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基础部分
以下矩阵乘积的结果为[ 1 − 1 2 2 1 3 3 1 4 ] \begin{bmatrix}1&-1&2\\2&1&3\\3&1&4\end{bmatrix} 1 2 3 − 1 1 1 2 3 4 的是( ).
A. [ 1 0 0 2 1 0 3 4 3 2 ] [ 1 − 1 2 0 3 − 1 0 0 − 2 3 ] \begin{bmatrix}1&0&0\\2&1&0\\3&\frac{4}{3}&2\end{bmatrix}\begin{bmatrix}1&-1&2\\0&3&-1\\0&0&-\frac{2}{3}\end{bmatrix} 1 2 3 0 1 3 4 0 0 2 1 0 0 − 1 3 0 2 − 1 − 3 2
B. [ 1 0 0 2 − 1 0 3 4 3 1 ] [ 1 − 1 2 0 − 3 − 1 0 0 − 2 3 ] \begin{bmatrix}1&0&0\\2&-1&0\\3&\frac{4}{3}&1\end{bmatrix}\begin{bmatrix}1&-1&2\\0&-3&-1\\0&0&-\frac{2}{3}\end{bmatrix} 1 2 3 0 − 1 3 4 0 0 1 1 0 0 − 1 − 3 0 2 − 1 − 3 2
C. [ 1 0 0 2 1 0 3 4 3 1 ] [ 1 0 0 0 3 0 0 0 − 2 3 ] [ 1 − 1 2 0 1 − 1 3 0 0 1 ] \begin{bmatrix}1&0&0\\2&1&0\\3&\frac{4}{3}&1\end{bmatrix}\begin{bmatrix}1&0&0\\0&3&0\\0&0&-\frac{2}{3}\end{bmatrix}\begin{bmatrix}1&-1&2\\0&1&-\frac{1}{3}\\0&0&1\end{bmatrix} 1 2 3 0 1 3 4 0 0 1 1 0 0 0 3 0 0 0 − 3 2 1 0 0 − 1 1 0 2 − 3 1 1
D. [ 1 0 0 2 − 1 0 3 4 3 1 ] [ 1 − 1 2 0 1 − 1 3 0 0 1 ] \begin{bmatrix}1&0&0\\2&-1&0\\3&\frac{4}{3}&1\end{bmatrix}\begin{bmatrix}1&-1&2\\0&1&-\frac{1}{3}\\0&0&1\end{bmatrix} 1 2 3 0 − 1 3 4 0 0 1 1 0 0 − 1 1 0 2 − 3 1 1
设A = [ 1 1 0 1 ] A=\begin{bmatrix}1&1\\0&1\end{bmatrix} A = [ 1 0 1 1 ] ,B = [ − 1 1 0 − 1 ] B=\begin{bmatrix}-1&1\\0&-1\end{bmatrix} B = [ − 1 0 1 − 1 ] ,则A 9 − B 9 = A^9-B^9= A 9 − B 9 =
A = [ 0 0 − 1 0 1 0 1 0 0 ] A=\begin{bmatrix}0&0&-1\\0&1&0\\1&0&0\end{bmatrix} A = 0 0 1 0 1 0 − 1 0 0 ,则A 13 = A^{13}= A 13 = _____.
B = [ 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 ] B=\begin{bmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{bmatrix} B = 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 ,A = E + B + B 2 + B 3 A=E+B+B^2+B^3 A = E + B + B 2 + B 3 ,则A − 1 = A^{-1}= A − 1 = _____.
