matrix inverse structure

Question

Let $A$ be an arbitrary invertible matrix, whose 2nd row is all zeros except the 2nd entry of the 2nd row is a nonzero element.

$$A= \left[ \begin{matrix} * & * & * & * & \cdots & * \\ 0 & * & 0 & 0 & \ldots & 0 \\ * & * & * & * & \ldots & * \\ * & * & * & * & \cdots & * \\ \vdots & \vdots & \vdots & \vdots & \cdots & \vdots \\ * & * & * & * & * & * \\ \end{matrix} \right] $$

Then, how to prove that $A^{-1}$ is a matrix whose 2nd row also is all zeros except the 2nd entry of the 2nd row is a nonzero element?

$$A^{-1}= \left[ \begin{matrix} * & * & * & * & \cdots & * \\ 0 & * & 0 & 0 & \ldots & 0 \\ * & * & * & * & \ldots & * \\ * & * & * & * & \cdots & * \\ \vdots & \vdots & \vdots & \vdots & \cdots & \vdots \\ * & * & * & * & * & * \\ \end{matrix} \right] $$

Answer 1

Hint: Consider the method of computing $A^{-1}$ through the adjugate matrix of $A$.

Answer 2

We know that $AA^{-1}=I$. In general, if you compute $AB$ with $A$ as above, then entries in the second row of the product are $$ a_{22}b_{2i} $$ in column $i$ (use the across-down multiplication method and observe that almost all of the summands are zero). In order for $B$ to be the inverse of $A$, then $$ a_{22}b_{2i}=0 $$ when $i\not=2$, and $$ a_{22}b_{2i}=1 $$ when $i=2$ (this are the entries of the second row of the identity matrix).

Answer 3

To prove this I will be using a generalized $A_{n \times n}$ matrix for which $A_{2(1,\dots,n)} = (0,a_{22},0,\dots,0)$ where $a_{22} \ne 0$. To calculate the inverse of a matrix the simplest method is to use Gauss-Jordan elimination (Guass elimination for short). Using this method we can get: $$[A_{n \times n}|I_{n \times n}] \xrightarrow{\text{Gauss elimination}} [I_{n \times n}|A_{n \times n}^{-1}]$$ Now we can start transforming our $A_{n \times n}$ matrix: $$[A_{n \times n}|I_{n \times n}] = \left[\begin{array}{@{}c|c@{}} \begin{matrix} a_{11} & a_{12} & a_{13} & \dots & a_{1n} \\ 0_{21} & a_{22} & 0_{23} & \dots & 0_{2n} \\ \vdots & & & & \vdots\\ a_{n1} & a_{n2} & a_{n3} & \dots & a_{nn} \end{matrix} & \begin{matrix} 1_{11} & 0_{12} & 0_{13} & \dots & 0_{1n} \\ 0_{21} & 1_{22} & 0_{23} & \dots & 0_{2n} \\ \vdots & & & & \vdots\\ 0_{n1} & 0_{n2} & 0_{n3} & \dots & 1_{nn} \end{matrix} \end{array}\right] \xrightarrow{\text{row}_2/a_{22}} \\ \xrightarrow{\ \ \ \ \ } \left[\begin{array}{@{}c|c@{}} \begin{matrix} a_{11} & a_{12} & a_{13} & \dots & a_{1n} \\ 0_{21} & 1_{22} & 0_{23} & \dots & 0_{2n} \\ \vdots & & & & \vdots\\ a_{n1} & a_{n2} & a_{n3} & \dots & a_{nn} \end{matrix} & \begin{matrix} 1_{11} & 0_{12} & 0_{13} & \dots & 0_{1n} \\ 0_{21} & \frac{1}{a_{22}} & 0_{23} & \dots & 0_{2n} \\ \vdots & & & & \vdots\\ 0_{n1} & 0_{n2} & 0_{n3} & \dots & 1_{nn} \end{matrix} \end{array}\right] \xrightarrow{\begin{matrix}\text{row}_1-a_{12}\text{row}_2 \\ \text{row}_3-a_{12}\text{row}_2 \\ \vdots \\ \text{row}_n-a_{n2}\text{row}_2\end{matrix}} \\ \xrightarrow{\ \ \ \ \ } \left[\begin{array}{@{}c|c@{}} \begin{matrix} a_{11} & 0_{12} & a_{13} & \dots & a_{1n} \\ 0_{21} & 1_{22} & 0_{23} & \dots & 0_{2n} \\ \vdots & & & & \vdots\\ a_{n1} & 0_{n2} & a_{n3} & \dots & a_{nn} \end{matrix} & \begin{matrix} 1_{11} & 0_{12} & 0_{13} & \dots & 0_{1n} \\ 0_{21} & \frac{1}{a_{22}} & 0_{23} & \dots & 0_{2n} \\ \vdots & & & & \vdots\\ 0_{n1} & 0_{n2} & 0_{n3} & \dots & 1_{nn} \end{matrix} \end{array}\right] :=[{\widetilde{A}}_{n \times n}|{\widetilde{I}}_{n \times n}] $$ Let $${\widetilde{\widetilde{A}}}_{(n-1) \times (n-1)} := {\widetilde{A}}_{n \times n} \backslash \{\widetilde{A}\text{'s 2nd row}\} \backslash \{\widetilde{A}\text{'s 2nd column}\}$$ $${\widetilde{\widetilde{I}}}_{(n-1) \times (n-1)} := {\widetilde{I}}_{n \times n} \backslash \{\widetilde{I}\text{'s 2nd row}\} \backslash \{\widetilde{I}\text{'s 2nd column}\}$$ We can see that ${\widetilde{\widetilde{I}}}_{(n-1) \times (n-1)} = I_{(n-1) \times (n-1)}$, since in $\widetilde{I}_{(n-1) \times (n-1)}$ only exactly the second row and the second column were modified. With the same reasoning we can see that $\widetilde{\widetilde{A}}_{(n-1) \times (n-1)} = A_{n \times n} \backslash \{A\text{'s 2nd row}\} \backslash \{A\text{'s 2nd column}\}.$

Let's invert $\widetilde{\widetilde{A}}_{(n-1) \times (n-1)}$ using the above mentioned (full) Gauss elimination: $$[\widetilde{\widetilde{A}}_{(n-1) \times (n-1)}|{\widetilde{\widetilde{I}}}_{(n-1) \times (n-1)}] \xrightarrow{\text{Gauss elimination}} [{\widetilde{\widetilde{I}}}_{(n-1) \times (n-1)}|\widetilde{\widetilde{A}}_{(n-1) \times (n-1)}^{-1}] \qquad (*)$$

We can see that $$\widetilde{\widetilde{A}}_{(n-1) \times (n-1)}^{-1} \bigcup \{A_{n \times n}\text{'s second row}\} \bigcup \{A_{n \times n}\text{'s second column}\} = A_{n \times n}^{-1}$$ Since throughout the Gauss elimination the second row and the second column don't need to be modified.

So $$A_{n \times n}^{-1} = \begin{bmatrix} s_{11} & s_{12} & s_{13} & \dots & s_{1n} \\ 0_{11} & \frac{1}{a_{22}} & 0_{13} & \dots & 0_{1n} \\ \vdots & & & & \vdots \\ s_{n1} & s_{n2} & s_{n3} & \dots & s_{nn} \\ \end{bmatrix}$$ Where $s_{ij}$ are some values that come from the $(*)$ Gauss elimination.