0
$\begingroup$

This question already has an answer here:

I've been doing some old Linear Algebra papers that my university library had on file just for some practise and came across this question which caused many a debate among my friends.

True or False: There exists a Matrix A $\in \mathbb{M}_{33}(\mathbb{Z})$ with determinant 2 such that:

$$ A \left[ {\begin{array}{cc} 2 \\ 1 \\ 4 \end{array} } \right] = \left[ {\begin{array}{cc} 4 \\ -8 \\ 16 \end{array} } \right]$$

I started by writing out each of the $[a]_{ij}$ entries and multiplying it out to look for a contradiction but that didn't get me very far. I then thought maybe since the determinant is 2 it may have something to do with the fact that $A^{-1}$ can be written as $\frac{1}{det(A)}B$ for some matrix B but a concrete answer still eludes me. A pointer in the right direction would be great!

Thanks!

$\endgroup$

marked as duplicate by user1551 linear-algebra Jun 23 '17 at 9:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3
$\begingroup$

Use Cramer's Rule: if $A$ is an $n\times n$ matrix and $\det(A)\ne0$ then the solution of $$A\pmatrix{x_1\cr\vdots\cr x_n\cr}={\bf b}$$ is given by $$x_k=\frac{\det{A_k}}{\det A}\ ,$$ where $A_k$ is $A$ with the $k$th column replaced by $\bf b$. If $A$ exists under the conditions of your problem we have $$1=x_2=\frac{\det A_2}{\det A}\ .$$ But $A_2$ is an integer matrix in which every entry of the second column is a multiple of $4$, so $\det A_2$ is a multiple of $4$; and $\det A=2$; so $x_2$ is a multiple of $2$. This is a contradiction.

$\endgroup$
1
$\begingroup$

define $$ B = \left( \begin{array}{rr} 2 & -1 & 0 \\ 1 & 0 & 0 \\ 4 & 0 & 1 \end{array} \right) $$

Then $\det B = 1,$ so $\det AB = 2.$ However, from the left hand column, we must have $\det AB$ divisible by $4.$

Contradiction

$$ AB = \left( \begin{array}{rr} 4 & ? & ? \\ -8 & ? & ? \\ 16 & ? & ? \end{array} \right) $$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.