已知A = [ 1 − 1 1 0 ] A=\begin{bmatrix}1&-1\\1&0\end{bmatrix} A = [ 1 1 − 1 0 ] ,若( P A ) 2 = P A (PA)^2=PA ( P A ) 2 = P A ,P P P 为可逆矩阵,则P = P= P =
设A , B A,B A , B 都是3阶矩阵,若∣ A ∣ = − 3 |A|=-3 ∣ A ∣ = − 3 ,∣ B ∣ = 4 |B|=4 ∣ B ∣ = 4 ,C = [ 2 A ∗ ( A B ) ∗ O B − 1 ] C=\begin{bmatrix}2A^*&(AB)^*\\O&B^{-1}\end{bmatrix} C = [ 2 A ∗ O ( A B ) ∗ B − 1 ] ,则∣ C ∣ = − ‾ |C|=\underline{-} ∣ C ∣ = −
A , B A,B A , B 是n n n 阶矩阵,A ∗ , B ∗ A^*,B^* A ∗ , B ∗ 分别是A , B A,B A , B 对应的伴随矩阵,则分块矩阵C = [ O A B O ] C=\begin{bmatrix}O&A\\B&O\end{bmatrix} C = [ O B A O ] 的伴随矩阵C ∗ = C^*= C ∗ = ( )
A. [ O ∣ A ∣ A ∗ ∣ B ∣ B ∗ O ] \begin{bmatrix}O&|A|A^*\\|B|B^*&O\end{bmatrix} [ O ∣ B ∣ B ∗ ∣ A ∣ A ∗ O ]
B. [ O ( − 1 ) n ∣ A ∣ B ∗ ( − 1 ) n ∣ B ∣ A ∗ O ] \begin{bmatrix}O&(-1)^n|A|B^*\\(-1)^n|B|A^*&O\end{bmatrix} [ O ( − 1 ) n ∣ B ∣ A ∗ ( − 1 ) n ∣ A ∣ B ∗ O ]
C. [ O ∣ B ∣ A ∗ ∣ A ∣ B O ] \begin{bmatrix}O&|B|A^*\\|A|B&O\end{bmatrix} [ O ∣ A ∣ B ∣ B ∣ A ∗ O ]
D. [ O ( − 1 ) n ∣ B ∣ B ∗ ( − 1 ) n ∣ A ∣ A ∗ O ] \begin{bmatrix}O&(-1)^n|B|B^*\\(-1)^n|A|A^*&O\end{bmatrix} [ O ( − 1 ) n ∣ A ∣ A ∗ ( − 1 ) n ∣ B ∣ B ∗ O ]
设矩阵A = [ 1 2 4 0 1 5 0 0 3 ] A=\begin{bmatrix}1&2&4\\0&1&5\\0&0&3\end{bmatrix} A = 1 0 0 2 1 0 4 5 3 ,B = [ 1 0 0 3 1 0 4 5 1 ] B=\begin{bmatrix}1&0&0\\3&1&0\\4&5&1\end{bmatrix} B = 1 3 4 0 1 5 0 0 1 ,则∣ A − 1 B ∗ − A ∗ B − 1 ∣ = |A^{-1}B^*-A^*B^{-1}|= ∣ A − 1 B ∗ − A ∗ B − 1 ∣ = _____.
设A , B A,B A , B 为3阶矩阵,且A B = [ 0 1 0 1 0 0 0 0 1 ] AB=\begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix} A B = 0 1 0 1 0 0 0 0 1 ,则必有( ).
A. 互换矩阵A − 1 A^{-1} A − 1 的第1,2行得矩阵B B B
B. 互换矩阵A − 1 A^{-1} A − 1 的第1,2列得矩阵B − 1 B^{-1} B − 1
C. 互换矩阵A A A 的第1,2行得矩阵B − 1 B^{-1} B − 1
D. 互换矩阵A A A 的第1,2列得矩阵B − 1 B^{-1} B − 1
设A A A 为3阶矩阵,将A A A 的第1行加到第2行得到B B B ,再将B B B 的第2列的-1倍加到第1列得到C C C ,记P = [ 1 1 0 0 1 0 0 0 1 ] P=\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix} P = 1 0 0 1 1 0 0 0 1 ,则C = C= C = ( )
A. P − 1 A P P^{-1}AP P − 1 A P B. P A P − 1 PAP^{-1} P A P − 1 C. P T A P P^TAP P T A P D. P T A ( P T ) − 1 P^TA(P^T)^{-1} P T A ( P T ) − 1
设3阶矩阵A A A 与B B B 等价,则下列结论正确的是( )
A. 存在可逆矩阵P P P ,使得P A = B PA=B P A = B
B. 存在可逆矩阵Q Q Q 使得A Q = B AQ=B A Q = B
C. 若r ( A ) = 2 r(A)=2 r ( A ) = 2 ,A A A 可经初等行变换化为矩阵B B B
D. 若r ( A ) = 3 r(A)=3 r ( A ) = 3 ,A A A 可经初等列变换化为矩阵B B B
设A = [ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ] A=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix} A = a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 ,B = [ a 21 a 22 a 23 a 11 a 12 a 13 a 31 + a 11 a 32 + a 12 a 33 + a 13 ] B=\begin{bmatrix}a_{21}&a_{22}&a_{23}\\a_{11}&a_{12}&a_{13}\\a_{31}+a_{11}&a_{32}+a_{12}&a_{33}+a_{13}\end{bmatrix} B = a 21 a 11 a 31 + a 11 a 22 a 12 a 32 + a 12 a 23 a 13 a 33 + a 13 ,P 1 = [ 0 1 0 1 0 0 0 0 1 ] P_1=\begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix} P 1 = 0 1 0 1 0 0 0 0 1 ,P 2 = [ 1 0 1 0 1 0 0 0 1 ] P_2=\begin{bmatrix}1&0&1\\0&1&0\\0&0&1\end{bmatrix} P 2 = 1 0 0 0 1 0 1 0 1 ,则( ).
