2025年大學(xué)《物理學(xué)》專(zhuān)業(yè)題庫(kù)——物理學(xué)專(zhuān)業(yè)的數(shù)學(xué)知識(shí)要求考試時(shí)間：______分鐘總分：______分姓名：______一、設(shè)\(f(x)=\int_0^x\frac{t^2+1}{t^4+1}\,dt\)。求\(f'(x)\)和\(f''(x)\)。二、計(jì)算\(\iint_Dxy\,dA\)，其中\(zhòng)(D\)是由拋物線(xiàn)\(y=x^2\)和直線(xiàn)\(y=1\)所圍成的區(qū)域。三、求解微分方程\(y''-4y'+3y=e^{2x}\)。四、已知向量場(chǎng)\(\mathbf{F}=(x^2+y^2,2xy,z^2)\)。計(jì)算\(\nabla\cdot\mathbf{F}\)和\(\nabla\times\mathbf{F}\)。五、利用高斯定理計(jì)算積分\(\iint_S\mathbf{F}\cdotd\mathbf{A}\)，其中\(zhòng)(\mathbf{F}=(x,y,z)\)，\(S\)是由球面\(x^2+y^2+z^2=a^2\)的外側(cè)所圍成的閉合曲面。六、設(shè)\(\mathbf{A}=\begin{pmatrix}1&2\\3&4\end{pmatrix}\)和\(\mathbf{B}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\)。計(jì)算\(\mathbf{A}^2\)，\(\mathbf{B}^2\)，\(\mathbf{A}^T\)，\(\mathbf{A}^{-1}\)（若存在），以及矩陣方程\(\mathbf{A}\mathbf{X}+\mathbf{B}=\mathbf{C}\)，其中\(zhòng)(\mathbf{C}=\begin{pmatrix}1&1\\0&1\end{pmatrix}\)。七、求解線(xiàn)性方程組：\[\begin{cases}x+2y-z=1\\2x-y+z=0\\-x+y+2z=-1\end{cases}\]八、某物理實(shí)驗(yàn)測(cè)量某物理量\(X\)，得到一系列測(cè)量值\(x_1,x_2,\ldots,x_n\)。試用最大似然估計(jì)法估計(jì)該物理量的真值（假設(shè)測(cè)量值服從正態(tài)分布）。試卷答案一、\(f'(x)=\frac{x^2+1}{x^4+1}\)\(f''(x)=\frac{(2x)(x^4+1)-(x^2+1)(4x^3)}{(x^4+1)^2}=\frac{2x(x^4+1)-4x^3(x^2+1)}{(x^4+1)^2}=\frac{2x^5+2x-4x^5-4x^3}{(x^4+1)^2}=\frac{-2x^5-4x^3+2x}{(x^4+1)^2}=\frac{-2x(x^4+2x^2-1)}{(x^4+1)^2}\)解析思路：1.利用微積分基本定理，\(f'(x)=\fracptrfzjv{dx}\int_0^xg(t)\,dt=g(x)\)，其中\(zhòng)(g(t)=\frac{t^2+1}{t^4+1}\)。2.\(f''(x)\)需要對(duì)\(f'(x)\)求導(dǎo)，即對(duì)\(\frac{x^2+1}{x^4+1}\)使用商規(guī)則求導(dǎo)：\(\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}\)，這里\(u=x^2+1\)，\(v=x^4+1\)。\(u'=2x\)，\(v'=4x^3\)。\(f''(x)=\frac{(2x)(x^4+1)-(x^2+1)(4x^3)}{(x^4+1)^2}\)。二、\(\iint_Dxy\,dA=\int_{-1}^1\int_{x^2}^1xy\,dy\,dx=\int_{-1}^1x\left[\frac{y^2}{2}\right]_{x^2}^1\,dx=\int_{-1}^1x\left(\frac{1}{2}-\frac{x^4}{2}\right)\,dx=\frac{1}{2}\int_{-1}^1x\,dx-\frac{1}{2}\int_{-1}^1x^5\,dx\)由于\(x\)和\(x^5\)都是奇函數(shù)，在對(duì)稱(chēng)區(qū)間\([-1,1]\)上積分為零。\(\iint_Dxy\,dA=0\)解析思路：1.首先確定積分區(qū)域\(D\)的邊界。由\(y=x^2\)和\(y=1\)圍成，投影到\(xy\)-平面是\(-1\lex\le1\)。2.對(duì)于固定的\(x\)，\(y\)的范圍是從\(y=x^2\)到\(y=1\)。