# What happens when we prepend a row and column to a regular matrix?

I am working with a matrix $$M \in \Bbb R^{n \times n}$$ that has non-zero determinant. I then prepend a row and a column to $$M$$, where each entry is positive, so that $$M$$ is now $$(n+1) \times (n+1)$$ and looks like $$\begin{pmatrix} a_{11} & b^T\\ c & M \end{pmatrix}$$ where $$a_{11} > 0$$ and $$b$$ and $$c$$ are vectors consisting of all positive numbers. Does this matrix also have a non-zero determinant?

• Take $n=1$ and let the extended matrix have all entries $=1$. Commented May 9, 2023 at 6:00
• Quick beginner guide for asking a well-received question Commented May 9, 2023 at 6:06
• I found a relevant post here at this link math.stackexchange.com/questions/3995443/… Commented May 9, 2023 at 6:37
• You are thinking like a programmer. Note that you claim that $$M = \begin{pmatrix} a_{11} & b^T\\ c & M \end{pmatrix}$$ Commented May 9, 2023 at 7:25

## 1 Answer

No, you can very easily generate a counterexample.

For instance, start with the $$2 \times 2$$ matrix $$M = \begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix}$$ which has $$\det M = 3 \neq 0$$. Then, your $$3 \times 3$$ matrix is $$M' = \begin{bmatrix}? & ? & ? \\ ? & 2 & 1 \\ ? & 1 & 2\end{bmatrix}.$$

Can you fill in the $$?$$ entries with positive real numbers to make the first and second columns the same? If so, then $$\det M' = 0$$.

• Thanks very much. If no columns are the same, is it then possible? The matrix I am working with is such that none of the ? entries in $M'$ are entries of $M$. Commented May 9, 2023 at 5:49
• Columns don't necessarily have to be the same for the determinant to be zero. Even if the new column is a linear combination of the existing columns, we can have a zero determinant. Consider the example: $[[2,1,1],[4,2,1],[2,1,2]]$ where column 1 is twice column 2. Commented May 9, 2023 at 6:52