$\mathrm{with}\left(\mathrm{Student}:-\mathrm{Basics}\right)\:$

$\mathrm{LongDivision}\left(48x^4+284x^3+620x^2+593x+210\,2x+3\right)$

$\begin{array}{cc}\stackrel{\phantom{\frac{x^2}{2}}}{2x+3}& \begin{array}{cccccc}& 24x^3& \+106x^2& \+151x& \+70& \\ \)\phantom{x^2}& \phantom{1}48x^4& \phantom{1}\+284x^3& \phantom{1}\+620x^2& \phantom{1}\+593x& \phantom{1}\+210\\ & \multicolumn{2}{c}{\frac{48x^4+72x^3}{\phantom{\.}}}& \\ & & \multicolumn{2}{c}{212x^3+620x^2}& & \\ & & \multicolumn{2}{c}{\frac{212x^3+318x^2}{\phantom{\.}}}& & \\ & & & \multicolumn{2}{c}{302x^2+593x}& & & \\ & & & \multicolumn{2}{c}{\frac{302x^2+453x}{\phantom{\.}}}& & & \\ & & & & \multicolumn{2}{c}{140x+210}& & & & \\ & & & & \multicolumn{2}{c}{\frac{140x+210}{\phantom{\.}}}& & & & \\ & & & & & 0\hfill & & & & & \end{array}\end{array}$

$\mathrm{LongDivision}\left(1001\,30\, \'\mathrm{decimaldigits}\'\=4\right)\;$

$\begin{array}{ccccccccccccccc}& & & & \underset{\—}{ }& \underset{\—}{ }& \underset{\—}{ }& \underset{\—}{ }& \underset{\—}{3}& \underset{\—}{3}& \underset{\—}{\.}& \underset{\—}{3}& \underset{\—}{6}& \underset{\—}{6}& \underset{\—}{6}\\ & 3& 0& & \)& & 1& 0& 0& 1& \.& 0& 0& 0& 0\\ & & & & & & & \underset{\—}{9}& \underset{\—}{0}\\ & & & & & & & 1& 0& 1\\ & & & & & & & & \underset{\—}{9}& \underset{\—}{0}\\ & & & & & & & & 1& 1& 0\\ & & & & & & & & & \underset{\—}{9}& \underset{\—}{0}\\ & & & & & & & & & 2& 0& 0\\ & & & & & & & & & \underset{\—}{1}& \underset{\—}{8}& \underset{\—}{0}\\ & & & & & & & & & & 2& 0& 0\\ & & & & & & & & & & \underset{\—}{1}& \underset{\—}{8}& \underset{\—}{0}\\ & & & & & & & & & & & 2& 0& 0\\ & & & & & & & & & & & \underset{\—}{1}& \underset{\—}{8}& \underset{\—}{0}\\ & & & & & & & & & & & & 2& 0\end{array}$