A. A P 1 9 P 2 T = B AP_1^9P_2^T=B A P 1 9 P 2 T = B B. A P 2 T P 1 9 = B AP_2^TP_1^9=B A P 2 T P 1 9 = B C. P 1 9 P 2 T A = B P_1^9P_2^TA=B P 1 9 P 2 T A = B D. P 2 T P 1 9 A = B P_2^TP_1^9A=B P 2 T P 1 9 A = B
将3阶方阵A A A 的第1行的2倍加到第2行得到矩阵B B B ,将3阶方阵C C C 的第3列的-3倍加到第1列得到矩阵D D D ,B D = [ 1 0 0 0 2 0 0 0 3 ] BD=\begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix} B D = 1 0 0 0 2 0 0 0 3 ,则A C = AC= A C = ( )
A. [ 1 0 0 2 2 0 − 9 0 3 ] \begin{bmatrix}1&0&0\\2&2&0\\-9&0&3\end{bmatrix} 1 2 − 9 0 2 0 0 0 3 B. [ 1 0 0 − 2 2 0 9 0 3 ] \begin{bmatrix}1&0&0\\-2&2&0\\9&0&3\end{bmatrix} 1 − 2 9 0 2 0 0 0 3 C. [ − 3 0 0 − 6 2 0 0 0 3 ] \begin{bmatrix}-3&0&0\\-6&2&0\\0&0&3\end{bmatrix} − 3 − 6 0 0 2 0 0 0 3 D. [ 1 0 0 − 2 2 0 − 1 0 3 ] \begin{bmatrix}1&0&0\\-2&2&0\\-1&0&3\end{bmatrix} 1 − 2 − 1 0 2 0 0 0 3
设A , B A,B A , B 是3阶矩阵,A A A 是非零矩阵,且满足A B = O AB=O A B = O ,B = [ 1 − 1 1 2 a 1 − a 2 a a − a a 2 − 2 ] B=\begin{bmatrix}1&-1&1\\2a&1-a&2a\\a&-a&a^2-2\end{bmatrix} B = 1 2 a a − 1 1 − a − a 1 2 a a 2 − 2 ,则( ).
A. a = − 1 a=-1 a = − 1 时,必有r ( A ) = 1 r(A)=1 r ( A ) = 1 B. a = 2 a=2 a = 2 时,必有r ( A ) = 2 r(A)=2 r ( A ) = 2 C. a = − 1 a=-1 a = − 1 时,必有r ( A ) = 2 r(A)=2 r ( A ) = 2 D. a = 2 a=2 a = 2 时,必有r ( A ) = 1 r(A)=1 r ( A ) = 1
设A , B , C A,B,C A , B , C 均是3阶方阵,满足A B = C AB=C A B = C ,其中B = [ 1 2 2 2 1 1 − 2 − 1 a ] B=\begin{bmatrix}1&2&2\\2&1&1\\-2&-1&a\end{bmatrix} B = 1 2 − 2 2 1 − 1 2 1 a ,C = [ 0 0 0 2 1 1 0 0 0 ] C=\begin{bmatrix}0&0&0\\2&1&1\\0&0&0\end{bmatrix} C = 0 2 0 0 1 0 0 1 0 ,则必有( ).
