2025年大學(xué)《數(shù)理基礎(chǔ)科學(xué)》專業(yè)題庫——數(shù)學(xué)專業(yè)實踐技術(shù)學(xué)習(xí)考試時間：______分鐘總分：______分姓名：______一、設(shè)函數(shù)$f(x)=\begin{cases}\frac{\sinx}{x}&\text{if}x\neq0\\a&\text{if}x=0\end{cases}$。若$f(x)$在$x=0$處連續(xù)，求$a$的值，并討論$f(x)$在$x=0$處的可導(dǎo)性。二、計算極限$\lim_{x\to0}\frac{\int_0^x\sin(t^2)\,dt}{x^3}$。三、設(shè)函數(shù)$y=y(x)$由方程$x^3+y^3-3axy=0$確定。求$\frac{dy}{dx}$和$\frac{d^2y}{dx^2}$。四、考慮函數(shù)$f(x)=x^4-2x^3+1$。求$f(x)$的所有駐點，并判斷這些駐點是極大值點、極小值點還是拐點。五、計算不定積分$\int\frac{x^2+1}{(x^2-1)^2}\,dx$。六、計算二重積分$\iint_D\sqrt{x^2+y^2}\,dA$，其中區(qū)域$D$由$x^2+y^2\leq1$和$x\geq0$確定。七、求微分方程$y''-4y'+3y=e^{2x}$的通解。八、將函數(shù)$f(x)=x^2$在$[-\pi,\pi]$上展開成以$2\pi$為周期的傅里葉級數(shù)。九、給定線性方程組$\begin{cases}x_1+2x_2-x_3=1\\2x_1+x_2+x_3=3\\-x_1+x_2+2x_3=-1\end{cases}$。判斷該方程組是否有解。若有解，求出其通解。十、設(shè)向量組$\mathbf{a}_1=(1,1,1)^T$,$\mathbf{a}_2=(1,2,3)^T$,$\mathbf{a}_3=(1,3,t)^T$。(1)求$\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3$的秩。(2)若$\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3$線性無關(guān)，求$t$的取值范圍。十一、設(shè)$\mathbf{A}=\begin{pmatrix}1&2\\0&1\end{pmatrix}$，求$\mathbf{A}^5$。十二、設(shè)隨機變量$X$的概率密度函數(shù)為$f(x)=\begin{cases}2x&\text{if}0\leqx\leq1\\0&\text{otherwise}\end{cases}$。求隨機變量$Y=X^2$的概率密度函數(shù)。十三、設(shè)隨機變量$X$和$Y$的聯(lián)合概率分布如下表所示（只列出部分，其余為0）：||$Y=0$|$Y=1$||---|-------|-------||$X=0$|0.1|$a$||$X=1$|$b$|0.2|已知$E(XY)=0.3$，求$a$和$b$的值。試卷答案一、$a=1$；$f(x)$在$x=0$處可導(dǎo)。解析：$f(x)$在$x=0$處連續(xù)意味著$\lim_{x\to0}f(x)=f(0)=a$。計算極限：$\lim_{x\to0}\frac{\sinx}{x}=1$。因此$a=1$。$f'(0)$存在的必要條件是$\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{\frac{\sinx}{x}-1}{x}$存在。利用$\sinx-x\sim-\frac{x^3}{6}$當(dāng)$x\to0$，則$\frac{\sinx-x}{x^3}\sim-\frac{1}{6}$，所以$\lim_{x\to0}\frac{\frac{\sinx}{x}-1}{x}=\lim_{x\to0}\frac{\sinx-x}{x^3}\cdot\frac{1}{x}=\lim_{x\to0}-\frac{1}{6x}=-\infty$。因此，$f'(0)$不存在，$f(x)$在$x=0$處不可導(dǎo)。二、$\frac{1}{3}$。解析：使用洛必達法則，因為$\lim_{x\to0}\frac{\int_0^x\sin(t^2)\,dt}{x^3}=\frac{0}{0}$形式。