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I was hoping someone could explain what exactly the following question is asking:

Find for which value $\ k _0 $ of $k$ is the following matrix invertible

A = $(\begin{matrix} 1 & k \\ 3 & 2 \end{matrix})$

Afterwards solve the system $A( \begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) $ for $k \ne k_0$.

I understand the first part, I found that A is invertible when $k \ne$ $ -3 \above 1pt 2 $

But I dont understand what the second part of the problem wants me to do. Am i supposed to multiply A by the matrix $(\begin{matrix} x \\ y \end{matrix})$ and then put it into an augmented matrix? I'm sorry if this seems like a basic question, but I am just really confused on how to answer this.

If someone could give be a push in the right direction I would really appreciate it. I just ask that you don't give me the answer. I am hoping to try and solve the rest by myself once I know what I need to do.

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    Welcome to Math SX! If $A$ is invertible, the system has only the trivial solution. – 2017-02-26
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    you have to solve this System $$x+yk=0$$ and $$3x+2y=0$$ – 2017-02-26
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    @Dr.SonnhardGraubner Did you get that system simply by multiplying A with $(\begin{matrix} x \\ y \end{matrix})$ ? – 2017-02-26
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    yes you must multiply $A$ with $$\binom {x}{y}$$ – 2017-02-26

2 Answers 2

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Hint:

$$ \begin{pmatrix} 1&k\\3&2 \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}= \begin{pmatrix} 0\\0 \end{pmatrix} $$ means: $$ \begin{pmatrix} x+ky\\3x+2y \end{pmatrix}= \begin{pmatrix} 0\\0 \end{pmatrix} $$

and this is equivalent to the system: $$ \begin{cases} x+ky=0\\ 3x+2y=0 \end{cases} $$

can you solve this?

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First of all: your matrix $A$ is invertible if and only if $k \neq \frac{2}{3}$ (perhaps you have made a computational mistake).

The second question asks you to solve the system $A\vec{x} = \vec{0}$, where $\vec{x} = (x,y)^t$ and $\vec{0} = (0,0)^t$ where you have to fill in $k = \frac{2}{3}$. Since this makes the matrix $A$ singular (i.e. non-invertible), you should find infinitely many solutions. (Note that you can fill in $k$, so that you can write down your solutions).

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    Thanks for your reply. You were right about the computational error. Though, I'm a bit confused why i would fill in k = $2 \above 1pt 3 $ because the question asks ' for $k \ne k_0$' – 2017-02-26
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    No problem. The first question shows that $A$ is invertible for every $k_0 \neq \frac{2}{3}$, so $A$ is invertible for every $k_0 \in ]-\infty, \frac{2}{3}[ \cup ]\frac{2}{3}, + \infty[$. The second question asks you to compute the solutions to the system for every $k \neq k_0$, hence only for $k = \frac{2}{3}$. – 2017-02-26