A. a = − 1 a=-1 a = − 1 时,r ( A ) = 1 r(A)=1 r ( A ) = 1 B. a = − 1 a=-1 a = − 1 时,r ( A ) = 2 r(A)=2 r ( A ) = 2 C. a ≠ − 1 a\neq-1 a = − 1 时,r ( A ) = 1 r(A)=1 r ( A ) = 1 D. a ≠ − 1 a\neq-1 a = − 1 时,r ( A ) = 2 r(A)=2 r ( A ) = 2
设2阶正交矩阵A A A 的主对角线元素满足a 11 + 2 = a 22 a_{11}+2=a_{22} a 11 + 2 = a 22 ,则A = A= A =
设A A A 是n n n 阶矩阵,满足( A − E ) 5 = 0 (A-E)^5=0 ( A − E ) 5 = 0 ,则A − 1 = A^{-1}= A − 1 =
设A A A 是n n n 阶矩阵,满足A 2 = A A^2=A A 2 = A ,则( A − 2 E ) 3 − 3 A = (A-2E)^3-3A= ( A − 2 E ) 3 − 3 A =
设A A A 是4阶矩阵,满足A 2 = 0 A^2=0 A 2 = 0 ,则r ( A ∗ ) = r(A^*)= r ( A ∗ ) = _____.
强化部分
设D = ∣ 1 − 3 1 − 2 2 − 5 − 2 − 2 0 − 4 5 1 − 3 9 − 6 7 ∣ D=\begin{vmatrix}1&-3&1&-2\\2&-5&-2&-2\\0&-4&5&1\\-3&9&-6&7\end{vmatrix} D = 1 2 0 − 3 − 3 − 5 − 4 9 1 − 2 5 − 6 − 2 − 2 1 7 ,M 3 j M_{3j} M 3 j 表示D D D 中第3行第j j j 列元素的余子式( j = 1 , 2 , 3 , 4 ) (j=1,2,3,4) ( j = 1 , 2 , 3 , 4 ) ,则M 31 + 3 M 32 − 2 M 33 + 2 M 34 = M_{31}+3M_{32}-2M_{33}+2M_{34}= M 31 + 3 M 32 − 2 M 33 + 2 M 34 = ( )
A. 0 B. 1 C. -2 D. -3
D 3 = ∣ 1 1 1 1 2 5 34 1 34 ∣ D_3=\begin{vmatrix}1&1&1\\1&2&5\\34&1&34\end{vmatrix} D 3 = 1 1 34 1 2 1 1 5 34 ,A i j A_{ij} A ij 表示元素a i j ( i , j = 1 , 2 , 3 ) a_{ij}(i,j=1,2,3) a ij ( i , j = 1 , 2 , 3 ) 的代数余子式,则5 A 11 + 2 A 12 + A 13 = 5A_{11}+2A_{12}+A_{13}= 5 A 11 + 2 A 12 + A 13 =
已知3阶行列式∣ A ∣ = − 9 |A|=-9 ∣ A ∣ = − 9 ,其第2行元素为1,1,2,第3行元素为2,2,1,则A 31 + A 32 − 3 A 33 A_{31}+A_{32}-3A_{33} A 31 + A 32 − 3 A 33
已知3阶行列式∣ A ∣ |A| ∣ A ∣ 的元素a i j a_{ij} a ij 均为实数,且a i j a_{ij} a ij 不全为0.若a i j = − A i j , i , j = 1 , 2 , 3 a_{ij}=-A_{ij},i,j=1,2,3 a ij = − A ij , i , j = 1 , 2 , 3 ,其中A i j A_{ij} A ij 是a i j a_{ij} a ij 的代数余子式,则∣ A ∣ = |A|= ∣ A ∣ =
矩阵运算
设A A A 是n n n 阶矩阵,则下列说法错误的是( ).