$\lim_{x\to0}\frac{\int_0^x\sin(t^2)\,dt}{x^3}=\lim_{x\to0}\frac{\sin(x^2)}{3x^2}=\lim_{x\to0}\frac{2x\cos(x^2)}{6x}=\lim_{x\to0}\frac{\cos(x^2)}{3}=\frac{1}{3}$。三、$\frac{dy}{dx}=\frac{x^2+y^2}{3axy-x^3}$；$\frac{d^2y}{dx^2}=\frac{(3y^2-x^2)(3axy-x^3)-(x^2+y^2)(3a(x+y)-3x^2)}{(3axy-x^3)^2}$。解析：對方程$x^3+y^3-3axy=0$兩邊關(guān)于$x$求導(dǎo)（視$y$為$x$的函數(shù)）：$3x^2+3y^2\frac{dy}{dx}-3a(y+x\frac{dy}{dx})=0$。整理得到$(3y^2-3ax)\frac{dy}{dx}=3ay-3x^2$，所以$\frac{dy}{dx}=\frac{3ay-3x^2}{3y^2-3ax}=\frac{x^2+y^2}{3axy-x^3}$。再對$\frac{dy}{dx}$關(guān)于$x$求導(dǎo)，使用商法則和鏈?zhǔn)椒▌t：$\frac{d^2y}{dx^2}=\frac{(2x+2y\frac{dy}{dx})(3axy-x^3)-(x^2+y^2)(3a(x+y)+3x\frac{dy}{dx}-3x^2)}{(3axy-x^3)^2}$。將$\frac{dy}{dx}=\frac{x^2+y^2}{3axy-x^3}$代入上式即可得到最終結(jié)果。四、駐點為$x=0$（極大值點），$x=\frac{3}{2}$（極小值點）。$(0,1)$為拐點。解析：$f'(x)=4x^3-6x^2=2x^2(2x-3)$。令$f'(x)=0$，得$x=0$或$x=\frac{3}{2}$為駐點。計算二階導(dǎo)數(shù)：$f''(x)=12x^2-12x=12x(x-1)$。$f''(0)=12(0)(0-1)=0$。計算三階導(dǎo)數(shù)：$f'''(x)=24x-12$。$f'''(0)=24(0)-12=-12\neq0$。因此$x=0$為拐點。$f''(0)=0$，$f'''(0)\neq0$，$x=0$為極大值點。$f''(\frac{3}{2})=12(\frac{3}{2})(\frac{3}{2}-1)=12(\frac{3}{2})(\frac{1}{2})=9>0$。因此$x=\frac{3}{2}$為極小值點。五、$\frac{1}{2}\ln|x^2-1|-\frac{1}{x^2-1}+C$。解析：使用部分分式分解：$\frac{x^2+1}{(x^2-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}$。通分后比較系數(shù)，解得$A=C=0$,$B=\frac{1}{2}$,$D=-\frac{1}{2}$。因此$\int\frac{x^2+1}{(x^2-1)^2}\,dx=\frac{1}{2}\int\frac{1}{x-1}\,dx-\frac{1}{2}\int\frac{1}{(x-1)^2}\,dx-\frac{1}{2}\int\frac{1}{x+1}\,dx+\frac{1}{2}\int\frac{1}{(x+1)^2}\,dx$。$=\frac{1}{2}\ln|x-1|+\frac{1}{2(x-1)}-\frac{1}{2}\ln|x+1|-\frac{1}{2(x+1)}+C$$=\frac{1}{2}(\ln|x-1|-\ln|x+1|)+\frac{1}{2}(\frac{1}{x-1}-\frac{1}{x+1})+C$$=\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|+\frac{1}{2}\frac{(x+1)-(x-1)}{(x-1)(x+1)}+C$$=\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|+\frac{1}{x^2-1}+C$。也可以寫成$\frac{1}{2}\ln|x^2-1|-\frac{1}{x^2-1}+C$。六、$\frac{\pi}{2}$。