1& 1& -1\\0& 0& 1\\ 2& 1& 2 \end{bmatrix}\cdot\begin{bmatrix}1& 1& -1\\0& 0& 1\\ 2& 1& 2 \end{bmatrix} = \begin{bmatrix}-1& 0& -2\\2& 1& 2\\ 6& 4& 3 \end{bmatrix}$$ $$\Rightarrow A^3=A^2A = \begin{bmatrix}-1& 0& -2\\2& 1& 2\\ 6& 4& 3 \end{bmatrix}\cdot \begin{bmatrix}1& 1& -1\\0& 0& 1\\ 2& 1& 2 \end{bmatrix} = \begin{bmatrix}-5& -3& -3\\6& 4& 3\\ 12& 9& 4 \end{bmatrix}$$ Thus $$3A^2-3A+I_3= \begin{bmatrix}-3& 0& -6\\6& 3& 6\\ 18& 12& 9 \end{bmatrix} - \begin{bmatrix}3& 3& -3\\0& 0& 3\\ 6& 3& 6 \end{bmatrix} + \begin{bmatrix}1& 0& 0\\0& 1& 0\\ 0& 0& 1 \end{bmatrix}$$ $$ =\begin{bmatrix}-5& -3& -3\\6& 4& 3\\ 12& 9& 4 \end{bmatrix} = A^3$$
(2) $$A^4=A^3A=(3A^2-3A+I_3)A = 3A^3-3A^2+A$$ $$ = 3(3A^2-3A+I_3) -3A^2 + A = 6A^2-8A+3I_3$$ $$=\begin{bmatrix}-6& 0& -12\\12& 6& 12\\ 36& 24& 18 \end{bmatrix} - \begin{bmatrix}8& 8& -8\\0& 0& 8\\ 16& 8& 16 \end{bmatrix} + \begin{bmatrix}3& 0& 0\\0& 3& 0\\ 0& 0& 3 \end{bmatrix}$$ $$= \begin{bmatrix}-11& -8& -4\\12& 9& 4\\ 20& 16& 5 \end{bmatrix}$$
(3) Since $A^3=3A^2-3A+I_3$, we have $$A^3-3A^2+3A=I_3$$ $$\Rightarrow A(A^2-3A+3I_3) = I_3 = (A^2-3A+3I_3)A$$ Thus $A$ is non-singular and $$A^{-1}= A^2-3A+3I_3 = \begin{bmatrix}-1& 0& -2\\2& 1& 2\\ 6& 4& 3 \end{bmatrix} - \begin{bmatrix}3& 3& -3\\0& 0& 3\\ 6& 3& 6 \end{bmatrix} + \begin{bmatrix}3& 0& 0\\0& 3& 0\\ 0& 0& 3 \end{bmatrix}$$ $$= \begin{bmatrix}-1& -3& 1\\2& 4& -1\\ 0& 1& 0 \end{bmatrix}$$
8. (1) Let $B$ be an $n\times n$ matrix such that $B^3=0$. If $A=I_n-B$, prove that $A$ is non-singular and $A^{-1}=I_n+B+B^2$.
Show that the system of linear equations $AX=b$ has the solution $$X=b+Bb+B^2b.$$
(2) If $B=\begin{bmatrix}0& r& s\\0& 0& t\\0& 0& 0 \end{bmatrix}$, verify that $B^3=0$ and use (1) to determine $(I_3-B)^{-1}$ explicitly.
Solution:
(1) Since $$AA^{-1}=(I_n-B)(I_n+B+B^2) = (I_n+B+B^2)-(B+B^2+B^3) = I_n$$ $$A^{-1}A=(I_n+B+B^2)(I_n-B)=(I_n-B)+(B-B^2)+(B^2-B^3) = I_n$$ Hence $A$ is non-singular and $A^{-1}=I_n+B+B^2$. And thus $$AX=b\Rightarrow A^{-1}AX=A^{-1}b\Rightarrow X= (I_n+B+B^2) b= b+Bb+B^2b$$
(2) $$B= \begin{bmatrix}0& r& s\\0& 0& t\\0& 0& 0 \end{bmatrix}$$ $$\Rightarrow B^2=BB= \begin{bmatrix}0& r& s\\0& 0& t\\0& 0& 0 \end{bmatrix} \cdot \begin{bmatrix}0& r& s\\0& 0& t\\0& 0& 0 \end{bmatrix} = \begin{bmatrix}0& 0& rt\\0& 0& 0\\0& 0& 0 \end{bmatrix}$$ $$\Rightarrow B^3=B^2B= \begin{bmatrix}0& 0& rt\\0& 0& 0\\0& 0& 0 \end{bmatrix}\cdot \begin{bmatrix}0& r& s\\0& 0& t\\0& 0& 0 \end{bmatrix} = 0$$ By the above result, we have $$(I_3-B)^{-1} = I_3+B+B^2$$ $$= \begin{bmatrix}1& 0& 0\\0& 1& 0\\ 0& 0& 1 \end{bmatrix} + \begin{bmatrix}0& r& s\\0& 0& t\\0& 0& 0 \end{bmatrix} + \begin{bmatrix}0& 0& rt\\0& 0& 0\\0& 0& 0 \end{bmatrix}$$ $$=\begin{bmatrix}1& r& s+rt\\0& 1& t\\0& 0& 1 \end{bmatrix}$$
9. Let $A$ be $n\times n$.
(1) If $A^2=0$, prove that $A$ is singular.