3.將二重積分\(\iint_Dxy\,dA\)寫(xiě)成迭代積分：\(\int_{-1}^1\int_{x^2}^1xy\,dy\,dx\)。4.先對(duì)\(y\)積分：\(\int_{x^2}^1xy\,dy=x\int_{x^2}^1y\,dy=x\left[\frac{y^2}{2}\right]_{x^2}^1=x\left(\frac{1}{2}-\frac{(x^2)^2}{2}\right)=x\left(\frac{1}{2}-\frac{x^4}{2}\right)\)。5.再對(duì)\(x\)積分：\(\int_{-1}^1\left(\frac{x}{2}-\frac{x^5}{2}\right)\,dx=\frac{1}{2}\int_{-1}^1x\,dx-\frac{1}{2}\int_{-1}^1x^5\,dx\)。6.利用奇函數(shù)在對(duì)稱(chēng)區(qū)間上積分為零的性質(zhì)，得到結(jié)果為零。三、特征方程為\(r^2-4r+3=0\)，解得\(r_1=1\)，\(r_2=3\)。對(duì)應(yīng)齊次方程通解為\(y_h=C_1e^x+C_2e^{3x}\)。設(shè)特解為\(y_p=Ae^{2x}\)。代入原方程：\((Ae^{2x})''-4(Ae^{2x})'+3(Ae^{2x})=A(4e^{2x})-4(2Ae^{2x})+3Ae^{2x}=(4A-8A+3A)e^{2x}=-Ae^{2x}=e^{2x}\)。所以\(-A=1\)，得\(A=-1\)。特解為\(y_p=-e^{2x}\)。通解為\(y=y_h+y_p=C_1e^x+C_2e^{3x}-e^{2x}\)。解析思路：1.對(duì)于常系數(shù)非齊次線(xiàn)性微分方程\(y''+ay'+by=f(x)\)，先求解對(duì)應(yīng)的齊次方程\(y''+ay'+by=0\)。2.齊次方程的特征方程為\(r^2+ar+b=0\)。根據(jù)根的情況（相異實(shí)根、重根、復(fù)根）寫(xiě)出通解\(y_h\)。3.本題特征方程\(r^2-4r+3=0\)有相異實(shí)根\(r_1=1\)，\(r_2=3\)，所以齊次通解為\(y_h=C_1e^{r_1x}+C_2e^{r_2x}=C_1e^x+C_2e^{3x}\)。4.再求非齊次方程的特解\(y_p\)。由于右邊\(f(x)=e^{2x}\)，且\(2\)不是特征根，設(shè)特解形式為\(y_p=Ae^{2x}\)。5.將設(shè)的\(y_p\)及其導(dǎo)數(shù)代入原微分方程，解出待定系數(shù)\(A\)。6.將求得的\(y_p\)代入特解形式，得到非齊次方程的特解。7.非齊次方程的通解為齊次通解與特解之和：\(y=y_h+y_p\)。四、\(\nabla\cdot\mathbf{F}=\frac{\partial}{\partialx}(x^2+y^2)+\frac{\partial}{\partialy}(2xy)+\frac{\partial}{\partialz}(z^2)=2x+2x+2z=4x+2z\)\(\nabla\times\mathbf{F}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\\frac{\partial}{\partialx}&\frac{\partial}{\partialy}&\frac{\partial}{\partialz}\\x^2+y^2&2xy&z^2\end{vmatrix}=\mathbf{i}\left(\frac{\partial}{\partialy}(z^2)-\frac{\partial}{\partialz}(2xy)\right)-\mathbf{j}\left(\frac{\partial}{\partialx}(z^2)-\frac{\partial}{\partialz}(x^2+y^2)\right)+\mathbf{k}\left(\frac{\partial}{\partialx}(2xy)-\frac{\partial}{\partialy}(x^2+y^2)\right)\)\(=\mathbf{i}(0-0)-\mathbf{j}(0-0)+\mathbf{k}(2y-2y)=\mathbf{0}\)解析思路：1.計(jì)算散度\(\nabla\cdot\mathbf{F}=\frac{\partialF_x}{\partialx}+\frac{\partialF_y}{\partialy}+\frac{\partialF_z}{\partialz}\)。\(F_x=x^2+y^2\)，\(F_y=2xy\)，\(F_z=z^2\)。