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{Basics}\right)\:$

$\mathrm{FactorSteps}\left(x^3+6x^2+12x+8\right)$

$\begin{array}{lll}& & x^3+6\cdot x^2+12\cdot x+8\\ \text{▫}& & \text{1. Trial Evaluations}\\ & \text{◦}& \text{Rewrite in standard form}\\ & & x^3+6x^2+12x+8\\ & \text{◦}& \text{The factors of the constant coefficient}8\text{are:}\\ & & C=\left\{1\,2\,4\,8\right\}\\ & \text{◦}& \text{Trial evaluations of}x\text{in}\text{±}C\text{find}x\text{=}\mathrm{-2}\text{satisfies the equation, so}x+2\text{is a factor}\\ & & \genfrac{}{}{0ex}{}{\left(x^3+6x^2+12x+8\right)}{\phantom{x=\mathrm{-2}}}|\genfrac{}{}{0ex}{}{\phantom{\left(x^3+6x^2+12x+8\right)}}{x=\mathrm{-2}}=0\\ & \text{◦}& \text{Divide by}x+2\\ & & \begin{array}{cc}\stackrel{\phantom{\frac{x^2}{2}}}{x+2}& \begin{array}{ccccc}& x^2& \+4x& \+4& \\ \)\phantom{x^2}& \phantom{1}{x}^3& \phantom{1}\+6x^2& \phantom{1}\+12x& \phantom{1}\+8\\ & \multicolumn{2}{c}{\frac{x^3+2x^2}{\phantom{\.}}}& \\ & & \multicolumn{2}{c}{4x^2+12x}& & \\ & & \multicolumn{2}{c}{\frac{4x^2+8x}{\phantom{\.}}}& & \\ & & & \multicolumn{2}{c}{4x+8}& & & \\ & & & \multicolumn{2}{c}{\frac{4x+8}{\phantom{\.}}}& & & \\ & & & & 0\hfill & & & & \end{array}\end{array}\\ & \text{◦}& \text{Quotient times divisor from long division}\\ & & \left(x^2+4x+4\right)\cdot \left(x+2\right)\\ \text{•}& & \text{2. Examine term:}\\ & & x^2+4x+4\\ \text{▫}& & \text{3. Apply the AC Method}\\ & \text{◦}& \text{Examine quadratic}\\ & & \left(x^2+4x+4\right)\\ & \text{◦}& \text{Look at the coefficients,}A{x}^2+Bx+C\\ & & \left[''A''=1\,''B''=4\,''C''=4\right]\\ & \text{◦}& \text{Find factors of |AC| = |}1\cdot 4\text{| =}4\\ & & \left\{1\,2\,4\right\}\\ & \text{◦}& \text{Find pairs of the above factors, which, when multiplied equal}4\\ & & \left\{1\cdot 4\,2\cdot 2\right\}\\ & \text{◦}& \text{Which pairs of these factors have a}\text{sum}\text{of B =}4\text{? Found:}\\ & & 2+2=4\\ & \text{◦}& \text{Split the middle term to use above pair}\\ & & x^2+\left(2x+2x\right)+4\\ & \text{◦}& \text{Factor}x\text{out of the first pair}\\ & & \left(x\cdot \left(x+2\right)\right)+\left(2x+4\right)\\ & \text{◦}& \text{Factor}2\text{out of the second pair}\\ & & x\cdot \left(x+2\right)+\left(2\cdot \left(x+2\right)\right)\\ & \text{◦}& x+2\text{is a common factor}\\ & & x\cdot \left(x+2\right)+2\cdot \left(x+2\right)\\ & \text{◦}& \text{Group common factor}\\ & & \left(x+2\right)\cdot \left(x+2\right)\\ & & \text{This gives:}\\ & & \left(x+2\right)\cdot \left(x+2\right)\\ \text{•}& & \text{4. This gives:}\\ & & \left(x+2\right)\cdot \left(x+2\right)\cdot \left(x+2\right)\end{array}$

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{Basics}\right)\:$

$\mathrm{SolveSteps}\left(5{\ⅇ}^{4x}=16\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & 5\cdot {\ⅇ}^{4\cdot x}=16\\ \text{▫}& & \text{Convert from exponential equation}\\ & \text{◦}& \text{Divide both sides by}5\\ & & \frac{5\cdot {\ⅇ}^{4\cdot x}}{5}=\frac{16}{5}\\ & \text{◦}& \text{Simplify}\\ & & {\ⅇ}^{4\cdot x}=\frac{16}{5}\\ & \text{◦}& \text{Apply ln to each side}\\ & & \ln \left({\ⅇ}^{4x}\right)=\ln \left(\frac{16}{5}\right)\\ & \text{◦}& \text{Apply ln rule: ln(e^b) = b}\\ & & 4x=\ln \left(\frac{16}{5}\right)\\ \text{•}& & \text{Divide both sides by}4\\ & & \frac{4\cdot x}{4}=\frac{\ln \left(\frac{16}{5}\right)}{4}\\ \text{•}& & \text{Exact solution}\\ & & x=\frac{\ln \left(\frac{16}{5}\right)}{4}\\ \text{•}& & \text{Approximate solution}\\ & & x=0.2907877025\end{array}$