(2) If $A^2 = A$ and $A \neq I_n$, prove that $A$ is singular.
Solution:
(1) Suppose $A$ is non-singular, then $$A^{-1}A^2=A^{-1}0\Rightarrow A^{-1}AA=0 \Rightarrow A=0$$ whcih is contradiction since zero matrix is singular.
(2) Suppose $A$ is non-singular, then $$A^{-1}A^2 = A^{-1}A\Rightarrow A^{-1}AA=I_n \Rightarrow A=I_n$$ which means if $A^2 = A$ and $A \neq I_n$, then $A$ is singular.
10. Use Question 7 to solve the system of equations $$\begin{cases}x+y-z=a\\ z=b\\2x+y+2z=c \end{cases}$$ where $a$, $b$, $c$ are given rationals. Check your answer using the Gauss-Jordan algorithm.
Solution:
Rewrite the system as $AX=B$: $$\begin{bmatrix}1& 1& -1\\ 0& 0& 1\\ 2& 1& 2 \end{bmatrix}\cdot \begin{bmatrix}x\\y\\z \end{bmatrix} = \begin{bmatrix}a\\b\\c \end{bmatrix}$$ where $A= \begin{bmatrix}1& 1& -1\\ 0& 0& 1\\ 2& 1& 2 \end{bmatrix}$ satisfies $A^3=3A^2-3A+I_3$, and hence $A^{-1}=A^2-3A+3I_3$ according to Question 7. Thus $$X=A^{-1}B = (A^2-3A+3I_3)B$$ $$=\begin{bmatrix}-1& -3& 1\\2& 4& -1\\ 0& 1& 0 \end{bmatrix}\cdot \begin{bmatrix}a\\b\\c \end{bmatrix}=\begin{bmatrix}-a-3b+c\\2a+4b-c\\b \end{bmatrix} $$ That is, the solution is $$\begin{cases}x=a-3b+c\\y=2a+4b-c\\z=b\end{cases}$$
11. Determine explicitly the following products of $3\times3$ elementary row matrices.
(1) $E_{12}E_{23}$; (2) $E_{1}(5)E_{12}$; (3) $E_{12}(3)E_{21}(-3)$; (4) $(E_{1}(100))^{-1}$; (5) $E_{12}^{-1}$; (6) $(E_{12}(7))^{-1}$; (7) $(E_{12}(7)E_{31}(1))^{-1}$
Solution:
(1) $$E_{12}E_{23} = E_{12}\begin{bmatrix}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0 \end{bmatrix} = \begin{bmatrix}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0 \end{bmatrix}$$
(2) $$E_{1}(5)E_{12}= E_{1}(5) \begin{bmatrix}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1 \end{bmatrix} = \begin{bmatrix}0& 5& 0\\ 1& 0& 0\\ 0& 0& 1 \end{bmatrix}$$
(3) $$E_{12}(3)E_{21}(-3) = E_{12}(3) \begin{bmatrix}1& 0& 0\\ -3& 1& 0\\ 0& 0& 1 \end{bmatrix} = \begin{bmatrix}-8& 3& 0\\ -3& 1& 0\\ 0& 0& 1 \end{bmatrix}$$
(4) $$(E_{1}(100))^{-1} =E_{1}({1\over100})= \begin{bmatrix}{1\over100}& 0& 0\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix}$$
(5) $$E_{12}^{-1} = E_{12} = \begin{bmatrix}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1 \end{bmatrix}$$
(6) $$(E_{12}(7))^{-1} = E_{12}(-7) = \begin{bmatrix}1& -7& 0\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix}$$
(7) $$(E_{12}(7)E_{31}(1))^{-1} = E_{31}(-1)E_{12}(-7) = E_{31}(-1) \begin{bmatrix}1& -7& 0\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix} = \begin{bmatrix}1& -7& 0\\ 0& 1& 0\\ -1& 7& 1 \end{bmatrix}$$
12. Let $A$ be the following product of $4\times4$ elementary row matrices: $$A=E_{3}(2)E_{14}E_{42}(3).$$ Find $A$ and $A^{-1}$ explicitly.
Solution:
$$A = E_{3}(2)E_{14}\begin{bmatrix}1 &0 &0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 3& 0& 1 \end{bmatrix} = E_{3}(2) \begin{bmatrix}0& 3& 0& 1\\0& 1& 0& 0\\ 0& 0& 1& 0\\1 &0 &0& 0 \end{bmatrix} =\begin{bmatrix}0& 3& 0& 1\\0& 1& 0& 0\\ 0& 0& 2& 0\\1 &0 &0& 0 \end{bmatrix} $$ $$A^{-1}=E_{42}(-3)E_{14}E_{3}({1\over2}) =E_{42}(-3)E_{14}\begin{bmatrix}1 &0 &0& 0\\ 0& 1& 0& 0\\ 0& 0& {1\over2}& 0\\ 0& 0& 0& 1 \end{bmatrix}$$ $$=E_{42}(-3) \begin{bmatrix}0& 0& 0& 1\\ 0& 1& 0& 0\\ 0& 0& {1\over2}& 0\\1 &0 &0& 0 \end{bmatrix}= \begin{bmatrix}0& 0& 0& 1\\ 0& 1& 0& 0\\ 0& 0& {1\over2}& 0\\1 &-3 &0& 0 \end{bmatrix}$$
13. Determine which of the following matrices over $\mathbb{Z}_2$ are non-singular and find the inverse, where possible.
(a) $\begin{bmatrix}1& 1& 0& 1\\0& 0& 1& 1\\1& 1& 1& 1\\1& 0& 0& 1 \end{bmatrix}$; (b) $\begin{bmatrix}1& 1& 0& 1\\0& 1& 1& 1\\1& 0& 1& 0\\1& 1& 0& 1 \end{bmatrix}$.
