Determine values of k for a matrix to have a unique solution

Question

I have the following system and need to find for what values of $k$ does the system have

i) a unique solution
ii) no solution
iii) an infinite number of solutions

$(k^3+3k)x + (k+5)y + (k+3)z = k^5+(k+3)^2 ky + z = 3 (k^3+k^2-6k)z = k(k^2-9)$

Putting this into a matrix I get

$M =\left[ \begin{array}{cc} k^3 + 3k & k+5 & k+3 & k^5+(k+3)^2 \\ 0 & k & 1 & 3 \\ 0 & 0 & k^3+k^2-6k & k(k^2-9)\\ \end{array}\right]$

I understand for ii) we need the bottom row to read $0 \ 0 \ 0 \ k(k^2-9)$ which it does for values $k=2$.

For iii) we need the bottom row to read $0 \ 0 \ 0 \ 0$ which we do when $k=0$ and $k=-3$.

Because the first column only contains a value for the top row, I can't use elementary row operations to chance the value in $a_{11}$, so I don't know how I can make it into the identity matrix?

Answer 1

$\textbf{HINT:}$ Check the determinant. if $\det=0$ you will have infinite solutions, also you can check the last file, if $a_{32}=0$, then check $a_{34}:$

• if $a_{34}\neq0$ then there are no solutions.
• if $a_{34}=0$ there are infinite solutions.

Answer 2

Coefficient matrix is given by $A = \begin{pmatrix} k^3+3k & k+5 & k+3\\ 0 & k & 1 \\ 0 & 0 & k^3+k^2-6k \end{pmatrix}$

And, augmented matrix is given by $[A:b] = \begin{pmatrix} k^3+3k & k+5 & k+3 & k^5+(k+3)^2\\ 0 & k & 1 & 3\\ 0 & 0 & k^3+k^2-6k & k(k^2-9) \end{pmatrix}$

(1) UNIQUE SOLUTION, if $rank(A) = rank([A:b]) =$ number of variables (here, number of variables is $3$). Find out the values of $k$ for which $det(A) \neq 0$. Values of $k$ are = $\mathbb{C}-\{0, \pm\sqrt{3} i, 2, -3\}$, where $\mathbb{C}$ denotes the set of complex numbers.

(2) NO SOLUTION, if $rank(A) \neq rank([A:b])$. Check that, for $k = \pm\sqrt{3} i, 2$, $rank(A) = 2$ but $rank([A:b]) = 3$.

(3) INFINITELY MANY SOLUTION, if $rank(A) = rank ([A:b]) <$ number of variables. Check that, for $k= 0, -3$, $rank(A) = rank ([A:b]) <$ number of variables.