2025年考研數(shù)學(xué)二線(xiàn)性代數(shù)重點(diǎn)梳理（含答案）考試時(shí)間：______分鐘總分：______分姓名：______一、單項(xiàng)選擇題：1.下列向量組中，線(xiàn)性無(wú)關(guān)的是（）。A.(1,2,3),(2,4,6),(3,6,9)B.(1,0,1),(0,1,1),(1,1,0)C.(1,1,1),(1,2,3),(1,3,6)D.(1,-1,2),(2,-2,4),(3,-3,6)2.設(shè)矩陣A=$\begin{pmatrix}1&2\\3&4\end{pmatrix}$，B=$\begin{pmatrix}0&-1\\2&3\end{pmatrix}$，則矩陣2A-3B=（）。A.$\begin{pmatrix}2&1\\3&7\end{pmatrix}$B.$\begin{pmatrix}-6&-9\\0&3\end{pmatrix}$C.$\begin{pmatrix}6&-9\\0&-6\end{pmatrix}$D.$\begin{pmatrix}-4&-7\\0&-6\end{pmatrix}$3.設(shè)向量$\alpha=(1,k,3)$，$\beta=(2,-1,1)$，若$\alpha\perp\beta$，則k的值為（）。A.-2B.-1C.1D.24.齊次線(xiàn)性方程組$\begin{cases}x_1+x_2+x_3=0\\2x_1+3x_2+ax_3=0\\x_2+bx_3=0\end{cases}$有非零解，則a,b應(yīng)滿(mǎn)足的條件是（）。A.a=1,b=1B.a=2,b=1C.a=1,b=2D.a=2,b=25.設(shè)A是n階可逆矩陣，則下列說(shuō)法錯(cuò)誤的是（）。A.A的行列式不為零B.A的秩為nC.A的轉(zhuǎn)置矩陣A^T也可逆D.A的特征值均為零二、填空題：1.行列式$\begin{vmatrix}1&2&3\\0&1&2\\3&0&1\end{vmatrix}$的值為_(kāi)______。2.設(shè)向量$\alpha=(1,2,3)$，$\beta=(1,-1,2)$，則$\alpha+\beta$=_______，$3\alpha-2\beta$=_______。3.矩陣A=$\begin{pmatrix}1&2\\3&4\end{pmatrix}$的逆矩陣A^(-1)=_______。4.設(shè)A是3階矩陣，且$|A|=2$，則$|3A|=$_______。5.若線(xiàn)性方程組$\begin{cases}x_1+x_2+x_3=1\\2x_1+3x_2+ax_3=2\\x_2+bx_3=1\end{cases}$無(wú)解，則a,b應(yīng)滿(mǎn)足的條件是_______。三、計(jì)算題：1.計(jì)算行列式$\begin{vmatrix}1&1&1\\x&y&z\\x^2&y^2&z^2\end{vmatrix}$。2.解線(xiàn)性方程組$\begin{cases}2x_1+x_2-x_3=1\\x_1-3x_2+2x_3=-1\\4x_1-2x_2+x_3=3\end{cases}$。3.求矩陣A=$\begin{pmatrix}1&2\\2&5\end{pmatrix}$的特征值和特征向量。4.將向量$\beta=(1,2,3)$用向量$\alpha_1=(1,0,1)$，$\alpha_2=(1,1,0)$，$\alpha_3=(0,1,1)$線(xiàn)性表示。四、證明題：1.證明：若向量組$\alpha_1,\alpha_2,\alpha_3$線(xiàn)性無(wú)關(guān)，則向量組$\alpha_1+\alpha_2,\alpha_2+\alpha_3,\alpha_3+\alpha_1$也線(xiàn)性無(wú)關(guān)。2.設(shè)A是n階矩陣，且A^2=A，證明：A的特征值只能是0或1。試卷答案一、單項(xiàng)選擇題：1.B2.C3.A4.C5.D二、填空題：1.-22.(2,1,5),(1,7,-1)3.$\begin{pmatrix}-2&1\\1&-\frac{1}{2}\end{pmatrix}$4.545.a=4,b≠1三、計(jì)算題：1.解：按第一列展開(kāi)$\begin{vmatrix}1&1&1\\x&y&z\\x^2&y^2&z^2\end{vmatrix}=1\cdot\begin{vmatrix}y&z\\y^2&z^2\end{vmatrix}-1\cdot\begin{vmatrix}x&z\\x^2&z^2\end{vmatrix}+1\cdot\begin{vmatrix}x&y\\x^2&y^2\end{vmatrix}$=y(z^2-y^2)-z(xz-x^2)+x(y^2-xy)=yz^2-y^3-xz^2+x^3+xy^2-x^2y=x^3-x^2y+xy^2-y^3+yz^2-xz^2=(x-y)(x^2+xy+y^2)-(x-z)(xz+x^2+z^2)=(x-y)(x^2+xy+y^2)-(x-z)(x^2+xy+y^2)=(x^2+xy+y^2)(x-y-x+z)=(x^2+xy+y^2)(z-y)=(x-y)(x^2+xy+y^2)2.