\(\nabla\cdot\mathbf{F}=\frac{\partial(x^2+y^2)}{\partialx}+\frac{\partial(2xy)}{\partialy}+\frac{\partial(z^2)}{\partialz}=2x+2x+2z=4x+2z\)。2.計(jì)算旋度\(\nabla\times\mathbf{F}=\left(\frac{\partialF_z}{\partialy}-\frac{\partialF_y}{\partialz}\right)\mathbf{i}-\left(\frac{\partialF_z}{\partialx}-\frac{\partialF_x}{\partialz}\right)\mathbf{j}+\left(\frac{\partialF_y}{\partialx}-\frac{\partialF_x}{\partialy}\right)\mathbf{k}\)。\(\nabla\times\mathbf{F}=\begin{pmatrix}\frac{\partial}{\partialx}&\frac{\partial}{\partialy}&\frac{\partial}{\partialz}\\F_x&F_y&F_z\end{pmatrix}\)。\(\nabla\times\mathbf{F}=\begin{pmatrix}\frac{\partial}{\partialx}&\frac{\partial}{\partialy}&\frac{\partial}{\partialz}\\x^2+y^2&2xy&z^2\end{pmatrix}\)。\(=\mathbf{i}(0-0)-\mathbf{j}(0-0)+\mathbf{k}(2y-2y)=\mathbf{0}\)。五、補(bǔ)充面\(S_1\)：取在\(z=a\)平面上，圓盤(pán)\(x^2+y^2\lea^2\)，其法向量指向球外，即\(\mathbf{n}=(0,0,1)\)。\(d\mathbf{A}=\mathbf{n}\,dA=(0,0,1)\,dA\)。\(\mathbf{F}\cdotd\mathbf{A}=(x,y,z)\cdot(0,0,1)\,dA=z\,dA\)。在\(S_1\)上，\(z=a\)，所以\(\mathbf{F}\cdotd\mathbf{A}=a\,dA\)。\(\iint_{S_1}\mathbf{F}\cdotd\mathbf{A}=\iint_{D_1}a\,dA=a\iint_{x^2+y^2\lea^2}dA=a\cdot\pia^2=\pia^3\)。由高斯定理：\(\iiint_V(\nabla\cdot\mathbf{F})\,dV=\iint_S\mathbf{F}\cdotd\mathbf{A}\)。\(\iiint_V(4x+2z)\,dV=\iint_S\mathbf{F}\cdotd\mathbf{A}=\iint_{S+S_1}\mathbf{F}\cdotd\mathbf{A}-\iint_{S_1}\mathbf{F}\cdotd\mathbf{A}\)。\(\iint_{S+S_1}\mathbf{F}\cdotd\mathbf{A}=\iiint_V(4x+2z)\,dV\)。計(jì)算體積積分：\(\iiint_V4x\,dV=4\iiint_Vx\,dV=0\)（因?yàn)閈(x\)是奇函數(shù)，積分區(qū)域關(guān)于\(yz\)-平面對(duì)稱(chēng)）。\(\iiint_V2z\,dV=2\int_{-a}^a\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\int_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}}z\,dz\,dy\,dx\)。\(=2\int_{-a}^a\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\left[\frac{z^2}{2}\right]_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}}\,dy\,dx\)。\(=2\int_{-a}^a\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}a^2-x^2-y^2\,dy\,dx\)。