$\mathrm{SolveSteps}\left(\left[12x+y=18\,7x-8y=32\right]\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left[12\cdot x+y=18\,7\cdot x-8\cdot y=32\right]\\ \text{•}& & \text{Pick the 2nd}\text{equation to solve for}y\\ & & 7\cdot x-8\cdot y=32\\ \text{▫}& & \text{A}\text{: isolate for}y\\ & \text{◦}& \text{Subtract}7\cdot x\text{from both sides}\\ & & 7\cdot x-8\cdot y-7\cdot x=32-7\cdot x\\ & \text{◦}& \text{Simplify}\\ & & \left(\mathrm{-8}\right)\cdot y=32-7\cdot x\\ & \text{◦}& \text{Divide both sides by}\mathrm{-8}\\ & & \frac{\left(\mathrm{-8}\right)\cdot y}{\mathrm{-8}}=\frac{32-7\cdot x}{\mathrm{-8}}\\ & \text{◦}& \text{Simplify}\\ & & y=\mathrm{-4}+\frac{7x}{8}\\ & \text{◦}& \text{Solution}\\ & & y=-4+\frac{7x}{8}\\ \text{•}& & \text{Substitute the value of}y=-4+\frac{7x}{8}\text{into the 1st}\text{equation of the system}\\ & & 12\cdot x+\left(-4+\frac{7x}{8}\right)=18\\ \text{▫}& & \text{Solve for}x\\ & \text{◦}& \text{Evaluate subtraction and addition}\\ & & \frac{103x}{8}-4=18\\ & \text{◦}& \text{Add}4\text{to both sides}\\ & & \frac{103}{8}\cdot x-4+4=18+4\\ & \text{◦}& \text{Simplify}\\ & & \frac{103}{8}\cdot x=22\\ & \text{◦}& \text{Divide both sides by}\frac{103}{8}\\ & & \frac{x\cdot \left(\frac{103}{8}\right)}{\frac{103}{8}}=\frac{22}{\frac{103}{8}}\\ & \text{◦}& \text{Simplify}\\ & & x=\frac{22}{\frac{103}{8}}\\ & \text{◦}& \text{Rewrite division as multiplication by reciprocal}\\ & & x=22\cdot \left(\frac{8}{103}\right)\\ & \text{◦}& \text{Multiply fraction and reduce by gcd}\\ & & x=\frac{176}{103}\\ \text{•}& & \text{Substitute}x=\frac{176}{103}\text{into equation}\text{A}\\ & & y=\mathrm{-4}+\frac{7}{8}\cdot \left(\frac{176}{103}\right)\\ \text{▫}& & \text{Solve for}y\\ & \text{◦}& \text{Evaluate multiplication and division}\\ & & y=\mathrm{-4}+\frac{154}{103}\\ & \text{◦}& \text{Evaluate subtraction and addition}\\ & & y=-\frac{258}{103}\\ \text{•}& & \text{Solution}\\ & & \left[x=\frac{176}{103}\,y=-\frac{258}{103}\right]\end{array}$