A. 对任意的n n n 维列向量ξ , \xi, ξ , 有A ξ = 0 A\xi=0 A ξ = 0 ,则A = 0 A=0 A = 0
B. 对任意的n n n 维列向量ξ , \xi, ξ , 有ξ T A ξ = 0 \xi^TA\xi=0 ξ T A ξ = 0 ,则A = 0 A=0 A = 0
C. 对任意的n n n 阶矩阵B B B ,有A T B = O A^TB=O A T B = O 则A = O A=O A = O
D. 对任意的n n n 阶矩阵B B B ,有B T A B = O B^TAB=O B T A B = O ,则A = O A=O A = O
A = [ 1 1 − 1 0 1 1 0 0 1 ] A=\begin{bmatrix}1&1&-1\\0&1&1\\0&0&1\end{bmatrix} A = 1 0 0 1 1 0 − 1 1 1 ,则A 10 = A^{10}= A 10 = _____.
[ 1 0 − 1 1 ] 3 [ 1 2 − 1 3 ] [ 0 1 1 0 ] 5 = \begin{bmatrix}1&0\\-1&1\end{bmatrix}^3\begin{bmatrix}1&2\\-1&3\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}^5= [ 1 − 1 0 1 ] 3 [ 1 − 1 2 3 ] [ 0 1 1 0 ] 5 =
设A , B , C A,B,C A , B , C 均是3阶矩阵,且满足A B = B 2 − B C AB=B^2-BC A B = B 2 − B C ,其中B = [ 1 1 1 0 1 1 0 0 1 ] B=\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix} B = 1 0 0 1 1 0 1 1 1 ,C = [ 0 1 1 0 0 1 0 0 1 ] C=\begin{bmatrix}0&1&1\\0&0&1\\0&0&1\end{bmatrix} C = 0 0 0 1 0 0 1 1 1 ,则A 99 = A^{99}= A 99 =
设A A A 为n ( n ≥ 2 ) n(n\geq2) n ( n ≥ 2 ) 阶实对称矩阵,且满足E − 2 A + A 2 − 2 A 3 = 0 E-2A+A^2-2A^3=0 E − 2 A + A 2 − 2 A 3 = 0 ,其中E E E 为n n n 阶单位矩阵,则A = A= A =
设A A A 是n n n 阶矩阵,E + A E+A E + A 是可逆矩阵,则下列等式不成立的是( ).
A. ( A + E ) ( A − E ) = ( A − E ) ( A + E ) (A+E)(A-E)=(A-E)(A+E) ( A + E ) ( A − E ) = ( A − E ) ( A + E )
B. ( A + E ) − 1 ( A − E ) = ( A − E ) ( A + E ) − 1 (A+E)^{-1}(A-E)=(A-E)(A+E)^{-1} ( A + E ) − 1 ( A − E ) = ( A − E ) ( A + E ) − 1
C. ( A + E ) ∗ ( A − E ) = ( A − E ) ( A + E ) ∗ (A+E)^*(A-E)=(A-E)(A+E)^* ( A + E ) ∗ ( A − E ) = ( A − E ) ( A + E ) ∗
D. ( A + E ) T ( A − E ) = ( A − E ) ( A + E ) T (A+E)^T(A-E)=(A-E)(A+E)^T ( A + E ) T ( A − E ) = ( A − E ) ( A + E ) T
设A A A 为2阶方阵,α \alpha α 为2维非零列向量,且α \alpha α 不是A A A 的特征向量,P = [ α , A α ] P=[\alpha,A\alpha] P = [ α , A α ] ,A 2 α + A α − 2 α = 0 A^2\alpha+A\alpha-2\alpha=0 A 2 α + A α − 2 α = 0 ,若矩阵B B B 满足A P = P B AP=PB A P = P B ,则B = B= B = ( ).
A. [ − 1 0 1 2 ] \begin{bmatrix}-1&0\\1&2\end{bmatrix} [ − 1 1 0 2 ] B. [ 0 1 2 − 1 ] \begin{bmatrix}0&1\\2&-1\end{bmatrix} [ 0 2 1 − 1 ] C. [ − 1 1 0 2 ] \begin{bmatrix}-1&1\\0&2\end{bmatrix} [ − 1 0 1 2 ] D. [ 0 2 1 − 1 ] \begin{bmatrix}0&2\\1&-1\end{bmatrix} [ 0 1 2 − 1 ]
设A A A 为n n n 阶可逆矩阵,α \alpha α 为n n n 维列向量,b b b 为常数.记分块矩阵P = [ E 0 − α T A ∗ ∣ A ∣ ] P=\begin{bmatrix}E&0\\-\alpha^TA^*&|A|\end{bmatrix} P = [ E − α T A ∗ 0 ∣ A ∣ ] ,Q = [ A α α T b ] Q=\begin{bmatrix}A&\alpha\\\alpha^T&b\end{bmatrix} Q = [ A α T α b ] ,其中A ∗ A^* A ∗ 是矩阵A A A 的伴随矩阵,E E E 为n n n 阶单位矩阵.