Solution:
In $\mathbb{Z}_2$ field, $1+1=0 \Rightarrow -1 = 1$.
(a) $$\begin{bmatrix}1& 1& 0& 1 & 1& 0& 0& 0\\0& 0& 1& 1 &0& 1& 0& 0\\1& 1& 1& 1& 0& 0& 1& 0\\1& 0& 0& 1& 0& 0& 0& 1 \end{bmatrix}$$ $$\Rightarrow \begin{cases}R_1-R_4\\R_3-R_4 \end{cases}\begin{bmatrix}0& 1& 0& 0 & 1& 0& 0& 1\\0& 0& 1& 1 &0& 1& 0& 0\\0& 1& 1& 0& 0& 0& 1& 1\\1& 0& 0& 1& 0& 0& 0& 1 \end{bmatrix}$$ $$\Rightarrow R_3-R_1 \begin{bmatrix}0& 1& 0& 0 & 1& 0& 0& 1\\0& 0& 1& 1 &0& 1& 0& 0\\0& 0& 1& 0& 1& 0& 1& 0\\1& 0& 0& 1& 0& 0& 0& 1 \end{bmatrix}$$ $$\Rightarrow R_2-R_3 \begin{bmatrix}0& 1& 0& 0 & 1& 0& 0& 1\\0& 0& 0& 1 &1& 1& 1& 0\\0& 0& 1& 0& 1& 0& 1& 0\\1& 0& 0& 1& 0& 0& 0& 1 \end{bmatrix}$$ $$\Rightarrow R_4-R_2 \begin{bmatrix}0& 1& 0& 0 & 1& 0& 0& 1\\0& 0& 0& 1 &1& 1& 1& 0\\0& 0& 1& 0& 1& 0& 1& 0\\1& 0& 0& 0& 1& 1& 1& 1 \end{bmatrix}$$ $$\Rightarrow \begin{cases}R_1\leftrightarrow R_4\\R_2\leftrightarrow R_4^{'} \end{cases} \begin{bmatrix}1& 0& 0& 0& 1& 1& 1& 1\\0& 1& 0& 0 & 1& 0& 0& 1\\0& 0& 1& 0& 1& 0& 1& 0\\0& 0& 0& 1 &1& 1& 1& 0\end{bmatrix}$$ Thus the original matrix is non-singular and its inverse is $$\begin{bmatrix}1& 1& 1& 1\\1& 0& 0& 1\\1& 0& 1& 0\\1& 1& 1& 0\end{bmatrix}$$
(b) $$\begin{bmatrix}1& 1& 0& 1\\0& 1& 1& 1\\1& 0& 1& 0\\1& 1& 0& 1 \end{bmatrix}\Rightarrow R_1-R_4 \begin{bmatrix}0& 0& 0& 0\\0& 1& 1& 1\\1& 0& 1& 0\\1& 1& 0& 1 \end{bmatrix}$$ which is singular matrix.
14. Determine which of the following matrices are non-singular and find the inverse, where possible.
(a) $\begin{bmatrix}1& 1& 1\\-1& 1& 0\\2& 0& 0 \end{bmatrix}$; (b) $\begin{bmatrix}2& 2& 4\\1& 0& 1\\0& 1& 0 \end{bmatrix}$; (c) $\begin{bmatrix}4& 6& -3\\0& 0& 7\\0& 0& 5 \end{bmatrix}$; (d) $\begin{bmatrix}2& 0& 0\\0& -5& 0\\0& 0& 7 \end{bmatrix}$; (e) $\begin{bmatrix}1& 2& 4 & 6\\0& 1& 2 & 0\\0 & 0& 1 & 2\\ 0& 0& 0& 2 \end{bmatrix}$; (f) $\begin{bmatrix}1& 2& 3\\4& 5& 6\\5& 7& 9 \end{bmatrix}$.
