# Find determinant of the matrix $P$.

Consider $$J$$ to be an $$n\times n$$ matrix whose entries are all $$1s$$ .

If $$P$$ is an $$n\times n$$ matrix such that

$$P=$$ $$\begin{bmatrix} v_1| v_2|,\ldots |v_{n-1} |v_n\end{bmatrix}$$

where the columns $$v_i=e_i-e_n$$ for $$1\le i\le n-1$$ and $$v_n=\sum e_i$$ where $$e_i$$ is the ith column of the Identity Matrix

Note that $$v_i,1\le i\le i-1$$ are the eigen vectors corresponding to $$0$$ of $$J$$ and $$v_n$$ is an eigen vector corresponding to $$n$$ of $$J$$

Find $$\det P$$.

It is very difficult to expand by Laplace Expansion

Is there any efficient way to to find the determinant?

• How about doing it by induction and expanding with respect to the first line? – Mindlack Dec 17 '18 at 8:56
• What does the matrix $J$ have to do with anything? – BigbearZzz Dec 17 '18 at 9:04
• @BigbearZzz;Sorry forgot to write that – user596656 Dec 17 '18 at 9:07

Use row operations to simplify the matrix in an upper triangular form. The form of $$P$$ is, $$P=\begin{pmatrix} 1 &0&0& ...&1 \\ 0&1&0&...&1\\ \vdots\\ -1&-1&-1&...&1 \end{pmatrix},$$ So carry out the row operation (which doesn't change the determinant) $$R_n=R_n+R_k$$ for $$1 \leq k \leq n-1$$. The resulting matrix is of the form $$P'=\begin{pmatrix} 1 &0&0& ...&1 \\ 0&1&0&...&1\\ \vdots\\ 0&0&0&...&1+(n-1) \end{pmatrix},$$ so $$\det P=\det P'=1 \cdot 1 \cdot...\cdot1 \cdot n=n$$.
• It will be $n$ in the last entry of the matrix ,not $n+1$ – user596656 Dec 17 '18 at 9:23