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Find the determinant of the following matrix: $$\begin{bmatrix} 1& 1& 1& 1& 1\\ 3 & 3 &3 &3 &2\\ 4& 4& 4& 3& 3\\ 5& 5& 4& 4& 4\\ 6& 5& 5& 5 &5\end{bmatrix}$$

Laplace doesn't seem like the best method here, can we somehow turn this into a triangular matrix so that the determinant is the product of the elements on the main diagonal? I multiplied the first row by $(-3)$ and added it to he second one, then by $(-4)$ and added it to the third one, by $(-5)$ and added it to the fourth one, and by $(-6)$ and added it to last one. $$\begin{vmatrix} 1& 1& 1& 1& 1\\ 3 & 3 &3 &3 &2\\ 4& 4& 4& 3& 3\\ 5& 5& 4& 4& 4\\ 6& 5& 5& 5 &5 \end{vmatrix}=\begin{vmatrix} 1& 1& 1& 1& 1\\ 0& 0 &0 &0 &-1\\ 0& 0& 0& -1& -1\\ 0& 0& -1& -1& -1\\ 0& -1& -1& -1 &-1 \end{vmatrix}$$

What should I do now?

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    You can use row exchange and make it into upper triangular matrix. – 2017-01-21
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    That's all. The answer is $1$. – 2017-01-21
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    You may expand by second column. – 2017-01-21

1 Answers 1

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Now you can expand by the first collumn, and you get

$$\begin{vmatrix} 0 & 0 &0 &-1 \\ 0 & 0 &-1 &-1 \\ 0 &-1 &-1 &-1 \\ -1 &-1&-1&-1 \end{vmatrix}$$

which is triangular, so the determinant equals $(-1)^4=1$.

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    I know this wasn't the question, but how would you find the adjunct of this matrix? Can we somehow use this triangular matrix that you got? – 2017-01-21
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    @Lewis I don't know about that, but my first guess would be no... – 2017-01-21