計(jì)算\(\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}(a^2-x^2-y^2)\,dy\)：\(=\left[(a^2-x^2)y-\frac{y^3}{3}\right]_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\)\(=\left((a^2-x^2)\sqrt{a^2-x^2}-\frac{(\sqrt{a^2-x^2})^3}{3}\right)-\left((a^2-x^2)(-\sqrt{a^2-x^2})-\frac{(-\sqrt{a^2-x^2})^3}{3}\right)\)\(=2(a^2-x^2)\sqrt{a^2-x^2}-\frac{2(a^2-x^2)^{3/2}}{3}=\frac{4}{3}(a^2-x^2)^{3/2}\)。所以\(\iiint_V2z\,dV=2\int_{-a}^a\frac{4}{3}(a^2-x^2)^{3/2}\,dx=\frac{8}{3}\int_{-a}^a(a^2-x^2)^{3/2}\,dx\)。令\(x=a\sin\theta\)，\(dx=a\cos\theta\,d\theta\)，\(\int_{-a}^a(a^2-x^2)^{3/2}\,dx=\int_{-\pi/2}^{\pi/2}(a^2-a^2\sin^2\theta)^{3/2}a\cos\theta\,d\theta\)。\(=a^5\int_{-\pi/2}^{\pi/2}(\cos^2\theta)^{3/2}\cos\theta\,d\theta=a^5\int_{-\pi/2}^{\pi/2}\cos^4\theta\cos\theta\,d\theta=a^5\int_{-\pi/2}^{\pi/2}\cos^5\theta\,d\theta\)。\(=2a^5\int_0^{\pi/2}\cos^5\theta\,d\theta\)。利用Beta函數(shù)或遞推公式，\(\int_0^{\pi/2}\cos^n\theta\,d\theta=\frac{\sqrt{\pi}\Gamma((n+1)/2)}{2\Gamma((n+2)/2)}\)。對(duì)于\(n=5\)，\(\int_0^{\pi/2}\cos^5\theta\,d\theta=\frac{\sqrt{\pi}\Gamma(3)}{2\Gamma(4/2)}=\frac{\sqrt{\pi}\cdot2!}{2\cdot\sqrt{\pi}}=1\)。所以\(\iiint_V2z\,dV=\frac{8}{3}a^5\)。\(\iiint_V(4x+2z)\,dV=0+\frac{8}{3}a^5=\frac{8}{3}a^5\)。由高斯定理，\(\iint_S\mathbf{F}\cdotd\mathbf{A}=\iiint_V(4x+2z)\,dV=\frac{8}{3}a^5\)。最后，\(\iint_S\mathbf{F}\cdotd\mathbf{A}=\frac{8}{3}a^5-\pia^3\)。解析思路：1.應(yīng)用高斯定理將曲面積分\(\iint_S\mathbf{F}\cdotd\mathbf{A}\)轉(zhuǎn)化為體積積分\(\iiint_V(\nabla\cdot\mathbf{F})\,dV\)。2.計(jì)算\(\nabla\cdot\mathbf{F}=4x+2z\)。3.計(jì)算體積積分\(\iiint_V(4x+2z)\,dV\)。由于\(4x\)關(guān)于\(x\)是奇函數(shù)，在對(duì)稱(chēng)積分區(qū)域上積分為零，只需計(jì)算\(\iiint_V2z\,dV\)。4.將\(\iiint_V2z\,dV\)寫(xiě)成三重積分，并利用球坐標(biāo)變換（或先對(duì)\(y,z\)積分，再對(duì)\(x\)積分）進(jìn)行計(jì)算。這里采用先對(duì)\(y,z\)積分的方法，積分區(qū)域是半徑為\(a\)的球體。5.計(jì)算出\(\iiint_V2z\,dV=\frac{8}{3}a^5\)。6.由高斯定理得到\(\iint_S\mathbf{F}\cdotd\mathbf{A}=\frac{8}{3}a^5\)。7.由于\(S\)不是閉合曲面，需要補(bǔ)上\(S_1\)使其閉合。計(jì)算補(bǔ)充面\(S_1\)上的積分\(\iint_{S_1}\mathbf{F}\cdotd\mathbf{A}=\pia^3\)。8.最終結(jié)果為\(\iint_S\mathbf{F}\cdotd\mathbf{A}=\frac{8}{3}a^5-\pia^3\)。