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If $f(x)$ = det($xI-A$), what is the determinant of the matrix $aA-bI$. I need to express my answer in terms of $f(x), a$ and $b$.

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    Do you know the size of the matrix $A$? – 2017-02-23
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    @B.Mehta it is just an $n$x$n$ matrix – 2017-02-23

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$$\begin{align}\det(aA-bI) &= \det\left(a\left(A-\frac ba I\right)\right) \\ &= (-a)^n\det\left(\frac ba I-A\right) \\ &= (-a)^n f\left(\frac ba\right)\end{align}$$

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    Thanks! This makes a lot of sense, for some reason I was trying to figure out $det (xI-(aA-bI))$, which was quite bothersome but I believe I was doing the wrong question because they weren't all that explicit in their question. – 2017-02-23
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    Note you can rewrite that as $$ \det(xI - (aA - bI)) = \det((x+b)I - aA) $$ – 2017-02-23
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    The new characteristic polynomial will be $a^n f(\frac{x + b}{a})$ except in the case of $a = 0$, where we get $f(x+b)$. – 2017-02-23