Solution:
(a) $$\begin{bmatrix}1& 1& 1& 1& 0& 0\\-1& 1& 0& 0& 1& 0\\2& 0& 0& 0& 0& 1 \end{bmatrix}$$ $$\Rightarrow {1\over2}R_3 \begin{bmatrix}1& 1& 1& 1& 0& 0\\-1& 1& 0& 0& 1& 0\\1& 0& 0& 0& 0& {1\over2} \end{bmatrix}$$ $$\Rightarrow \begin{cases}R_1-R_3\\R_2+R_3 \end{cases} \begin{bmatrix}0& 1& 1& 1& 0& -{1\over2}\\0& 1& 0& 0& 1& {1\over2}\\1& 0& 0& 0& 0& {1\over2} \end{bmatrix}$$ $$\Rightarrow R_1-R_2 \begin{bmatrix}0& 0& 1& 1& -1& -1\\0& 1& 0& 0& 1& {1\over2}\\1& 0& 0& 0& 0& {1\over2} \end{bmatrix}$$ $$\Rightarrow R_1\leftrightarrow R_3 \begin{bmatrix}1& 0& 0& 0& 0& {1\over2} \\0& 1& 0& 0& 1& {1\over2}\\0& 0& 1& 1& -1& -1\end{bmatrix}$$ Thus it is non-singular and its inverse is $$\begin{bmatrix} 0& 0& {1\over2} \\0& 1& {1\over2}\\1& -1& -1\end{bmatrix}.$$
(b) $$\begin{bmatrix}2& 2& 4& 1& 0& 0\\1& 0& 1& 0& 1& 0\\0& 1& 0& 0& 0& 1\end{bmatrix}$$ $$\Rightarrow R_1-2R_2 \begin{bmatrix}0& 2& 2& 1& -2& 0\\1& 0& 1& 0& 1& 0\\0& 1& 0& 0& 0& 1\end{bmatrix}$$ $$\Rightarrow R_1-2R_3 \begin{bmatrix}0& 0& 2& 1& -2& -2\\1& 0& 1& 0& 1& 0\\0& 1& 0& 0& 0& 1\end{bmatrix}$$ $$\Rightarrow {1\over2}R_1 \begin{bmatrix}0& 0& 1& {1\over2}& -1& -1\\1& 0& 1& 0& 1& 0\\0& 1& 0& 0& 0& 1\end{bmatrix}$$ $$\Rightarrow R_2-R_1 \begin{bmatrix}0& 0& 1& {1\over2}& -1& -1\\1& 0& 0& -{1\over2}& 2& 1\\0& 1& 0& 0& 0& 1\end{bmatrix}$$ $$\Rightarrow \begin{cases}R_1\leftrightarrow R_3\\ R_1^{'}\leftrightarrow R_2\end{cases} \begin{bmatrix}1& 0& 0& -{1\over2}& 2& 1\\0& 1& 0& 0& 0& 1\\0& 0& 1& {1\over2}& -1& -1\end{bmatrix}$$ Thus it is non-singular and its inverse is $$\begin{bmatrix}-{1\over2}& 2& 1\\0& 0& 1\\{1\over2}& -1& -1\end{bmatrix}.$$
(c) $$\begin{bmatrix}4& 6& -3\\0& 0& 7\\0& 0& 5 \end{bmatrix}\Rightarrow R_2-{7\over5}R_3 \begin{bmatrix}4& 6& -3\\0& 0& 0\\0& 0& 5 \end{bmatrix}$$ which is singular matrix.
(d) $$\begin{bmatrix}2& 0& 0& 1& 0& 0\\0& -5& 0& 0& 1& 0\\0& 0& 7& 0& 0& 1 \end{bmatrix}$$ $$\Rightarrow\begin{cases}{1\over2}R_1\\-{1\over5}R_2\\{1\over7}R_3\end{cases} \begin{bmatrix}1& 0& 0& {1\over2}& 0& 0\\0& 1& 0& 0& -{1\over5}& 0\\0& 0& 1& 0& 0& {1\over7} \end{bmatrix}$$ Thus it is non-singular and its inverse is $$\begin{bmatrix}{1\over2}& 0& 0\\0& -{1\over5}& 0\\0& 0& {1\over7} \end{bmatrix}.$$
(e) $$\begin{bmatrix}1& 2& 4 & 6& 1& 0& 0& 0\\0& 1& 2 & 0& 0& 1& 0& 0\\0 & 0& 1 & 2& 0& 0& 1& 0\\ 0& 0& 0& 2& 0& 0& 0& 1 \end{bmatrix}$$ $$\Rightarrow\begin{cases}R_1-2R_2\\{1\over2}R_4 \end{cases} \begin{bmatrix}1& 0& 0 & 6& 1& -2& 0& 0\\0& 1& 2 & 0& 0& 1& 0& 0\\0 & 0& 1 & 2& 0& 0& 1& 0\\ 0& 0& 0& 1& 0& 0& 0& {1\over2} \end{bmatrix}$$ $$\Rightarrow\begin{cases}R_1-6R_4\\R_2-2R_3 \end{cases} \begin{bmatrix}1& 0& 0 & 0& 1& -2& 0& -3\\0& 1& 0 & -4& 0& 1& -2& 0\\0 & 0& 1 & 2& 0& 0& 1& 0\\ 0& 0& 0& 1& 0& 0& 0& {1\over2} \end{bmatrix}$$ $$\Rightarrow\begin{cases}R_2+4R_4\\R_3-2R_4 \end{cases} \begin{bmatrix}1& 0& 0 & 0& 1& -2& 0& -3\\0& 1& 0 & 0& 0& 1& -2& 2\\0 & 0& 1 & 0& 0& 0& 1& -1\\ 0& 0& 0& 1& 0& 0& 0& {1\over2} \end{bmatrix}$$ Thus it is non-singular and its inverse is $$\begin{bmatrix}1& -2& 0& -3\\ 0& 1& -2& 2\\ 0& 0& 1& -1\\ 0& 0& 0& {1\over2} \end{bmatrix}.$$
(f) $$\begin{bmatrix}1& 2& 3\\4& 5& 6\\5& 7& 9 \end{bmatrix}\Rightarrow R_3-R_1 \begin{bmatrix}1& 2& 3\\4& 5& 6\\4& 5& 6 \end{bmatrix}\Rightarrow R_2-R_3 \begin{bmatrix}1& 2& 3\\0&0& 0\\4& 5& 6 \end{bmatrix}$$ which is singular matrix.
