system of equations using Gauss Jordan method

Question

I'm trying to discuss a system of equations that depends on a parameter $k$ using the method Gauss-Jordan but the more I calculate the more entangled it becomes. While the solution in the book seems so easy and simple.

\begin{equation} \begin{cases} kx-y-z+3t=0\\2x-kz+2t=k+4\\kx+y+t=-k\\x+y-z=2 \end{cases} \end{equation} My attempt is:

\begin{vmatrix} k & -1& -1& 3& 0 \\2 & 0& -k& 2& k+4 \\k & 1& 0& 1& -k \\ 1 & 1 & -1& 0& 2\end{vmatrix} $ \longrightarrow$ changing the order of first and fourth row \begin{vmatrix} 1 & 1& -1& 0& 2 \\2 & 0& -k& 2& k+4 \\k & 1& 0& 1& -k \\ k & -1 & -1& 3& 0\end{vmatrix}

$\xrightarrow{R_2 \to -2R_1+R_2,\hspace{2mm}R_3 \to -kR_1+R_3,\hspace{2mm}R_4 \to -kR_1+R_4}$

\begin{vmatrix} 1 & 1& -1& 0& 2 \\0 & -2& 2-k& 2& k \\0 & 1-k& k& 1& -3k \\ 0 & -1-k & -1+k& 3& -2k\end{vmatrix}

$\xrightarrow{R_2 \to R_2*(-1/2)}$

\begin{vmatrix} 1 & 1& -1& 0& 2 \\0 & 1& (k-2)/2& -1& -k/2 \\0 & 1-k& k& 1& -3k \\ 0 & -1-k & -1+k& 3& -2k\end{vmatrix}

$\xrightarrow{R_1 \to -R_2+R_1,\hspace{2mm}R_3 \to -(1-k)R_2+R_3,\hspace{2mm}R_4 \to -(-1-k)R_2+R_4}$

\begin{vmatrix} 1 & 0& -k/2& 1& (k+4)/2 \\0 & 1& (k-2)/2& -1& -k/2 \\0 & 0& (k^2-k+2)/2& -k& (-k^2-5k)/2 \\ 0 & 0 & (k^2+k-4)/2& 2-k& (-k^2-5k)/2\end{vmatrix}

$\xrightarrow{R_3 \to R_3*(2/(k^2-k+2)),\hspace{2mm}R_4 \to R_4*(2/(k^2+k-4))}$

\begin{vmatrix} 1 & 0& -k/2& 1& (k+4)/2 \\0 & 1& (k-2)/2& -1& -k/2 \\0 & 0& 1& -2k/(k^2-k+2)& (-k^2-5k)/(k^2-k+2) \\ 0 & 0 & 1& 2(2-k)/(k^2+k-4)& (-k^2-5k)/(k^2+k-4)\end{vmatrix}

$\xrightarrow{R_1 \to R_3*(k/2)+R_1,\hspace{2mm}R_2 \to -(k-2/2)R_3,\hspace{2mm}R_4 \to -R_3+R_4}$

\begin{vmatrix} 1 & 0& 0& -k+2/k^2-k+2& -(k^2-k-1)/(k^2-k+2) \\0 & 1& 0& -k-2/k^2-k+2& (2k^2-6k)/(k^2-k+2) \\0 & 0& 1& -2k/(k^2-k+2)& (-k^2-5k)/(k^2-k+2) \\ 0 & 0 & 0& 8(k-1)^2/(k^2-k+2)(k^2+k-4)& (2k^3+4k^2-30k)/(k^2-k+2)(k^2+k-4)\end{vmatrix}

$\xrightarrow{R_4 \to R_48(k-1)^2/(k^2-k+2)(k^2+k-4)}$

\begin{vmatrix} 1 & 0& 0& -k+2/k^2-k+2& -(k^2-k-1)/(k^2-k+2) \\0 & 1& 0& -k-2/k^2-k+2& (2k^2-6k)/(k^2-k+2) \\0 & 0& 1& -2k/(k^2-k+2)& (-k^2-5k)/(k^2-k+2) \\ 0 & 0 & 0& 1& 2k(k^2+2k-15)/8(k-1)^2 \end{vmatrix}

$\xrightarrow{R_1 \to R_4*(k-2/k^2-k+2)+R_1,\hspace{2mm}R_2 \to R_4*(k+2/k^2-k+2)+R_2,\hspace{2mm}R_3 \to R_4*(2k/k^2-k+2)+R_3}$

\begin{vmatrix} 1 & 0& 0& 0& (-6k^4+8*k^3-22k^2+36k+8)/8(k-1)^2*(k^2-k+2) \\0 & 1& 0& 0& (-14k^4+88*k^3-134k^2-12k)/8(k-1)^2*(k^2-k+2) \\0 & 0& 1& 0& (-4k^4-16*k^3+12k^2-40k)/8(k-1)^2*(k^2-k+2) \\ 0 & 0 & 0& 1& 2k(k^2+2k-15)/8(k-1)^2 \end{vmatrix}

Answer 1

Here is one solution (did it two other ways also - one using Cramer's Rule and one using a different set of steps for Gaussian Elimination than is shown below). $R_x$ means Row - $x$ and these are the row steps in order.