$\mathrm{with}\left( \mathrm{Student}:-\mathrm{Calculus1} \right)\:$

$\mathrm{ShowSolution}\left(\int \sin {\left(x\right)}^2\ⅆx\right)$

$\begin{array}{lll}& & \text{Integration Steps}\\ & & \int {\sin \left(x\right)}^2\ⅆx\\ \text{▫}& & \text{1. Rewrite}\\ & \text{◦}& \text{Equivalent expression}\\ & & {\sin \left(x\right)}^2=\frac{1}{2}-\frac{\cos \left(2x\right)}{2}\\ & & \text{This gives:}\\ & & \int \left(\frac{1}{2}-\frac{\cos \left(2x\right)}{2}\right)\ⅆx\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{sum}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{sum}}\text{rule}\\ & & \int \left(f\left(x\right)+g\left(x\right)\right)\ⅆx=\int f\left(x\right)\ⅆx+\int g\left(x\right)\ⅆx\\ & & f\left(x\right)=\frac{1}{2}\\ & & g\left(x\right)=-\frac{\cos \left(2x\right)}{2}\\ & & \text{This gives:}\\ & & \int \frac{1}{2}\ⅆx+\int -\frac{\cos \left(2x\right)}{2}\ⅆx\\ \text{▫}& & \text{3. Apply the}\mathbf{\text{constant}}\text{rule to the term}\int \frac{1}{2}\ⅆx\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant}}\text{rule}\\ & & \int C\ⅆx=Cx\\ & \text{◦}& \text{This means}\\ & & \int \frac{1}{2}\ⅆx=\frac{x}{2}\\ & & \text{We can now rewrite the integral as:}\\ & & \frac{x}{2}+\int -\frac{\cos \left(2x\right)}{2}\ⅆx\\ \text{▫}& & \text{4. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int -\frac{\cos \left(2x\right)}{2}\ⅆx\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int Cf\left(x\right)\ⅆx=C\left(\int f\left(x\right)\ⅆx\right)\\ & \text{◦}& \text{This means:}\\ & & \int -\frac{\cos \left(2x\right)}{2}\ⅆx=-\frac{\left(\int \cos \left(2x\right)\ⅆx\right)}{2}\\ & & \text{We can rewrite the integral as:}\\ & & \frac{x}{2}-\frac{\left(\int \cos \left(2x\right)\ⅆx\right)}{2}\\ \text{▫}& & \text{5. Apply a change of variables to rewrite the integral in terms of}u\\ & \text{◦}& \text{Let}u\text{be}\\ & & u=2x\\ & \text{◦}& \text{Isolate equation for}x\\ & & x=\frac{u}{2}\\ & \text{◦}& \text{Differentiate both sides}\\ & & \mathrm{dx}=\frac{\mathrm{du}}{2}\\ & \text{◦}& \text{Substitute the values for x and dx back into the original}\\ & & \int \cos \left(2x\right)\ⅆx=\int \frac{\cos \left(u\right)}{2}\ⅆu\\ & & \text{This gives:}\\ & & \frac{x}{2}-\frac{\left(\int \frac{\cos \left(u\right)}{2}\ⅆu\right)}{2}\\ \text{▫}& & \text{6. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int \frac{\cos \left(u\right)}{2}\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int Cf\left(u\right)\ⅆu=C\left(\int f\left(u\right)\ⅆu\right)\\ & \text{◦}& \text{This means:}\\ & & \int \frac{\cos \left(u\right)}{2}\ⅆu=\frac{\left(\int \cos \left(u\right)\ⅆu\right)}{2}\\ & & \text{We can rewrite the integral as:}\\ & & \frac{x}{2}-\frac{\left(\int \cos \left(u\right)\ⅆu\right)}{4}\\ \text{▫}& & \text{7. Evaluate the integral of}\cos \text{(}u\text{)}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{cos}\text{rule}\\ & & \int \cos \left(u\right)\ⅆu=\sin \left(u\right)\\ & & \text{This gives:}\\ & & \frac{x}{2}-\frac{\sin \left(u\right)}{4}\\ \text{▫}& & \text{8. Revert change of variable}\\ & \text{◦}& \text{Variable we defined in step}5\\ & & u=2x\\ & & \text{This gives:}\\ & & \frac{x}{2}-\frac{\sin \left(2x\right)}{4}\end{array}$

$\mathrm{ShowSolution}\left( \mathrm{Limit}\left( \frac{\sin \left(x\right)}{x}\, x\=0 \right) \right)$

$\mathrm{Diff}\left(x^2\cdot \sin \left(x\right)\, x\right)$

$\frac{\ⅆ}{\ⅆx}\left(x^2\sin \left(x\right)\right)$

$\mathrm{ShowSolution}\left( \right)$

$\mathrm{with}\left( \mathrm{Student}:-\mathrm{ODEs} \right)\:$

$\mathrm{ode1}≔t^2\left(z\left(t\right)+1\right)+{z\left(t\right)}^2\left(t-1\right)\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right)=0$

$\mathrm{ode1}≔t^2\left(z\left(t\right)+1\right)+{z\left(t\right)}^2\left(t-1\right)\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right)=0$