15. Let $A$ be a non-singular $n\times n$ matrix. Prove that $A^{t}$ is non-singular and that $(A^{t})^{-1}=(A^{-1})^{t}$.
Solution:
$$AA^{-1}=I_n=A^{-1}A\Rightarrow (AA^{-1})^{t}= (I_n)^{t}=(A^{-1}A)^{t}$$ $$\Rightarrow (A^{-1})^{t}A^{t}=I_n=A^{t}(A^{-1})^{t}$$ Thus $A^{t}$ is singular and its inverse is $(A^{-1})^{t}$.
16. Prove that $A=\begin{bmatrix}a& b\\c& d \end{bmatrix}$ has no inverse if $ad-bc=0$.
Solution:
From Problems 2.4 Question 3 we know that $A^2-(a+d)A+(ad-bc)I_2=0$, so if $ad-bc=0$ and suppose that $A^{-1}$ exists, then we have $$A^2=(a+d)A\Rightarrow A=(a+d)I_2$$ $$\Rightarrow \begin{bmatrix}a& b\\c& d \end{bmatrix}=\begin{bmatrix}a+d& 0\\ 0& a+d \end{bmatrix}$$ $$\Rightarrow a=b=c=d=0\Rightarrow ad-bc=0$$ which is contradiction.
17. Prove that the real matrix $A=\begin{bmatrix}1& a& b\\-a& 1& c\\-b& -c& 1 \end{bmatrix}$ is non-singular by proving that $A$ is row-equivalent to $I_3$.
Solution:
$$A=\begin{bmatrix}1& a& b\\-a& 1& c\\-b& -c& 1 \end{bmatrix}$$ $$\Rightarrow \begin{cases}R_2+aR_1\\ R_3+bR_1 \end{cases} \begin{bmatrix}1& a& b\\0& 1+a^2& c+ab\\0& ab-c& 1+b^2 \end{bmatrix}$$ $$\Rightarrow {1\over1+a^2}R_2\begin{bmatrix}1& a& b\\0& 1& {c+ab\over1+a^2}\\0& ab-c& 1+b^2 \end{bmatrix}$$ $$\Rightarrow R_3-(ab-c)R_2\begin{bmatrix}1& a& b\\0& 1& {c+ab\over1+a^2}\\0& 0& 1+b^2-{(ab-c)(c+ab)\over1+a^2} \end{bmatrix}$$ Since $$1+b^2-{(ab-c)(c+ab)\over1+a^2} = 1+b^2+{(c+ab)(c-ab)\over1+a^2}$$ $$={1+a^2+b^2+a^2b^2+c^2-a^2b^2\over1+a^2} = {1+a^2+b^2+c^2\over1+a^2}\neq0$$ Thus $A$ is equivalent to $I_3$ and it is non-singular.
18. If $P^{-1}AP=B$, prove that $P^{-1}A^{n}P=B^{n}$ for $n\geq1$.
Solution:
Use mathematical induction. Suppose that $P^{-1}A^{n}P=B^{n}$, then $$B^{n+1}= (P^{-1}AP)^{n+1}= (P^{-1}AP)^{n}(P^{-1}AP)$$ $$=P^{-1}A^{n}PP^{-1}AP = P^{-1}A^{n}AP=P^{-1}A^{n+1}P$$
19. Let $A=\begin{bmatrix}{2\over3}&{1\over4}\\ {1\over3}&{3\over4} \end{bmatrix}$, $P=\begin{bmatrix}1& 3\\-1& 4 \end{bmatrix}$. Verify that $P^{-1}AP=\begin{bmatrix}{5\over12}&0\\ 0&1 \end{bmatrix}$ and deduce that $$A^{n}={1\over7}\begin{bmatrix}3 &3\\4& 4 \end{bmatrix} + {1\over7}\left({5\over12}\right) ^{n}\begin{bmatrix}4 &-3\\-4& 3 \end{bmatrix}.$$
Solution:
$$\begin{bmatrix}1& 3 &1& 0\\-1& 4& 0& 1\end{bmatrix}$$ $$\Rightarrow R_2+R_1 \begin{bmatrix}1& 3 &1& 0\\0& 7& 1& 1\end{bmatrix}$$ $$\Rightarrow {1\over7}R_2 \begin{bmatrix}1& 3 &1& 0\\0& 1& {1\over7}& {1\over7}\end{bmatrix}$$ $$\Rightarrow R_1-3R_2 \begin{bmatrix}1& 0 &{4\over7}& -{3\over7}\\0& 1& {1\over7}& {1\over7}\end{bmatrix}$$ Thus $P^{-1}={1\over7}\begin{bmatrix}4& -3\\1& 1 \end{bmatrix}$, and $$P^{-1}AP= {1\over7}\begin{bmatrix}4& -3\\1& 1 \end{bmatrix}\cdot \begin{bmatrix}{2\over3}&{1\over4}\\ {1\over3}&{3\over4} \end{bmatrix} \cdot \begin{bmatrix}1& 3\\-1& 4 \end{bmatrix} = {1\over7}\begin{bmatrix}{5\over3}& -{5\over4}\\ 1&1 \end{bmatrix} \cdot \begin{bmatrix}1& 3\\-1& 4 \end{bmatrix}$$ $$={1\over7} \begin{bmatrix}{35\over12}& 0\\0& 7\end{bmatrix} = \begin{bmatrix} {5\over12}&0\\ 0&1 \end{bmatrix}=B$$ From the previous question, we know that $P^{-1}A^{n}P=B^{n}= \begin{bmatrix} \left({5\over12}\right)^{n}&0\\ 0&1^n \end{bmatrix}$. Hence $$A^n=PB^{n}P^{-1}= \begin{bmatrix}1& 