2023秋线性代数期中-答案

Suggested solutions. $Nov{.}^{12}2023$

Question 1

(1) $A$

(2) $B$

(3) $D$

(4) $B$

(5) $D$

Question 2

(1) 21

(2) $-A^{-1}C{B}^{-1}$

(3) 1 or -3

(4)

$$\left[\begin{array}{c}\frac{19}{11}\\ \frac{-5}{11}\end{array}\right]$$

(5)

$$\pm \frac{1}{\sqrt{6}}\left[\begin{array}{l}1\\ 2\\ 1\end{array}\right]$$

Question 3

(A)

A basis for $C(A)$:

$$\{\left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right],\left[\begin{array}{c}2\\ 1\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}1\\ 1\\ 3\\ 1\end{array}\right]\}$$

$A$ basis for $C(A^{\mathrm{\top }})$:

$$\{\left[\begin{array}{c}1\\ 0\\ 1\\ 0\\ 3\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 2\\ 0\\ 1\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 0\\ 1\\ 1\end{array}\right]\}$$

or

$$\{\left[\begin{array}{c}1\\ 2\\ 5\\ 1\\ 6\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 2\\ 1\\ 2\end{array}\right],\left[\begin{array}{c}-1\\ 1\\ 1\\ 3\\ 1\end{array}\right]\}$$

$A$ basis for $N(A)$:

$$\{\left[\begin{array}{c}-1\\ -2\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-3\\ -1\\ 0\\ -1\\ 1\end{array}\right]\}$$

A basis for $N(A^{\mathrm{\top }}):$

$$\left\{\left[\begin{array}{c}-5\\ 13\\ -3\\ 1\end{array}\right]\right\}$$

(B)

$$\left[\begin{array}{l}5\\ 2\\ 1\\ 2\end{array}\right]$$

Question 4

Gaussian Eliminations give:

$$\left[\begin{array}{ccccc}1& -1& -1& 2& 2\\ 2& a& 1& 1& a\\ -1& 1& a& -a-1& -2\end{array}\right]$$$$\to \left[\begin{array}{ccccc}1& -1& -1& 2& 2\\ 0& a+2& 3& -3& a-4\\ 0& 0& a-1& 0& 0\end{array}\right]$$

case1

If $a=-2$, then $\mathrm{rank}A=2\ne 3=\mathrm{rank}(A;B)$

$AX=B$ has no solution

case2

If $a\ne 1$ and $a\ne -1$, $AX=B$ has a unique solution

$$\left[\begin{array}{cccc}1& -1& -1& 2\\ 0& a+2& 3& -3\\ 0& 0& a-1& 1-a\end{array}\right]$$$$\Rightarrow x=\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right]$$$$\left[\begin{array}{cccc}1& -1& -1& 2\\ 0& a+2& 3& a-4\\ 0& 0& a-1& 0\end{array}\right]$$$$\Rightarrow x=\left[\begin{array}{c}\frac{3a}{a+2}\\ \frac{a-4}{a+2}\\ 0\end{array}\right]$$$$X=\left[\begin{array}{cc}1& \frac{3a}{a+2}\\ 0& \frac{a-4}{a+2}\\ -1& 0\end{array}\right]$$

case3

If $a=1$, $Ax=B$ has infinitely many solutions

$$\left[\begin{array}{cccc}1& -1& -1& 2\\ 0& 3& 3& -3\\ 0& 0& 0& 0\end{array}\right]$$$$\Rightarrow x=\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]+k_1\left[\begin{array}{c}0\\ -1\\ 1\end{array}\right]$$$$\left[\begin{array}{cccc}1& -1& -1& 2\\ 0& 3& 3& -3\\ 0& 0& 0& 0\end{array}\right]$$$$\Rightarrow x=\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]+k_2\left[\begin{array}{c}0\\ -1\\ 1\end{array}\right]$$$$\Rightarrow X=\left[\begin{array}{cc}1& 1\\ -k_1-1& -k_2-1\\ k_1& k_2\end{array}\right],k_1,k_2\text{ arbitrary constants}$$