(1) 计算并化简P Q PQ P Q ;(2) 证明:矩阵Q Q Q 可逆的充分必要条件是α T A − 1 α ≠ b \alpha^TA^{-1}\alpha\neq b α T A − 1 α = b
求与A = [ 1 1 0 1 ] A=\begin{bmatrix}1&1\\0&1\end{bmatrix} A = [ 1 0 1 1 ] 可交换的全部2阶矩阵.
设A = [ 2 − 2 − 4 − 1 3 4 1 − 2 − 3 ] A=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix} A = 2 − 1 1 − 2 3 − 2 − 4 4 − 3 ,问是否存在非单位矩阵B B B ,使得A B = A AB=A A B = A ?若不存在,请说明理由;若存在,求出所有满足A B = A AB=A A B = A 的B B B
矩阵的秩
设A A A 是3阶非零矩阵,满足A 2 = A A^2=A A 2 = A ,且A ≠ E A\neq E A = E ,则必有( ).
A. r ( A ) = 1 r(A)=1 r ( A ) = 1 B. r ( A − E ) = 2 r(A-E)=2 r ( A − E ) = 2 C. [ r ( A ) − 1 ] [ r ( A − E ) − 2 ] = 0 [r(A)-1][r(A-E)-2]=0 [ r ( A ) − 1 ] [ r ( A − E ) − 2 ] = 0 D. [ r ( A ) − 1 ] [ r ( A − E ) − 1 ] = 0 [r(A)-1][r(A-E)-1]=0 [ r ( A ) − 1 ] [ r ( A − E ) − 1 ] = 0
若A , A ∗ , B A,A^*,B A , A ∗ , B 都是n ( n > 2 ) n(n>2) n ( n > 2 ) 阶非零矩阵,且A ∗ A^* A ∗ 是A A A 的伴随矩阵,A B = O AB=O A B = O ,则r ( B ) = r(B)= r ( B ) = ( )
A. 1 B. n − 1 n-1 n − 1 C. n n n D. n − 1 n-1 n − 1 或n n n
设矩阵A = [ 1 − 1 1 − 2 2 1 1 − 1 k ] A=\begin{bmatrix}1&-1&1\\-2&2&1\\1&-1&k\end{bmatrix} A = 1 − 2 1 − 1 2 − 1 1 1 k ,r ( ( 3 E − A ) 2 ) < r ( 3 E − A ) r((3E-A)^2)<r(3E-A) r (( 3 E − A ) 2 ) < r ( 3 E − A ) ,其中E E E 是3阶单位矩阵,则常数k = k= k = ( )
A. 3 B. 4 C. 5 D. 6
设A , B A,B A , B 为n n n 阶实矩阵,则下列结论不成立的是( ).
A. r ( [ A A B ] ) = r ( A ) r\left(\begin{bmatrix}A&AB\end{bmatrix}\right)=r(A) r ( [ A A B ] ) = r ( A )
B. r ( [ A B T A B T B ] ) = r ( A B T ) r\left(\begin{bmatrix}AB^T&AB^TB\end{bmatrix}\right)=r(AB^T) r ( [ A B T A B T B ] ) = r ( A B T )
C. r ( [ B A B T B A ] ) = r ( A B T ) r\left(\begin{bmatrix}BA\\B^TBA\end{bmatrix}\right)=r(AB^T) r ( [ B A B T B A ] ) = r ( A B T )
D. r ( [ A T A B T A ] ) = r ( A ) r\left(\begin{bmatrix}A^TA\\B^TA\end{bmatrix}\right)=r(A) r ( [ A T A B T A ] ) = r ( A )
已知A A A 为n n n 阶矩阵,E E E 为n n n 阶单位矩阵,记矩阵[ O A A T E ] , [ O A T A A T E ] , [ A T E A T A A T A ] \begin{bmatrix}O&A\\A^T&E\end{bmatrix},\begin{bmatrix}O&A^TA\\A^T&E\end{bmatrix},\begin{bmatrix}A^T&E\\A^TA&A^TA\end{bmatrix} [ O A T A E ] , [ O A T A T A E ] , [ A T A T A E A T A ] 的秩分别为r 1 , r 2 , r 3 r_1,r_2,r_3 r 1 , r 2 , r 3 ,则( ).