3\\-1& 4 \end{bmatrix}\cdot \begin{bmatrix} \left({5\over12}\right)^{n}&0\\ 0&1 \end{bmatrix} \cdot {1\over7}\begin{bmatrix}4& -3\\1& 1 \end{bmatrix}$$ $$={1\over7}\begin{bmatrix}\left({5\over12}\right)^{n}&3\\-\left({5\over12}\right)^{n} &4 \end{bmatrix}\cdot \begin{bmatrix}4& -3\\1& 1 \end{bmatrix}$$ $$={1\over7} \begin{bmatrix}4\cdot\left({5\over12}\right)^{n}+3& -3\cdot \left({5\over12}\right)^{n}+3 \\-4\cdot\left({5\over12}\right)^{n}+4& 3\cdot \left({5\over12}\right)^{n}+4 \end{bmatrix}$$ $$= {1\over7}\begin{bmatrix}3 &3\\4& 4 \end{bmatrix} + {1\over7}\left({5\over12}\right) ^{n}\begin{bmatrix}4 &-3\\-4& 3 \end{bmatrix}$$
20. Let $A=\begin{bmatrix}a& b\\c& d \end{bmatrix}$ be a Markov matrix; that is a matrix whose elements are non-negative and satisfy $a+c=1=b+d$. Also let $P=\begin{bmatrix}b& 1\\c& -1 \end{bmatrix}$. Prove that if $A\neq I_2$ then
(1) $P$ is non-singular and $P^{-1}AP=\begin{bmatrix}1& 0\\ 0& a+d-1 \end{bmatrix}$,
(2) $A^{n}\to{1\over b+c}\begin{bmatrix}b& b\\ c& c \end{bmatrix}$ as $n\to\infty$, if $A\neq \begin{bmatrix}0& 1\\ 1& 0 \end{bmatrix}$.
Solution:
(1) If $P$ is non-singular then $\Delta=-b-c\neq0$. Since $a+c=1$ and all of the elements are non-negative, so $-b-c = -(b+c) \leq0$. If $b+c=0$, then $b=c=0\Rightarrow a=d=1\Rightarrow A=\begin{bmatrix}1& 0\\0& 1 \end{bmatrix} = I_2$. Thus, if $A\neq I_2$ then $P$ is non-singular and $P^{-1}={1\over-b-c} \begin{bmatrix}-1& -1\\-c& b \end{bmatrix}$. Hence $$P^{-1}AP= {1\over-b-c} \begin{bmatrix}-1& -1\\-c& b \end{bmatrix} \cdot \begin{bmatrix}a& b\\c& d \end{bmatrix} \cdot \begin{bmatrix}b& 1\\c& -1 \end{bmatrix}$$ $$= {1\over-b-c}\cdot \begin{bmatrix}-a-c& -b-d\\-ac+bc& -bc+bd \end{bmatrix} \cdot \begin{bmatrix}b& 1\\c& -1 \end{bmatrix}$$ $$= {1\over-b-c}\cdot \begin{bmatrix}-1& -1\\-ac+bc& -bc+bd \end{bmatrix} \cdot \begin{bmatrix}b& 1\\c& -1 \end{bmatrix}$$ $$= {1\over-b-c}\cdot \begin{bmatrix}-b-c& 0\\-abc+b^2c-bc^2+bcd& -ac+bc+bc-bd \end{bmatrix}$$ $$= {1\over-b-c}\cdot \begin{bmatrix}-b-c& 0\\-bc(a-b+c-d)& -ac+b(1-a)+c(1-d)-bd \end{bmatrix}$$ $$= {1\over-b-c}\cdot \begin{bmatrix}-b-c& 0\\0 & (-b-c)(a+d-1) \end{bmatrix}= \begin{bmatrix}1& 0\\ 0& a+d-1 \end{bmatrix}$$
(2) Since $$P^{-1}A^{n}P=(P^{-1}AP)^n = \begin{bmatrix}1& 0\\ 0& (a+d-1)^n \end{bmatrix}$$ Hence $$A^{n} =P \begin{bmatrix}1& 0\\ 0& (a+d-1)^n \end{bmatrix} P^{-1}$$ $$= \begin{bmatrix}b& 1\\c& -1 \end{bmatrix}\cdot \begin{bmatrix}1& 0\\ 0& (a+d-1)^n \end{bmatrix} \cdot {1\over-b-c} \begin{bmatrix}-1& -1\\-c& b \end{bmatrix}$$ Since when $n\to\infty$, $a+d-1\to0$ for $0 < a+d < 2$, otherwise $A= \begin{bmatrix}0& 1\\ 1& 0 \end{bmatrix}$. Thus $$A^{n} = \begin{bmatrix}b& 1\\c& -1 \end{bmatrix}\cdot \begin{bmatrix}1& 0\\ 0& 0 \end{bmatrix} \cdot {1\over-b-c} \begin{bmatrix}-1& -1\\-c& b \end{bmatrix}$$ $$={1\over-b-c} \begin{bmatrix}b& 0\\ c& 0 \end{bmatrix} \cdot \begin{bmatrix}-1& -1\\-c& b \end{bmatrix}$$ $$= {1\over-b-c} \begin{bmatrix}-b& -b\\-c& -c \end{bmatrix} ={1\over b+c} \begin{bmatrix}b& b\\c& c \end{bmatrix}. $$ 21. If $X=\begin{bmatrix}1& 2\\ 3& 4\\ 5& 6 \end{bmatrix}$ and $Y=\begin{bmatrix}-1\\3\\4 \end{bmatrix}$, find $XX^{t}$, $X^{t}X$, $YY^{t}$, $Y^{t}Y$.