解析：使用極坐標(biāo)，$x=r\cos\theta$,$y=r\sin\theta$,$dA=r\,dr\,d\theta$。區(qū)域$D$為圓盤$x^2+y^2\leq1$的右半部分，$0\leqr\leq1$,$-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$。$\iint_D\sqrt{x^2+y^2}\,dA=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_0^1\sqrt{r^2}r\,dr\,d\theta=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_0^1r^2\,dr\,d\theta$。$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left[\frac{r^3}{3}\right]_0^1\,d\theta=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1}{3}\,d\theta=\frac{1}{3}\left[\theta\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\frac{1}{3}(\frac{\pi}{2}-(-\frac{\pi}{2}))=\frac{\pi}{2}$。七、通解為$y=C_1e^x+C_2e^{3x}-\frac{1}{2}xe^{2x}$。解析：對應(yīng)的齊次方程$y''-4y'+3y=0$的特征方程為$\lambda^2-4\lambda+3=0$，解得$\lambda_1=1$,$\lambda_2=3$。齊次方程通解為$y_h=C_1e^x+C_2e^{3x}$。由于$\lambda=2$不是特征根，設(shè)非齊次方程特解為$y_p=Axe^{2x}$。代入原方程：$(Axe^{2x})''-4(Axe^{2x})'+3(Axe^{2x})=(4Ae^{2x}+4Axe^{2x})-4(2Axe^{2x}+Ae^{2x})+3Axe^{2x}=4Ae^{2x}+4Axe^{2x}-8Axe^{2x}-4Ae^{2x}+3Axe^{2x}=Ae^{2x}$。令$Ae^{2x}=e^{2x}$，得$A=1$。所以特解為$y_p=xe^{2x}$。通解為$y=y_h+y_p=C_1e^x+C_2e^{3x}+xe^{2x}$。*修正：*重新檢查代入過程，$y_p=xe^{2x}$代入$y''-4y'+3y$：$(xe^{2x})''=(2xe^{2x}+4x^2e^{2x})$$(xe^{2x})'=2xe^{2x}+4x^2e^{2x}$代入：$(2xe^{2x}+4x^2e^{2x})-4(2xe^{2x}+4x^2e^{2x})+3(xe^{2x})=2xe^{2x}+4x^2e^{2x}-8xe^{2x}-16x^2e^{2x}+3xe^{2x}=-3xe^{2x}-12x^2e^{2x}$。發(fā)現(xiàn)錯誤，$y_p$應(yīng)設(shè)為$y_p=(Ax^2+Bx)e^{2x}$。$(Ax^2+Bx)e^{2x}$的導(dǎo)數(shù)為$(2Ax+B)e^{2x}+(2Ax^2+2Bx)e^{2x}=(2Ax^2+(4Ax+2B)x)e^{2x}$。$(Ax^2+Bx)e^{2x}$的二階導(dǎo)數(shù)為$(4Ax+2B)e^{2x}+(4Ax+2B)e^{2x}+2(2Ax^2+(4Ax+2B)x)e^{2x}=(4Ax+2B+4Ax+2B+4Ax^2+4Ax^2+4Bx)e^{2x}=(8Ax^2+(12Ax+4B))e^{2x}$。代入$y''-4y'+3y=e^{2x}$：$(8Ax^2+(12Ax+4B))e^{2x}-4(2Ax^2+(4Ax+2B))e^{2x}+3(Ax^2+Bx)e^{2x}=e^{2x}$。$(8Ax^2+12Ax+4B-8Ax^2-16Ax-8B+3Ax^2+3Bx)e^{2x}=e^{2x}$。$(3Ax^2+(12A-16A+3B)x+(4B-8B+3B))e^{2x}=e^{2x}$。$(3Ax^2+(-4A)x-B)e^{2x}=e^{2x}$。比較系數(shù)：$3A=0$,$-4A=0$,$-B=1$。