A. r 1 = r 2 ≥ r 3 r_1=r_2\geq r_3 r 1 = r 2 ≥ r 3 B. r 1 = r 2 ≤ r 3 r_1=r_2\leq r_3 r 1 = r 2 ≤ r 3 C. r 1 = r 3 ≥ r 2 r_1=r_3\geq r_2 r 1 = r 3 ≥ r 2 D. r 1 = r 3 ≤ r 2 r_1=r_3\leq r_2 r 1 = r 3 ≤ r 2
已知n n n 阶矩阵A , B , C A,B,C A , B , C 满足A B C = O ABC=O A B C = O ,E E E 为n n n 阶单位矩阵,记矩阵[ O A B C E ] , [ A B C O E ] , [ E A B A B O ] \begin{bmatrix}O&A\\BC&E\end{bmatrix},\begin{bmatrix}AB&C\\O&E\end{bmatrix},\begin{bmatrix}E&AB\\AB&O\end{bmatrix} [ O B C A E ] , [ A B O C E ] , [ E A B A B O ] 的秩分别为r 1 , r 2 , r 3 r_1,r_2,r_3 r 1 , r 2 , r 3 ,则( ).
A. r 1 ≤ r 2 ≤ r 3 r_1\leq r_2\leq r_3 r 1 ≤ r 2 ≤ r 3 B. r 1 ≤ r 3 ≤ r 2 r_1\leq r_3\leq r_2 r 1 ≤ r 3 ≤ r 2 C. r 3 ≤ r 1 ≤ r 2 r_3\leq r_1\leq r_2 r 3 ≤ r 1 ≤ r 2 D. r 2 ≤ r 1 ≤ r 3 r_2\leq r_1\leq r_3 r 2 ≤ r 1 ≤ r 3
设A , B , C A,B,C A , B , C 均为n n n 阶矩阵,r ( A B ) ≤ r ( B A ) r(AB)\leq r(BA) r ( A B ) ≤ r ( B A ) ,记[ O A B B B C ] , [ B B C A B O ] , [ B A B A C O B ] \begin{bmatrix}O&AB\\B&BC\end{bmatrix},\begin{bmatrix}B&BC\\AB&O\end{bmatrix},\begin{bmatrix}BA&BAC\\O&B\end{bmatrix} [ O B A B B C ] , [ B A B B C O ] , [ B A O B A C B ] 的秩分别为r 1 , r 2 , r 3 r_1,r_2,r_3 r 1 , r 2 , r 3 ,则( ).
A. r 2 ≤ r 3 ≤ r 1 r_2\leq r_3\leq r_1 r 2 ≤ r 3 ≤ r 1 B. r 2 ≤ r 1 ≤ r 3 r_2\leq r_1\leq r_3 r 2 ≤ r 1 ≤ r 3 C. r 1 ≤ r 2 ≤ r 3 r_1\leq r_2\leq r_3 r 1 ≤ r 2 ≤ r 3 D. r 3 ≤ r 2 ≤ r 1 r_3\leq r_2\leq r_1 r 3 ≤ r 2 ≤ r 1
设A A A 为n n n 阶矩阵,r ( A ) = r r(A)=r r ( A ) = r ,E r E_r E r 为r r r 阶单位矩阵,则"A 2 = A A^2=A A 2 = A "是"存在列满秩矩阵C n × r C_{n\times r} C n × r ,使得A = C B A=CB A = C B ,B C = E r BC=E_r B C = E r "的( ).
A. 充分非必要条件 B. 必要非充分条件 C. 充分必要条件 D. 既非充分又非必要条件