# What is the correct way to solve $|2K^3-2K^4|$ determinant?

Given - $$K_{3\times3} = \begin{bmatrix} 1&1&1 \\ 3&2&1 \\ 1&2&1 \end{bmatrix}$$ $$|K| = 2$$ Find - $$|2K^3-2K^4|$$

I tried this:

Since $|A+B|=|A|+|B|$ ( $\Leftarrow$ This is the main mistake ) - $$|2K^3-2K^4|=|2K^3+(-2K^4)|=|2K^3|+|(-2K^4)|$$

Now using $|\alpha A_{n\times n}|=\alpha ^n|A|$ - $$=2^3|K^3|+(-2^4)(K^4)|=8*8+(-16)*16=-192$$

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But $\det(A+B)$ usually doesn't equal $\det A+\det B$, as any example will show you. –  Gerry Myerson Jul 30 '12 at 8:44
@GerryMyerson: In which specific cases it does equal? –  MichaelS Jul 30 '12 at 8:48
MichaelS: Do your own work! Try almost any non diagonal A and B and check. –  Did Jul 30 '12 at 8:51

Now that you've edited $K$ into the question, we can get somewhere!
Use $2K^3-2K^4=(2)(K^3)(I-K)$, and $\det cA=c^nA$, and $\det A^r=(\det A)^r$, and then you just have to calculate $\det(I-K)$ directly.
Why couldn't we do the same trick without knowing how K is build? We know K is invertible ($det(K) \neq 0$) and we know that its size is $3\times 3$. Isn't it enough? –  MichaelS Jul 30 '12 at 9:07
It's enough to do everything except calculate $\det(I-K)$. Just knowing $K$ is $3\times3$ and has determinant 2 is not enough to pin down $\det(I-K)$. See for yourself! Construct some examples! –  Gerry Myerson Jul 30 '12 at 12:43
Of course.. Totally forgot about $I-K$. Thanks! –  MichaelS Jul 30 '12 at 14:53
$$\det(a\cdot M^k\cdot(I-M))=a^{\mathrm{size}(M)}\cdot[\det(M)]^k\cdot\det(I-M)$$