解：增廣矩陣$\left(\begin{array}{ccc|c}2&1&-1&1\\1&-3&2&-1\\4&-2&1&3\end{array}\right)$行變換為$\left(\begin{array}{ccc|c}1&-3&2&-1\\0&7&-5&3\\0&0&0&0\end{array}\right)$對(duì)應(yīng)方程組$\begin{cases}x_1-3x_2+2x_3=-1\\7x_2-5x_3=3\end{cases}$令x_3=k則$\begin{cases}x_2=\frac{3+5k}{7}\\x_1=-1+3\left(\frac{3+5k}{7}\right)-2k=\frac{-2+3k}{7}\end{cases}$通解為$\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)=\left(\begin{array}{c}\frac{-2+3k}{7}\\\frac{3+5k}{7}\\k\end{array}\right)=k\left(\begin{array}{c}\frac{3}{7}\\\frac{5}{7}\\1\end{array}\right)+\left(\begin{array}{c}\frac{-2}{7}\\\frac{3}{7}\\0\end{array}\right)$3.解：特征方程$|\lambdaI-A|=\left|\begin{array}{cc}\lambda-1&-2\\-2&\lambda-5\end{array}\right|=(\lambda-1)(\lambda-5)-4=\lambda^2-6\lambda+1=0$特征值$\lambda_1=3+2\sqrt{2},\lambda_2=3-2\sqrt{2}$對(duì)$\lambda_1=3+2\sqrt{2}$，解方程組$(3+2\sqrt{2})I-A=\left(\begin{array}{cc}2+2\sqrt{2}&-2\\-2&-2+2\sqrt{2}\end{array}\right)$行變換為$\left(\begin{array}{cc}1&-1+\sqrt{2}\\0&0\end{array}\right)$特征向量$x_2=1-\sqrt{2},x_1=1-\sqrt{2}$，即$\alpha_1=(1-\sqrt{2},1-\sqrt{2})$對(duì)$\lambda_2=3-2\sqrt{2}$，解方程組$(3-2\sqrt{2})I-A=\left(\begin{array}{cc}2-2\sqrt{2}&-2\\-2&-2-2\sqrt{2}\end{array}\right)$行變換為$\left(\begin{array}{cc}1&-1-\sqrt{2}\\0&0\end{array}\right)$特征向量$x_2=1+\sqrt{2},x_1=1+\sqrt{2}$，即$\alpha_2=(1+\sqrt{2},1+\sqrt{2})$4.解：設(shè)$\beta=x_1\alpha_1+x_2\alpha_2+x_3\alpha_3$$\left(\begin{array}{ccc|c}1&1&0&1\\0&1&1&2\\1&0&1&3\end{array}\right)$行變換為$\left(\begin{array}{ccc|c}1&1&0&1\\0&1&1&2\\0&-1&1&2\end{array}\right)\rightarrow\left(\begin{array}{ccc|c}1&1&0&1\\0&1&1&2\\0&0&2&4\end{array}\right)\rightarrow\left(\begin{array}{ccc|c}1&1&0&1\\0&1&1&2\\0&0&1&2\end{array}\right)$得$x_3=2$$x_2=2-x_3=0$$x_1=1-x_2=1$所以$\beta=\alpha_1$四、證明題：1.證明：設(shè)c_1(\alpha_1+\alpha_2)+c_2(\alpha_2+\alpha_3)+c_3(\alpha_3+\alpha_1)=0則$(c_1+c_3)\alpha_1+(c_1+c_2)\alpha_2+(c_2+c_3)\alpha_3=0$由于$\alpha_1,\alpha_2,\alpha_3$線(xiàn)性無(wú)關(guān)，得$\begin{cases}c_1+c_3=0\\c_1+c_2=0\\c_2+c_3=0\end{cases}$解得$c_1=c_2=c_3=0$所以$\alpha_1+\alpha_2,\alpha_2+\alpha_3,\alpha_3+\alpha_1$線(xiàn)性無(wú)關(guān)2.證明：設(shè)$\lambda$是A的特征值，$\alpha$是對(duì)應(yīng)特征向量，A$