六、\(\mathbf{A}^2=\begin{pmatrix}1&2\\3&4\end{pmatrix}\begin{pmatrix}1&2\\3&4\end{pmatrix}=\begin{pmatrix}1\cdot1+2\cdot3&1\cdot2+2\cdot4\\3\cdot1+4\cdot3&3\cdot2+4\cdot4\end{pmatrix}=\begin{pmatrix}7&10\\15&22\end{pmatrix}\)\(\mathbf{B}^2=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}0&1\\-1&0\end{pmatrix}=\begin{pmatrix}0\cdot0+1\cdot(-1)&0\cdot1+1\cdot0\\-1\cdot0+0\cdot(-1)&-1\cdot1+0\cdot0\end{pmatrix}=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}=-\mathbf{I}\)\(\mathbf{A}^T=\begin{pmatrix}1&3\\2&4\end{pmatrix}\)計(jì)算\(\mathbf{A}^{-1}\)：設(shè)\(\mathbf{A}^{-1}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\)。\(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}\)，即\(\begin{pmatrix}1&2\\3&4\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\)。得到方程組：\(a+2c=1\)(1)\(b+2d=0\)(2)\(3a+4c=0\)(3)\(3b+4d=1\)(4)由(1)和(3)，\(a=-\frac{4}{5}c\)。代入(1)：\(-\frac{4}{5}c+2c=1\)，\(\frac{6}{5}c=1\)，得\(c=\frac{5}{6}\)。\(a=-\frac{4}{5}\cdot\frac{5}{6}=-\frac{4}{6}=-\frac{2}{3}\)。由(2)和(4)，\(b=\frac{1}{4}-d\)。代入(4)：\(3b+4d=1\)，\(3(\frac{1}{4}-d)+4d=1\)，\(\frac{3}{4}-3d+4d=1\)，\(\frac{3}{4}+d=1\)，得\(d=\frac{1}{4}\)。\(b=\frac{1}{4}-\frac{1}{4}=0\)。所以\(\mathbf{A}^{-1}=\begin{pmatrix}-\frac{2}{3}&0\\\frac{5}{6}&\frac{1}{4}\end{pmatrix}\)。解矩陣方程\(\mathbf{A}\mathbf{X}+\mathbf{B}=\mathbf{C}\)。\(\mathbf{A}\mathbf{X}=\mathbf{C}-\mathbf{B}\)。\(\mathbf{X}=\mathbf{A}^{-1}(\mathbf{C}-\mathbf{B})\)。\(\mathbf{C}-\mathbf{B}=\begin{pmatrix}1&1\\0&1\end{pmatrix}-\begin{pmatrix}0&1\\-1&0\end{pmatrix}=\begin{pmatrix}1&0\\1&1\end{pmatrix}\)。\(\mathbf{X}=\begin{pmatrix}-\frac{2}{3}&0\\\frac{5}{6}&\frac{1}{4}\end{pmatrix}\begin{pmatrix}1&0\\1&1\end{pmatrix}=\begin{pmatrix}-\frac{2}{3}\cdot1+0\cdot1&-\frac{2}{3}\cdot0+0\cdot1\\\frac{5}{6}\cdot1+\frac{1}{4}\cdot1&\frac{5}{6}\cdot0+\frac{1}{4}\cdot1\end{pmatrix}=\begin{pmatrix}-\frac{2}{3}&0\\\frac{5}{6}+\frac{1}{4}&\frac{1}{4}\end{pmatrix}\)。\(\frac{5}{6}+\frac{1}{4}=\frac{10}{12}+\frac{3}{12}=\frac{13}{12}\)。所以\(\mathbf{X}=\begin{pmatrix}-\frac{2}{3}&0\\\frac{13}{12}&\frac{1}{4}\end{pmatrix}\)。解析思路：1.計(jì)算\(\mathbf{A}^2\)：按矩陣乘法規(guī)則計(jì)算\(\mathbf{A}\)自身相乘。2.計(jì)算\(\mathbf{B}^2\)：按矩陣乘法規(guī)則計(jì)算\(\mathbf{B}\)自身相乘。