• $R_3 = R_3 - R_1$
• $R_1 = \dfrac{R_1}{k}$
• $R_2 = R_2 - 2 R_1$
• $R_4 = R_4 - R_1$
• $R_2 = \dfrac{k}{2} R_2$, the matrix after this step is

$$\begin{bmatrix} 1 & -\dfrac{1}{k} & -\dfrac{1}{k} & \dfrac{3}{k} & 0 \\ 0 & 1 & -\dfrac{k^2}{2} + 1 & k - 3 & \dfrac{k}{2} (k + 4) \\ 0 & 2 & 1 & -2 & -k \\ 0 & \dfrac{1}{k}(k + 1) & \dfrac{1}{k} (-k + 1) & -\dfrac{3}{k} & 2 \end{bmatrix}$$

• $R_1 = R_1 - \left(-\dfrac{1}{k} R_2\right)$
• $R_3 = R_3 - 2 R_2$
• $R_4 = R_4 - \left(\dfrac{1}{k}(k+1)\right)R_2$
• $R_3 = \left(\dfrac{1}{k^2-1}\right)R_3$
• $R_1 = R_1 - \left(-\dfrac{k}{2}\right)R_3$, the matrix after this step is

$$\begin{bmatrix} 1 & 0 & 0 & \dfrac{2k-1}{k^2-1} &-\dfrac{k^2+k+4}{2k^2 -2} \\ 0 & 1 & -\dfrac{k^2}{2} + 1 & k - 3 & \dfrac{k}{2} (k + 4) \\ 0 & 0 & 1 & \dfrac{2(-k+2)}{k^2-1)} & -\dfrac{k(k+5)}{k^2-1} \\ 0 & 0 & \dfrac{k^2}{2} + \dfrac{k}{2}-2 & -k+2 & -\dfrac{k}{2}(k+5) \end{bmatrix}$$

• $R_2 = R_2 -\left(-\dfrac{k^2}{2} + 1 \right)R_3$
• $R_4 = R_4 -\left(\dfrac{k^2}{2} + \dfrac{k}{2} - 2 \right)R_3$
• $R_4 = \left(\dfrac{k^2-1}{k^2-5k+6}\right)R_4$
• $R_1 = R_1 -\left(\dfrac{2k-1}{k^2-1}\right)R_4$, the matrix after this step is

$$\begin{bmatrix} 1 & 0 & 0 & 0 &-\dfrac{3k+8}{2(k-4)} \\ 0 & 1 & 0 & -\dfrac{k^2+k-1}{k^2-1} & \dfrac{k(-k^2+k+6)}{2k^2-2}\\ 0 & 0 & 1 & \dfrac{2(-k+2)}{k^2-1)} & -\dfrac{k(k+5)}{k^2-1} \\ 0 & 0 & 0 & 1 & \dfrac{k(k+5)}{2(k-2)} \end{bmatrix}$$

• $R_2 = R_2 -\left(\dfrac{-k^2+k-1}{k^2-1}\right)R_4$, the matrix after this step is

$$\begin{bmatrix} 1 & 0 & 0 & 0 &-\dfrac{3k+8}{2(k-4)} \\ 0 & 1 & 0 & 0 & \dfrac{7k}{2k^2-2}\\ 0 & 0 & 1 & \dfrac{2(-k+2)}{k^2-1)} & -\dfrac{k(k+5)}{k^2-1} \\ 0 & 0 & 0 & 1 & \dfrac{k(k+5)}{2(k-2)} \end{bmatrix}$$

• $R_3 = R_3 - \left(\dfrac{2(-k+2)}{k^2-1}\right)R_4$, , the matrix after this step is the final result and is

$$\begin{bmatrix} 1 & 0 & 0 & 0 & -\dfrac{3 k+8}{2 (k-2)} \\ 0 & 1 & 0 & 0 & \dfrac{7 k}{2 (k-2)} \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & \dfrac{k(k+5)}{2 (k-2)} \\ \end{bmatrix}$$

Answer 2

$ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} $ $\xrightarrow{R_3 \to aR_3-gR_1,\hspace{2mm}R_2 \to aR_2-dR_1}\begin{bmatrix} a & b & c \\ 0 & e' & f' \\ 0 & h' & i' \end{bmatrix} $

$ \begin{bmatrix} a & b & c \\ 0 & e' & f' \\ 0 & h' & i' \end{bmatrix} $ $\xrightarrow{R_3 \to e'R_3-h'R_1}\begin{bmatrix} a & b & c \\ 0 & e' & f' \\ 0 & 0 & i'' \end{bmatrix}$

$\begin{bmatrix} a & b & c \\ 0 & e' & f' \\ 0 & 0 & i'' \end{bmatrix}$ $\xrightarrow{R_2 \to i''R_2-f'R_3,\hspace{2mm}R_1 \to i''R_1-cR_3}$ $\begin{bmatrix} a' & b' & 0 \\ 0 & e'' & 0 \\ 0 & 0 & i'' \end{bmatrix}$

$ \begin{bmatrix} a' & b' & 0 \\ 0 & e'' & 0 \\ 0 & 0 & i'' \end{bmatrix} $ $\xrightarrow{R_1 \to e''R_1-b'R_2}\begin{bmatrix} a'' & 0 & 0 \\ 0 & e'' & 0 \\ 0 & 0 & i'' \end{bmatrix}$

Now, using these Row operations, you can get a 3*3 matrix into the required from. Try forming the operations for 4*4 and see if it helps.