# Thread: if det(aM + K) = det(bM + K) where a, b are scalars and M, K are matrices, is a = b?

1. ## if det(aM + K) = det(bM + K) where a, b are scalars and M, K are matrices, is a = b?

hello,

i have a problem. if i know that

$
\textnormal{det}(a \mathbf{M} + \mathbf{K}) = 0
$

$
\textnormal{det}(b \mathbf{M} + \mathbf{K}) = 0
$

where $a$ and $b$ are scalars and $\mathbf{M}$ and $\mathbf{K}$ are matrices,
does this mean that $a = b$?

also, $\mathbf{M}$ is diagonal, if that helps.

cheers for reading and sorry if this is a dumb question,
jack.

2. Originally Posted by thepartymushroom
hello,

i have a problem. if i know that

$
\textnormal{det}(a \mathbf{M} + \mathbf{K}) = 0
$

$
\textnormal{det}(b \mathbf{M} + \mathbf{K}) = 0
$

where $a$ and $b$ are scalars and $\mathbf{M}$ and $\mathbf{K}$ are matrices,
does this mean that $a = b$?

also, $\mathbf{M}$ is diagonal, if that helps.

cheers for reading and sorry if this is a dumb question,
jack.
Look at the 2x2 case. Assuming a=1 then we need to solve the equation
(b-1)(n(p+bm+m)+sm) = 0 where p is the top left coordinate of the matrix K, s the bottom left. m, n are the coordinates of M. Assuming b is not 1 then we want to find n, m, b, p, s such that n(p+bm+m)+sm=0. So, let s=0=n.

That is, $M =
\left( \begin{array}{ccc}
m & 0 \\
0 & 0 \end{array} \right)$
and $K =
\left( \begin{array}{ccc}
p & q \\
r & 0 \end{array} \right)$

is a counter-example.

3. thanks, this helped get me thinking... what if the diagonal entries are positive? what if a and b are both negative? etc. seems like a bit of a long shot though.

thanks again,
jack.

p.s. i assume you meant "s is the bottom right".