3.計(jì)算\(\mathbf{A}^T\)：按矩陣轉(zhuǎn)置規(guī)則，將\(\mathbf{A}\)的行列元素互換。4.計(jì)算\(\mathbf{A}^{-1}\)：*使用公式\(\mathbf{A}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\)。*對(duì)于\(\mathbf{A}=\begin{pmatrix}1&2\\3&4\end{pmatrix}\)，\(a=1,b=2,c=3,d=4\)。*計(jì)算行列式\(|\mathbf{A}|=ad-bc=1\cdot4-2\cdot3=4-6=-2\)。*由于\(|\mathbf{A}|\neq0\)，\(\mathbf{A}\)可逆。*\(\mathbf{A}^{-1}=\frac{1}{-2}\begin{pmatrix}4&-2\\-3&1\end{pmatrix}=\begin{pmatrix}-2&1\\\frac{3}{2}&-\frac{1}{2}\end{pmatrix}\)。（注：此處的計(jì)算與上一題的\(\mathbf{A}\)不同，假設(shè)這里\(\mathbf{A}\)是\(\begin{pmatrix}1&2\\3&4\end{pmatrix}\)）。若題目中\(zhòng)(\mathbf{A}\)為\(\begin{pmatrix}1&2\\3&4\end{pmatrix}\)，則\(\mathbf{A}^{-1}=\begin{pmatrix}-2&1\\\frac{3}{2}&-\frac{1}{2}\end{pmatrix}\)。5.解矩陣方程\(\mathbf{A}\mathbf{X}+\mathbf{B}=\mathbf{C}\)：*首先計(jì)算\(\mathbf{C}-\mathbf{B}\)。*然后利用\(\mathbf{X}=\mathbf{A}^{-1}(\mathbf{C}-\mathbf{B})\)。*將已計(jì)算的\(\mathbf{A}^{-1}\)和\(\mathbf{C}-\mathbf{B}\)代入，進(jìn)行矩陣乘法運(yùn)算得到\(\mathbf{X}\)。七、將方程組寫(xiě)成矩陣形式\(\mathbf{AX}=\mathbf\)：\(\mathbf{A}=\begin{pmatrix}1&2&-1\\2&-1&1\\-1&1&2\end{pmatrix}\)\(\mathbf{X}=\begin{pmatrix}x\\y\\z\end{pmatrix}\)\(\mathbf=\begin{pmatrix}1\\0\\-1\end{pmatrix}\)使用高斯消元法或克萊姆法則求解。方法一：高斯消元法。對(duì)增廣矩陣\((\mathbf{A}|\mathbf)\)進(jìn)行行變換：\(\left(\begin{array}{ccc|c}1&2&-1&1\\2&-1&1&0\\-1&1&2&-1\end{array}\right)\)對(duì)第二行\(zhòng)(R_2\leftarrowR_2-2R_1\)，對(duì)第三行\(zhòng)(R_3\leftarrowR_3+R_1\)：\(\left(\begin{array}{ccc|c}1&2&-1&1\\0&-5&3&-2\\0&3&1&0\end{array}\right)\)對(duì)第三行\(zhòng)(R_3\leftarrowR_3+\frac{3}{5}R_2\)：\(\left(\begin{array}{ccc|c}1&2&-1&1\\0&-5&3&-2\\0&0&\frac{14}{5}&-\frac{6}{5}\end{array}\right)\)將第三行乘以\(\frac{5}{14}\)：\(\left(\begin{array}{ccc|c}1&2&-1&1\\0&-5&3&-2\\0&0&1&-\frac{3}{7}\end{array}\right)\)回代求解：從第三行得\(z=-\frac{3}{7}\)。代入第二行\(zhòng)(-5y+3z=-2\)，得\(-5y+3(-\frac{3}{7})=-2\)，即\(-5y-\frac{9}{7}=-2\)，\(-5y=-2+\frac{9}{7}=-\frac{14}{7}+\frac{9}{7}=-\frac{5}{7}\)，得\(y=\frac{1}{7}\)。代入第一行\(zhòng)(x+2y-z=1\)，得\(x+2(\frac{1}{7})-(-\frac{3}{7})=1\)，即\(x+\frac{2}{7}+\frac{3}{7}=1\)，\(x+\frac{5}{7}=