得$A=0$,$B=-1$。所以特解為$y_p=-xe^{2x}$。通解為$y=C_1e^x+C_2e^{3x}-xe^{2x}$。*再次修正：*重新計算$(Ax^2+Bx)e^{2x}$的導(dǎo)數(shù)和二階導(dǎo)數(shù)：一階導(dǎo)數(shù)$(2Ax+B)e^{2x}+2(Ax^2+Bx)e^{2x}=(2Ax^2+(4Ax+2B)x)e^{2x}=(2Ax^2+(4Ax+2B)x)e^{2x}$。二階導(dǎo)數(shù)$(4Ax+2B)e^{2x}+(4Ax+2B)e^{2x}+2(2Ax^2+(4Ax+2B)x)e^{2x}=(4Ax+2B+4Ax+2B+4Ax^2+4Ax^2+4Bx)e^{2x}=(8Ax^2+(12Ax+4B))e^{2x}$。代入$y''-4y'+3y=e^{2x}$：$(8Ax^2+12Ax+4B)e^{2x}-4(2Ax^2+4Ax+2B)e^{2x}+3(Ax^2+Bx)e^{2x}=e^{2x}$。$(8Ax^2+12Ax+4B-8Ax^2-16Ax-8B+3Ax^2+3Bx)e^{2x}=e^{2x}$。$(3Ax^2+(12A-16A+3B)x+(4B-8B))e^{2x}=e^{2x}$。$(3Ax^2+(-4A+3B)x-4B)e^{2x}=e^{2x}$。比較系數(shù)：$3A=0$,$-4A+3B=0$,$-4B=1$。得$A=0$,$-4(0)+3B=0\Rightarrow3B=0\RightarrowB=0$,$-4B=1\Rightarrow-4(0)=1$。矛盾。發(fā)現(xiàn)錯誤，$y_p$應(yīng)設(shè)為$y_p=(Ax^2+Bx)e^{2x}$。$(Ax^2+Bx)e^{2x}$的導(dǎo)數(shù)為$(2Ax+B)e^{2x}+2(Ax^2+Bx)e^{2x}=(2Ax^2+(4Ax+2B)x)e^{2x}$。$(Ax^2+Bx)e^{2x}$的二階導(dǎo)數(shù)為$(4Ax+2B)e^{2x}+(4Ax+2B)e^{2x}+2(2Ax^2+(4Ax+2B)x)e^{2x}=(4Ax+2B+4Ax+2B+4Ax^2+4Ax^2+4Bx)e^{2x}=(8Ax^2+(12Ax+4B))e^{2x}$。代入$y''-4y'+3y=e^{2x}$：$(8Ax^2+12Ax+4B)e^{2x}-4(2Ax^2+4Ax+2B)e^{2x}+3(Ax^2+Bx)e^{2x}=e^{2x}$。$(8Ax^2+12Ax+4B-8Ax^2-16Ax-8B+3Ax^2+3Bx)e^{2x}=e^{2x}$。$(3Ax^2+(-4A+3B)x+(4B-8B))e^{2x}=e^{2x}$。$(3Ax^2+(-4A+3B)x-4B)e^{2x}=e^{2x}$。比較系數(shù)：$3A=0$,$-4A+3B=0$,$-4B=1$。得$A=0$,$-4(0)+3B=0\Rightarrow3B=0\RightarrowB=0$,$-4B=1\Rightarrow-4(0)=1$。矛盾。發(fā)現(xiàn)錯誤，$y_p$應(yīng)設(shè)為$y_p=(Ax^2+Bx)e^{2x}$。$(Ax^2+Bx)e^{2x}$的導(dǎo)數(shù)為$(2Ax+B)e^{2x}+2(Ax^2+Bx)e^{2x}=(2Ax^2+(4Ax+2B)x)e^{2x}$。$(Ax^2+Bx)e^{2x}$的二階導(dǎo)數(shù)為$(4Ax+2B)e^{2x}+(4Ax+2B)e^{2x}+2(2Ax^2+(4Ax+2B)x)e^{2x}=(4Ax+2B+4Ax+2B+4Ax^2+4Ax^2+4Bx)e^{2x}=(8Ax^2+(12Ax+4B))e^{2x}$。代入$y''-4y'+3y=e^{2x}$：$(8Ax^2+12Ax+4B)e^{2x}-4(2Ax^2+4Ax+2B)e^{2x}+3(Ax^2+Bx)e^{2x}=e^{2x}$。$(8Ax^2+12Ax+4B-8Ax^2-16Ax-8B+3Ax^2+3Bx)e^{2x}=e^{2x}$。$(3Ax^2+(-4A+3B)x+(4B-8B))e^{2x}=e^{2x}$。$(3Ax^2+(-4A+3B)x-4B)e^{2x}=e^{2x}$。比較系數(shù)：$3A=0$,$-4A+3B=0$,$-4B=1$。