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Problem 15: For which values of k will the following system have i. No solution ? ii. Exactly one solution ? iii. Infinitely many solutions ?

Problem 15: For which values of k will the following system have i. No solution ? ii. Exactly one solution ? iii. Infinitely many solutions ?

\begin{align*} 4x+y+(k^2-14)z=k+2\\ x+2y-3z=4\\ 3x-y+5z=2 \end{align*}

Solution:

Consider the augmented matrix of this system and apply row operations.

\[\left[\begin{array}{c|c} A & B \end{array} \right] = \begin{align*} \left[\begin{array}{rrr|r} 4 & 1 & k^2-14& k+2 \\ 1 & 2 & -3& 4 \\ 3 & -1 & 5& 2 \\ \end{array}\right] \xrightarrow{R1\leftrightarrow R2} \left[\begin{array}{rrr|r} 1 & 2 & -3& 4 \\ 4 & 1 & k^2-14& k+2 \\ 3 & -1 & 5& 2 \\ \end{array}\right] \end{align*}\] \[\xrightarrow{R2\leftrightarrow R3} \left[\begin{array}{rrr|r} 1 & 2 & -3& 4 \\ 3 & -1 & 5& 2 \\ 4 & 1 & k^2-14& k+2 \\ \end{array}\right] \xrightarrow{R2\rightarrow R_2-3R_1,R3\rightarrow R_3-4R_1 } \left[\begin{array}{rrr|r} 1 & 2 & -3& 4 \\ 0 & -7 & 14& -10 \\ 0 & -7 & k^2-2& k-14 \\ \end{array}\right]\] \[\xrightarrow{R2\rightarrow -\frac{1}{7}R_2} \left[\begin{array}{rrr|r} 1 & 2 & -3& 4 \\ 0 & 1 & -2& \frac{10}{7} \\ 0 & -7 & k^2-2& k-14 \\ \end{array}\right]\] \[\xrightarrow{R3\rightarrow R_3+7R_2} \left[\begin{array}{rrr|r} 1 & 2 & -3& 4 \\ 0 & 1 & -2& \frac{10}{7} \\ 0 & 0 & k^2-16& k-4 \\ \end{array}\right]\]

This matrix is in row-echelon form. There are three cases according to $k^2 – 16$ is zero or not.

Case i: If $k = 4$, the last row in REF consists entirely of zeros, that is, \begin{align*} \left[\begin{array}{rrr|r} 1 & 2 & -3& 4 \\ 0 & 1 & -2& \frac{10}{7} \\ 0 & 0 & 0& 0 \\ \end{array}\right] \end{align*}

we write down the corresponding system as \begin{align*} x+2y-3z=4\\ y-2z=\frac{10}{7} \end{align*} This shows that in case $k=4$, the system has infinitely many solutions

Case ii: If $k = -4$, \begin{align*} \left[\begin{array}{rrr|r} 1 & 2 & -3& 4 \\ 0 & 1 & -2& \frac{10}{7} \\ 0 & 0 & 0& 8 \\ \end{array}\right] \end{align*}

0 = 8 which is impossible. Hence in this case the system has no solution This shows that in case $k=-4$, the system has no solution

Case iii: If $k^2-16 \ne 0$; that is, if $k\ne \pm 4$, we can divide third row of the matrix by $k^2-16$ \begin{align*} \left[\begin{array}{rrr|r} 1 & 2 & -3& 4 \\ 0 & 1 & -2& \frac{10}{7} \\ 0 & 0 & 1& \frac{k-4}{k^2-16} \\ \end{array}\right] \left[\begin{array}{rrr|r} 1 & 2 & -3& 4 \\ 0 & 1 & -2& \frac{10}{7} \\ 0 & 0 & 1& \frac{1}{k+4} \\ \end{array}\right] \end{align*}

Then, the corresponding system of equations for this matrix is \begin{align*} x+2y-3z=4\\ y-2z=\frac{10}{7}\\ z=\frac{1}{k+4} \end{align*} in this case the system has only one solution. This shows that in case $k^2-16 \ne 0$, the system has only one solution

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