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Let $M$ be an invertible real matrix.

If I multiply one line ( or column ) by a scalar, what can I say about it's new inverse in regard of the old one?

Example : I have $M$ as $3 \times 3$ real matrix. Let's have $M'$ where the last line is multiplied by $-1$. Is there any information on $M'^{-1}$ from $M^{-1}$ ? Like we know that $det(M') = -1 \cdot det(M)$.

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3 Answers 3

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Put $$M = \begin{pmatrix} r_1 \\ r_2 \\ r_3 \end{pmatrix},$$ where $r_1,r_2,r_3$ are row vectors. Then $$M^{-1} = (c_1\,\,| \,\, c_2 \,\,| \,\, c_3) $$ where $c_1, c_2, c_3$ are column vectors such that $$r_ic_j = \delta_{ij} = \left \{ \begin{matrix} 1, & i = j, \\ 0, & i\neq j. \end{matrix} \right.$$

From this it is clear: if you swap $r_3$ for $-r_3$, then you simply need to swap $c_3$ for $-c_3$.

That is, if $$M' = \begin{pmatrix} r_1 \\ r_2 \\ -r_3 \end{pmatrix},$$ then $$(M')^{-1} = (c_1\,\,| \,\, c_2 \,\,| \,\, -c_3).$$

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  • $\begingroup$ Thank you, I add that in mind, but was not able to write it down. So I upvoted you. Anyway, I've also found an other way of solving this. $\endgroup$
    – PTRK
    Commented Aug 20, 2018 at 15:12
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From Wikipedia:

Viewing an $ n\times n$ matrix as being composed of $ n$ columns, the determinant is an n-linear function. This means that if the $ j^{th}$ column of a matrix $ A$ is written as a sum $ \mathbf {a} _{j}=\mathbf {v} +\mathbf {w} $ of two column vectors, and all other columns are left unchanged, then the determinant of $ A$ is the sum of the determinants of the matrices obtained from $ A$ by replacing the $ j^{th}$ column by $ \mathbf {v}$ (denoted $ A_{v}$) and then by $ \mathbf {w} $ (denoted $ A_{w}$) (and a similar relation holds when writing a column as a scalar multiple of a column vector).

That is, multiplying a row or a column of a matrix by a scalar implies that the determinant of the matrix is also multiplied by the same scalar.

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Working on it let me found the following :

Multiplying the n-th column of a matrix $M$ by $x$ is the same as doing $M.A$, where A is the identity, except on the n-th column, n-th row where it is $x$.

  • for multiplying a column $M'^{-1} = A^{-1}M^{-1}$.
  • for multiplying a row $M'^{-1} = M^{-1}A^{-1}$.

In my particular case, with -1, $A^{-1}=A$. So $M'^{-1} = AM^{-1}$ which translates to multiplying the n-th row of $M^{-1}$ by $-1$ which is the same answer as User8128.

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