$\mathrm{ODESteps}\left( \mathrm{ode1} \right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & t^2\left(z\left(t\right)+1\right)+{z\left(t\right)}^2\left(t-1\right)\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right)=0\\ \text{•}& & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{\ⅆ}{\ⅆt}z\left(t\right)\\ \text{•}& & \text{Separate variables}\\ & & \frac{\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right){z\left(t\right)}^2}{z\left(t\right)+1}=-\frac{t^2}{t-1}\\ \text{•}& & \text{Integrate both sides with respect to}t\\ & & \int \frac{\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right){z\left(t\right)}^2}{z\left(t\right)+1}\ⅆt=\int -\frac{t^2}{t-1}\ⅆt+\mathrm{\_C1}\\ \text{•}& & \text{Evaluate integral}\\ & & \frac{{z\left(t\right)}^2}{2}-z\left(t\right)+\ln \left(z\left(t\right)+1\right)=-\frac{t^2}{2}-t-\ln \left(t-1\right)+\mathrm{\_C1}\end{array}$

$\mathrm{ivp2}≔\left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{\ⅆ}{\ⅆx}y\left(x\right)-x{\ⅇ}^x=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{x=0}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{x=0}=0\,y\left(0\right)=1\right\}$

$\mathrm{ivp2}≔\left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{\ⅆ}{\ⅆx}y\left(x\right)-x{\ⅇ}^x=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=0\right\}}=0\,y\left(0\right)=1\right\}$

$\mathrm{ODESteps}\left(\mathrm{ivp2} \right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{\ⅆ}{\ⅆx}y\left(x\right)-x{\ⅇ}^x=0\,\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=0\right\}}=0\,y\left(0\right)=1\right\}\\ \text{•}& & \text{Isolate 2nd derivative}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\frac{\ⅆ}{\ⅆx}y\left(x\right)+x{\ⅇ}^x\\ \text{•}& & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE, ODE is linear}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-\frac{\ⅆ}{\ⅆx}y\left(x\right)=x{\ⅇ}^x\\ \text{•}& & \text{Characteristic polynomial of homogeneous ODE}\\ & & r^2-r=0\\ \text{•}& & \text{Factor the characteristic polynomial}\\ & & r\left(r-1\right)=0\\ \text{•}& & \text{Roots of the characteristic polynomial}\\ & & r=\left(0\,1\right)\\ \text{•}& & \text{1st solution of the homogeneous ODE}\\ & & y_1\left(x\right)=1\\ \text{•}& & \text{2nd solution of the homogeneous ODE}\\ & & y_2\left(x\right)={\ⅇ}^x\\ \text{•}& & \text{General solution of the ODE}\\ & & y\left(x\right)=\mathrm{\_C1}{y}_1\left(x\right)+\mathrm{\_C2}{y}_2\left(x\right)+y_p\left(x\right)\\ \text{•}& & \text{Substitute in solutions of the homogeneous ODE}\\ & & y\left(x\right)=\mathrm{\_C1}+\mathrm{\_C2}{\ⅇ}^x+y_p\left(x\right)\\ \text{▫}& & \text{Find a particular solution}{y}_p\left(x\right)\text{of the ODE}\\ & \text{◦}& \text{Use variation of paramaters to find}{y}_p\text{here}f\left(x\right)\text{is the forcing function}\\ & & \left[y_p\left(x\right)=-y_1\left(x\right)\left(\int \frac{y_2\left(x\right)f\left(x\right)}{W\left(y_1\left(x\right)\,y_2\left(x\right)\right)}\ⅆx\right)+y_2\left(x\right)\left(\int \frac{y_1\left(x\right)f\left(x\right)}{W\left(y_1\left(x\right)\,y_2\left(x\right)\right)}\ⅆx\right)\,f\left(x\right)=x{\ⅇ}^x\right]\\ & \text{◦}& \text{Wronskian of solutions of the homogeneous equation}\\ & & W\left(y_1\left(x\right)\,y_2\left(x\right)\right)=\left[?\right]\\ & \text{◦}& \text{Compute Wronskian}\\ & & W\left(y_1\left(x\right)\,y_2\left(x\right)\right)={\ⅇ}^x\\ & \text{◦}& \text{Substitute functions into equation for}{y}_p\left(x\right)\\ & & y_p\left(x\right)=-\left(\int x{\ⅇ}^x\ⅆx\right)+{\ⅇ}^x\left(\int x\ⅆx\right)\\ & \text{◦}& \text{Compute integrals}\\ & & y_p\left(x\right)=\frac{{\ⅇ}^x\left(x^2-2x+2\right)}{2}\\ \text{•}& & \text{Substitute particular solution into general solution to ODE}\\ & & y\left(x\right)=\mathrm{\_C1}+\mathrm{\_C2}{\ⅇ}^x+\frac{{\ⅇ}^x\left(x^2-2x+2\right)}{2}\\ \text{•}& & \text{Use initial condition}y\left(0\right)=1\\ & & 1=\mathrm{\_C1}+\mathrm{\_C2}+1\\ \text{•}& & \text{Compute derivative of the solution}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\mathrm{\_C2}{\ⅇ}^x+\frac{{\ⅇ}^x\left(x^2-2x+2\right)}{2}+\frac{{\ⅇ}^x\left(2x-2\right)}{2}\\ \text{•}& & \text{Use the initial condition}\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{\phantom{\left\{x=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}}{\left\{x=0\right\}}=0\\ & & 0=\mathrm{\_C2}\\ \text{•}& & \text{Solve for}\mathrm{\_C1}\text{and}\mathrm{\_C2}\\ & & \left\{\mathrm{\_C1}=0\,\mathrm{\_C2}=0\right\}\\ \text{•}& & \text{Solution to the IVP}\\ & & y\left(x\right)=\frac{{\ⅇ}^x\left(x^2-2x+2\right)}{2}\end{array}$

