Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
up vote 3 down vote accepted

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.

share|cite|improve this answer
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

$$\det(a\cdot M^k\cdot(I-M))=a^{\mathrm{size}(M)}\cdot[\det(M)]^k\cdot\det(I-M)$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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