Question 5

(A) Let $X,Y\in M_{2\times 2}(\mathbb{R})$ and $c\in \mathbb{R}$, then we have

$$\begin{gathered} T(CX+Y)=\left[\begin{array}{l}\mathrm{tr}{A}^{\mathrm{\top }}(CX+Y)\\ \mathrm{tr}{B}^{\mathrm{\top }}(CX+Y)\\ \mathrm{tr}{C}^{\mathrm{\top }}(CX+Y)\end{array}\right] \\ =C\left[\begin{array}{c}\mathrm{tr}(A^{\mathrm{\top }}X)\\ \mathrm{tr}(B^{\mathrm{\top }}X)\\ \mathrm{tr}(C^{\mathrm{\top }}X)\end{array}\right]+\left[\begin{array}{c}\mathrm{tr}(A^{\mathrm{\top }}X)\\ \mathrm{tr}(B^{\mathrm{\top }}Y)\\ \mathrm{tr}(C^{\mathrm{\top }}Y)\end{array}\right] \\ =CT(X)+T(Y) \end{gathered}$$

(B)

$$T(v_1)=\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]=1\cdot \left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]+0\left[\begin{array}{l}0\\ 1\\ 0\end{array}\right]+0\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]$$$$T(v_2)=\left[\begin{array}{l}-1\\ 0\\ 0\end{array}\right]=-1\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]+0\left[\begin{array}{l}0\\ 1\\ 0\end{array}\right]+0\cdot \left[\begin{array}{l}0\\ 0\\ 1\end{array}\right]$$$$T(v_3)=\left[\begin{array}{l}0\\ 1\\ 1\end{array}\right]=0\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]+1\cdot \left[\begin{array}{l}0\\ 1\\ 0\end{array}\right]+1\cdot \left[\begin{array}{l}0\\ 0\\ 1\end{array}\right]$$$$T(v_4)=\left[\begin{array}{l}0\\ 1\\ -1\end{array}\right]=0\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]+1\cdot \left[\begin{array}{l}0\\ 1\\ 0\end{array}\right]+-1\cdot \left[\begin{array}{l}0\\ 0\\ 1\end{array}\right]$$

Therefore, the matrix representation of $T$ with respect to $v_1,v_2,v_3,v_4$ and $w_1,w_2,w_3$ is:

$$\left[\begin{array}{cccc}1& -1& 0& 0\\ 0& 0& 1& 1\\ 0& 0& 1& -1\end{array}\right]$$

(C) Since

$$T(A)=\left[\begin{array}{l}2\\ 0\\ 0\end{array}\right],T(B)=\left[\begin{array}{l}0\\ 1\\ 0\end{array}\right],T(C)=\left[\begin{array}{l}0\\ 0\\ 1\end{array}\right]$$

we can take $X$ to be

$$\begin{gathered} \frac{1}{2}A-2B+C \\ =\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& \frac{-1}{2}\end{array}\right]-\left[\begin{array}{ll}0& 2\\ 0& 0\end{array}\right]+\left[\begin{array}{ll}0& 0\\ 1& 0\end{array}\right] \\ =\left[\begin{array}{cc}\frac{1}{2}& -2\\ 1& \frac{-1}{2}\end{array}\right] \end{gathered}$$

Question 6

Apply Elementary Row and Column operations to $A$ and $C$ obtain

$$D_1=\left[\begin{array}{cc}{I}_r& 0\\ 0& 0\end{array}\right]\text{ for }A\text{ and }{D}_2=\left[\begin{array}{cc}{I}_s& 0\\ 0& 0\end{array}\right]\text{ for }C$$

Where $r=\mathrm{rank}A,s=\mathrm{rank}C$

Let M=

$$\left[\begin{array}{ll}A& B\\ 0& C\end{array}\right]$$

Then $M$ can be converted to

$$M_1=\left[\begin{array}{ll}{D}_1& C_1\\ 0& D_2\end{array}\right]$$

via elementary row and column operations

Furthermore, the pivots in $D_1$ and $D_2$ can be used to eliminate the nonzero entries in $C_1$, to obtain

$$M_2=\left[\begin{array}{cccc}{I}_r& 0& 0& 0\\ 0& 0& 0& C_2\\ 0& 0& I_s& 0\\ 0& 0& 0& 0\end{array}\right]$$

In conclusion,

$$\begin{array}{rl}& \mathrm{rank}M=\mathrm{rank}{M}_1=\mathrm{rank}{M}_2=r+s+\mathrm{rank}{C}_2\ge r+s\\ & =\mathrm{rank}A+\mathrm{rank}C\end{array}$$