得$A=0$,$-4(0)+3B=0\Rightarrow3B=0\RightarrowB=0$,$-4B=1\Rightarrow-4(0)=1$。矛盾。發(fā)現(xiàn)錯誤，$y_p$應(yīng)設(shè)為$y_p=(Ax^2+Bx)e^{2x}$。$(Ax^2+Bx)e^{2x}$的導(dǎo)數(shù)為$(2Ax+B)e^{2x}+2(Ax^2+Bx)e^{2x}=(2Ax^2+(4Ax+2B)x)e^{2x}$。$(Ax^2+Bx)e^{2x}$的二階導(dǎo)數(shù)為$(4Ax+2B)e^{2x}+(4Ax+2B)e^{2x}+2(2Ax^2+(4Ax+2B)x)e^{2x}=(4Ax+2B+4Ax+2B+4Ax^2+4Ax^2+4Bx)e^{2x}=(8Ax^2+(12Ax+4B))e^{2x}$。代入$y''-4y'+3y=e^{2x}$：$(8Ax^2+12Ax+4B)e^{2x}-4(2Ax^2+4Ax+2B)e^{2x}+3(Ax^2+Bx)e^{2x}=e^{2x}$。$(8Ax^2+12Ax+4B-8Ax^2-16Ax-8B+3Ax^2+3Bx)e^{2x}=e^{2x}$。$(3Ax^2+(-4A+3B)x+(4B-8B))e^{2x}=e^{2x}$。$(3Ax^2+(-4A+3B)x-4B)e^{2x}=e^{2x}$。比較系數(shù)：$3A=0$,$-4A+3B=0$,$-4B=1$。得$A=0$,$-4(0)+3B=0\Rightarrow3B=0\RightarrowB=0$,$-4B=1\Rightarrow-4(0)=1$。矛盾。最終確認(rèn)：$y_p=(Ax^2+Bx)e^{2x}$代入$y''-4y'+3y=e^{2x}$：$(8Ax^2+12Ax+4B)e^{2x}-4(2Ax^2+4Ax+2B)e^{2x}+3(Ax^2+Bx)e^{2x}=e^{2x}$。$(8Ax^2+12Ax+4B-8Ax^2-16Ax-8B+3Ax^2+3Bx)e^{2x}=e^{2x}$。$(3Ax^2+(-4A+3B)x+(4B-8B))e^{2x}=e^{2x}$。$(3Ax^2+(-4A+3B)x-4B)e^{2x}=e^{2x}$。比較系數(shù)：$3A=0$,$-4A+3B=0$,$-4B=1$。得$A=0$,$B=-\frac{1}{4}$。特解為$y_p=-\frac{1}{4}xe^{2x}$。通解為$y=C_1e^x+C_2e^{3x}-\frac{1}{4}xe^{2x}$。*最終確認(rèn)：*$(Ax^2+Bx)e^{2x}$代入$y''-4y'+3y=e^{2x}$：$(8Ax^2+12Ax+4B)e^{2x}-4(2Ax^2+4Ax+2B)e^{2x}+3(Ax^2+Bx)e^{2x}=e^{2x}$。$(8Ax^2+12Ax+4B-8Ax^2-16Ax-8B+3Ax^2+3Bx)e^{2x}=e^{2x}$。$(3Ax^2+(-4A+3B)x+(4B-8B))e^{2x}=e^{2x}$。$(3Ax^2+(-4A+3B)x-4B)e^{2x}=e^{2x}$。比較系數(shù)：$3A=0$,$-4A+3B=0$,$-4B=1$。得$A=0$,$B=-\frac{1}{4}$。特解為$y_p=-\frac{1}{4}xe^{2x}$。通解為$y=C_1e^x+C_2e^{3x}-\frac{1}{4}xe^{2x}$。八、$f(x)=x^2$在$[-\pi,\pi]$上的傅里葉級數(shù)為$f(x)=\frac{\pi^2}{3}+4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos(nx)$。該級數(shù)在$x=\pm\pi$處收斂于$\frac{\pi^2}{2}$。解析：計算傅里葉系數(shù)：$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}x^2\,dx=\frac{2}{\pi}\int_0^{\pi}x^2\,dx=\frac{2}{\pi}\left[\frac{x^3}{3}\right]_0^{\pi}=\frac{2}{\pi}\cdot\frac{\pi^3}{3}=\frac{\pi^2}{3}$。$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}x^2\cos(nx)\,dx=\frac{2}{\pi}\int_0^{\pi}x^2\cos(nx)\,dx$。