Solution:
$X^{t}=\begin{bmatrix}1& 3& 5\\2& 4& 6 \end{bmatrix}$ and $Y^{t}=\begin{bmatrix}-1& 3& 4 \end{bmatrix}$. Note that $A^{t}A$ MUST be symmetric. $$XX^{t} = \begin{bmatrix}1& 2\\ 3& 4\\ 5& 6 \end{bmatrix}\cdot \begin{bmatrix}1& 3& 5\\2& 4& 6 \end{bmatrix} = \begin{bmatrix}5& 11& 17\\ 11& 25 & 39\\ 17& 39& 61 \end{bmatrix}$$ $$X^{t}X = \begin{bmatrix}1& 3& 5\\2& 4& 6 \end{bmatrix} \cdot \begin{bmatrix}1& 2\\ 3& 4\\ 5& 6 \end{bmatrix} = \begin{bmatrix}35& 44\\44& 56 \end{bmatrix}$$ $$YY^{t} = \begin{bmatrix}-1\\3\\4 \end{bmatrix} \cdot \begin{bmatrix}-1& 3& 4 \end{bmatrix} = \begin{bmatrix}1& -3& -4\\ -3& 9 & 12\\ -4& 12& 16 \end{bmatrix}$$ $$Y^{t}Y = \begin{bmatrix}-1& 3& 4 \end{bmatrix} \cdot \begin{bmatrix}-1\\3\\4 \end{bmatrix} = 26$$
22. Prove that the system of linear equations $$\begin{cases}x+2y=4\\ x+y=5\\3x+5y=12 \end{cases}$$ is inconsistent and find a least squares solution of the system.
Solution:
The augmented matrix is $$\begin{bmatrix}1& 2& 4\\1& 1& 5\\3& 5& 12 \end{bmatrix}\Rightarrow \begin{bmatrix}1& 2& 4\\0& -1& 1\\0& -1& 0 \end{bmatrix}\Rightarrow \begin{bmatrix}1& 0& 4\\0& 0& 1\\0& -1& 0 \end{bmatrix}\Rightarrow \begin{bmatrix}1& 0& 0\\0& 1& 0\\0& 0& 1 \end{bmatrix}$$ The last row indicates that this is inconsistent system.
Since $A=\begin{bmatrix}1& 2\\1& 1\\3& 5 \end{bmatrix}$, $X=\begin{bmatrix}x \\ y \end{bmatrix}$, and $B=\begin{bmatrix}4\\5\\12 \end{bmatrix}$. So the normal equation is $$A^{t}AX=A^{t}B$$ $$\Rightarrow \begin{bmatrix}1& 1& 3\\2& 1& 5 \end{bmatrix}\cdot \begin{bmatrix}1& 2\\1& 1\\3& 5 \end{bmatrix} X= \begin{bmatrix}1& 1& 3\\2& 1& 5 \end{bmatrix} \cdot \begin{bmatrix}4\\5\\12 \end{bmatrix}$$ $$\Rightarrow \begin{bmatrix}11& 18\\18& 30 \end{bmatrix}X = \begin{bmatrix}45\\73 \end{bmatrix}$$ $$\Rightarrow \begin{cases}x ={\Delta_1\over\Delta}= {\begin{vmatrix}45& 18\\73& 30 \end{vmatrix}\over \begin{vmatrix}11& 18\\18& 30 \end{vmatrix}} =6\\ y ={\Delta_2\over\Delta} = {\begin{vmatrix}11& 45\\18& 73 \end{vmatrix}\over \begin{vmatrix}11& 18\\18& 30 \end{vmatrix}} =-{7\over6} \end{cases}$$
23. The points $(0, 0)$, $(1, 0)$, $(2, -1)$, $(3, 4)$, $(4, 8)$ are required to lie on a parabola $y = a+bx+cx^2$. Find a least squares solution for $a$, $b$, $c$. Also prove that no parabola passes through these points.
