Answer: (i) The system has no unique solution when the determinant of the coefficient matrix is zero. This gives \(k=8\).
(ii) When \(k=8\), the solutions are \(\left(\begin{array}{l}x\\y\\z\end{array}\right)=\left(\begin{array}{l}0\\0\\\frac12\end{array}\right)+t\left(\begin{array}{c}1\\-3\\1\end{array}\right)\), where \(t\) is a scalar.
(i) The coefficient matrix is \(\begin{pmatrix}1&3&k\\4&-2&-10\\1&1&2\end{pmatrix}\). A unique solution exists only when its determinant is non-zero, so we set
\(\det\begin{pmatrix}1&3&k\\4&-2&-10\\1&1&2\end{pmatrix}=0\).
Expanding by row operations or directly calculating,
\(\det = 1\begin{vmatrix}-2&-10\\1&2\end{vmatrix}-3\begin{vmatrix}4&-10\\1&2\end{vmatrix}+k\begin{vmatrix}4&-2\\1&1\end{vmatrix}\)
\(=1((-2)(2)-(-10)(1)) -3(4\cdot 2-(-10)(1)) + k(4\cdot 1-(-2)(1))\)
\(=( -4+10 ) -3(8+10) + k(6)=6-54+6k=6k-48.\)
So \(6k-48=0\), hence \(k=8\).
(ii) Substitute \(k=8\) into the equations:
\(x+3y+8z=4\)
\(4x-2y-10z=-5\)
\(x+y+2z=1\)
Now eliminate \(x\) using the first and third equations. Subtract the third equation from the first:
\((x+3y+8z)-(x+y+2z)=4-1\)
\(2y+6z=3\)
\(y+3z=\frac32\).
Next use the third equation to express \(x\) in terms of \(y\) and \(z\):
\(x=1-y-2z\).
Since the system has infinitely many solutions, let \(z=t\). Then from \(y+3z=\frac32\),
\(y=\frac32-3t\).
Substitute into \(x=1-y-2z\):
\(x=1-\left(\frac32-3t\right)-2t=-\frac12+t\).
Therefore
\(\left(\begin{array}{l}x\\y\\z\end{array}\right)=\left(\begin{array}{c}-\frac12\\\frac32\\0\end{array}\right)+t\left(\begin{array}{c}1\\-3\\1\end{array}\right).\)
This is equivalent to \(\left(\begin{array}{l}x\\y\\z\end{array}\right)=\left(\begin{array}{c}0\\0\\\frac12\end{array}\right)+t\left(\begin{array}{c}1\\-3\\1\end{array}\right)\).