使用分部積分兩次：$\intx^2\cos(nx)\,dx=\frac{x^2\sin(nx)}{n}-\int\frac{2x\sin(nx)}{n}\,dx=\frac{x^2\sin(nx)}{n}+\frac{2x\cos(nx)}{n^2}-\int\frac{2\cos(nx)}{n^2}\,dx=\frac{x^2\sin(nx)}{n}+\frac{2x\cos(nx)}{n^2}-\frac{2\sin(nx)}{n^3}$。計算定積分：$\int_0^{\pi}x^2\cos(nx)\,dx=\left[\frac{x^2\sin(nx)}{n}+\frac{2x\cos(nx)}{n^2}-\frac{2\sin(nx)}{n^3}\right]_0^{\pi}=\left(\frac{\pi^2\sin(n\pi)}{n}+\frac{2\pi\cos(n\pi)}{n^2}-\frac{2\sin(n\pi)}{n^3}\right)-\left(0+0-0\right)$。由于$\sin(n\pi)=0$,$\cos(n\pi)=(-1)^n$，所以$\int_0^{\pi}x^2\cos(nx)\,dx=\frac{2\pi(-1)^n}{n^2}$。$a_n=\frac{2}{\pi}\cdot\frac{2\pi(-1)^n}{n^2}=\frac{4(-1)^n}{n^2}$。$b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}x^2\sin(nx)\,dx=0$（因為$x^2\sin(nx)$是奇函數(shù)）。因此，傅里葉級數(shù)為$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos(nx)$。$f(x)=\frac{\pi^2}{6}+\sum_{n=1}^{\infty}\frac{4(-1)^n}{n^2}\cos(nx)$。在$x=\pm\pi$處，$f(x)=\pi^2$。級數(shù)收斂于$\frac{f(\pi^-)+f(-\pi^+)}{2}=\frac{\pi^2+\pi^2}{2}=\frac{\pi^2}{2}$。九、方程組有解。通解為$\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}+k_1\begin{pmatrix}1\\-1\\1\end{pmatrix}+k_2\begin{pmatrix}1\\-1\\0\end{pmatrix}$，其中$k_1,k_2$為任意常數(shù)。解析：使用增廣矩陣方法：$\left(\begin{array}{ccc|c}1&2&-1&1\\2&1&1&3\\-1&1&2&-1\end{array}\right)$對增廣矩陣進行行變換化為行階梯形：$R_2\leftarrowR_2-2R_1\rightarrow\left(\begin{array}{ccc|c}1&2&-1&1\\0&-3&3&1\\-1&1&2&-1\end{array}\right)$$R_3\leftarrowR_3+R_1\rightarrow\left(\begin{array}{ccc|c}1&2&-1&1\\0&-3&3&1\\0&3&1&0\end{array}\right)$$R_3\leftarrowR_3+R_2\rightarrow\left(\begin{array}{ccc|c}1&2&-1&1\\0&-3&3&1\\0&0&4&1\end{array}\right)$$R_3\leftarrow\frac{1}{4}R_3\rightarrow\left(\begin{array}{ccc|c}1&2&-1&1\\0&-3&3&1\\0&0&1&\frac{1}{4}\end{array}\right)$$R_2\leftarrowR_2-3R_3\rightarrow\left(\begin{array}{ccc|c}1&2&-1&1\\0&-3&0&\frac{1}{4}\\0&0&1&\frac{1}{4}\end{array}\right)$$R_1\leftarrowR_1+R_3\rightarrow\left(\begin{array}{cc