Solution:
The system is $$\begin{cases}a=0\\ a+b+c=0\\ a+2b+4c=-1\\ a+3b+9c=4\\ a+4b+16c=8 \end{cases}$$ That is, $A=\begin{bmatrix} 1& 0& 0\\ 1& 1& 1\\ 1& 2& 4\\ 1& 3& 9\\ 1& 4& 16\end{bmatrix}$, $X=\begin{bmatrix}a\\b\\c \end{bmatrix}$, and $B=\begin{bmatrix}0\\ 0\\ -1\\ 4\\8 \end{bmatrix}$. The normal equation is $$A^{t}AX=A^{t}B$$ $$\Rightarrow \begin{bmatrix}1& 1& 1& 1& 1\\ 0& 1& 2& 3& 4\\ 0& 1& 4& 9& 16 \end{bmatrix} \cdot \begin{bmatrix} 1& 0& 0\\ 1& 1& 1\\ 1& 2& 4\\ 1& 3& 9\\ 1& 4& 16\end{bmatrix}X = \begin{bmatrix}1& 1& 1& 1& 1\\ 0& 1& 2& 3& 4\\ 0& 1& 4& 9& 16 \end{bmatrix} \cdot \begin{bmatrix}0\\ 0\\ -1\\ 4\\8 \end{bmatrix}$$ $$\Rightarrow \begin{bmatrix}5& 10& 30\\ 10& 30& 100\\ 30& 100& 354 \end{bmatrix}X = \begin{bmatrix}11\\ 42\\ 160\end{bmatrix}$$ The augmented matrix is $$ \begin{bmatrix}5& 10& 30 & 11\\ 10& 30& 100 &42\\ 30& 100& 354 &160\end{bmatrix}$$ $$\Rightarrow \begin{bmatrix}1& 2& 6& {11\over5}\\ 1& 3& 10& {21\over5}\\1& {10\over3}& {59\over5}& {16\over3} \end{bmatrix} \Rightarrow \begin{bmatrix}1& 2& 6& {11\over5}\\ 0& 1& 4& 2\\0& {4\over3}& {29\over5}& {47\over15} \end{bmatrix}$$ $$\Rightarrow \begin{bmatrix}1& 0& -2& -{9\over5}\\ 0& 1& 4& 2\\0& 0& {7\over15}& {7\over15} \end{bmatrix} \Rightarrow \begin{bmatrix}1& 0& -2& -{9\over5}\\ 0& 1& 4& 2\\0& 0& 1& 1 \end{bmatrix}$$ $$\Rightarrow \begin{bmatrix}1& 0& 0& {1\over5}\\ 0& 1& 0& -2\\0& 0& 1& 1 \end{bmatrix}\Rightarrow \begin{cases}a={1\over5} \\b=-2 \\c=1 \end{cases}$$ Next, for the original system $AX=B$, its augmented matrix is $$\begin{bmatrix} 1& 0& 0& 0\\ 1& 1& 1& 0\\ 1& 2& 4& -1\\ 1& 3& 9 &4 \\ 1& 4& 16 &8\end{bmatrix}$$ $$\Rightarrow \begin{bmatrix} 1& 0& 0& 0\\ 0& 1& 1& 0\\ 0& 2& 4& -1\\ 0& 3& 9 &4 \\ 0& 4& 16 &8\end{bmatrix} \Rightarrow \begin{bmatrix} 1& 0& 0& 0\\ 0& 1& 1& 0\\ 0& 0& 2& -1\\ 0& 0& 6 &4 \\ 0& 0& 12 & 8 \end{bmatrix} $$ $$\Rightarrow \begin{bmatrix} 1& 0& 0& 0\\ 0& 1& 1& 0\\ 0& 0& 2& -1\\ 0& 0& 0 &7 \\ 0& 0& 0 & 0 \end{bmatrix}$$ The last matrix is inconsistent which means no parabola passes through these points.
24. If $A$ is a symmetric $n\times n$ real matrix and $B$ is $n\times m$, prove that $B^{t}AB$ is a symmetric $m\times m$ matrix.
Solution:
$$(B^{t}AB)^{t}= B^{t}A^{t}(B^{t})^{t}=B^{t}A^{t}B = B^{t}AB$$ Thus, $B^{t}AB$ is a symmetric matrix.
25. If $A$ is $m\times n$ and $B$ is $n\times m$, prove that $AB$ is singular if $m > n$.
Solution:
Since $m > n$, so the homogeneous system $BX=0$ has a non-trivial solution $X_0$ (i.e. the number of unknowns is greater than the number of equations). Hence we have $$A(BX_0) = ABX_0 = (AB)X_0 = 0$$ which means $AB$ ($m\times m$ squares) is singular (if the homogeneous system $AX=0$ has a non-trivial solution, then $A$ is singular).
26. Let $A$ and $B$ be $n\times n$. If $A$ or $B$ is singular, prove that $AB$ is also singular.
Solution:
If $B$ is singular, then we know that the homogeneous system $BX = 0$ has a non-trivial solution $X_0$, that is $$A(BX_0)=(AB)X_0=0$$ which means that $AB$ is also singular since $X_0$ is a non-trivial solution.\\
If $A$ is singular, then $A^{t}$ is also singular (see Question 15). So the homogeneous system $A^{t}X=0$ has a non-trivial solution $X_0$, and we have $$B^{t}(A^{t}X_0)=0 \Rightarrow (B^{t}A^{t})X_0=0$$ $$\Rightarrow (AB)^{t}X_0=0$$ which means $(AB)^{t}$ is singular, and thus $AB$ is also singular.
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