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{LinearAlgebra}\right)\:$

$M ≔ \left[\begin{array}{ccc}3& -3& 7\\ 2& 4& 8\\ \mathrm{-2}& \mathrm{-4}& 6\end{array}\right]\:$

$\mathrm{InverseTutor}\left( M\, \mathrm{output}\=\mathrm{steps} \right)$

$\begin{array}{lll}& & \text{Compute the inverse of this matrix}\\ & & \left[?\right]\\ \text{•}& & \text{Matrix augmented with identity}\\ & & \left[?\right]\\ \text{•}& & \text{Multiply row 1 by 1/3}\\ & & \left[?\right]\\ \text{•}& & \text{Add -2 times row 1 to row 2}\\ & & \left[?\right]\\ \text{•}& & \text{Add 2 times row 1 to row 3}\\ & & \left[?\right]\\ \text{•}& & \text{Multiply row 2 by 1/6}\\ & & \left[?\right]\\ \text{•}& & \text{Add 1 times row 2 to row 1}\\ & & \left[?\right]\\ \text{•}& & \text{Add 6 times row 2 to row 3}\\ & & \left[?\right]\\ \text{•}& & \text{Multiply row 3 by 1/14}\\ & & \left[?\right]\\ \text{•}& & \text{Add -26/9 times row 3 to row 1}\\ & & \left[?\right]\\ \text{•}& & \text{Add -5/9 times row 3 to row 2}\\ & & \left[?\right]\\ \text{•}& & \text{We have found the inverse}\\ & & \left[?\right]\end{array}$

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{LinearAlgebra}\right)\:$

$M ≔ \left[\begin{array}{ccc}1& 2& 0\\ 2& 3& 2\\ 0& 2& 1\end{array}\right]\:$

$\mathrm{EigenvaluesTutor}\left( M\, \mathrm{output}\=\mathrm{steps} \right)$

$\begin{array}{lll}& & \text{Compute the Eigenvalues}\\ & & \left[?\right]\\ \text{•}& & \text{Calculate A=M-t*Id}\\ & & \left[?\right]\\ \text{•}& & \text{Find the determinant; this is also called the characteristic polynomial of M.}\\ & & -t^3+5t^2+t-5\\ \text{•}& & \text{Solve; the eigenvalues are the roots of the characteristic polynomial.}\\ & & \left[?\right]\end{array}$

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{LinearAlgebra}\right)\:$

$M ≔ \left[\begin{array}{ccc}1& 2& 0\\ 2& 3& 2\\ 0& 2& 1\end{array}\right]\:$

$\mathrm{EigenvectorsTutor}\left( M\, \mathrm{output}\=\mathrm{steps} \right)$