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I am trying to solve this question and things got really complicated while trying to find the inverse of the matrix, here's the question:
Given $ A = \left[\begin{array}{l}a&1&-1\\1&a&-1\\1&-1&a\end{array}\right]$, $a\in \mathbb{R}$, if $A$ is invertible, then $(A)_{11}*(A^{-1})_{11}=?$
I started to doubt myself since I reached $\left[\begin{array}{l}a&1&-1\\0&a-\frac{1}{a}&-1+\frac{1}{a}\\0&-1-\frac{1}{a}&a+\frac{1}{a}\end{array}\right]$.
And I couldn't figure out how to keep going, and started to doubt I'm missing something (or I'm just weak at finding inverse matrix).
Note I checked the determinant and got that it's invertible if and only if $a \ne 0,1,2$ so there's not problem with dividing by $a$.

Any help would be really appreciated, thanks in advance.

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  • $\begingroup$ Well, $(A)_{11}=a$ is obvious. $(A^{-1})_{11}$ is clearly $(a^2+1)/\det A$ by cofactors, so it remains to evaluate $\det A$. $\endgroup$ – user10354138 Mar 4 at 11:16
  • $\begingroup$ @user10354138 thank you for the help, could you elaborate a little on $(A^{-1})_{11}$? I can't understand how you can find it yet using the determinant. $\endgroup$ – Pwaol Mar 4 at 11:20
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    $\begingroup$ It amounts to this. $\endgroup$ – user10354138 Mar 4 at 11:23
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Since you are asked only to find the product $(A)_{11} \cdot \left( A^{-1} \right)_{11}$, you only need to find the $(1,1)$-entry of $A^{-1}$ and not the whole matrix. For this purpose, since the matrix is $3 \times 3$ it is useful to use the formula for the adjugate matrix. We have

$$ \det(A) = a \cdot (a^2 - 1) - (a - 1) + (-1 + a) = a \cdot (a^2 - 1), \\ \operatorname{adj}(A)_{11} = a^2 - 1 $$ and hence $$ \left( A^{-1} \right)_{11} = \frac{1}{\det(A)} \left( \operatorname{adj}(A) \right)_{11} = \frac{a^2 - 1}{a \cdot (a^2 - 1)} = \frac{1}{a}.$$


Alternatively, let's write

$$ A^{-1} = \begin{pmatrix} x & ? & ? \\ y & ? & ? \\ z & ? & ? \end{pmatrix}. $$

We need to find $x$. Since $A \cdot A^{-1} = I$, we must have $$ ax + y - z = 1, \\ x + ay - z = 0,\\ x - y + az = 0. $$

There are three equations for three unknowns (where $a$ is a parameter) and the augmented matrix associated to this system is $$ \left( \begin{array}{ccc|c} a & 1 & -1 & 1 \\ 1 & a & -1 & 0 \\ 1 & -1 & a & 0 \end{array} \right). $$

Performing elementary operations, we get the equivalent system $$ \left( \begin{array}{ccc|c} a & 1 & -1 & 1 \\ 1 & a & -1 & 0 \\ 1 & -1 & a & 0 \end{array} \right) \xrightarrow{R_1 \leftrightarrow R_2} \left( \begin{array}{ccc|c} 1 & a & -1 & 0 \\ a & 1 & -1 & 1 \\ 1 & -1 & a & 0 \end{array} \right) \xrightarrow{R_2 = R_2 - aR_1, R_3 = R_3 - R_1} \left( \begin{array}{ccc|c} 1 & a & -1 & 0 \\ 0 & 1 - a^2 & a - 1 & 1 \\ 0 & -1 - a & a + 1 & 0 \end{array} \right) \xrightarrow{ R_2 = R_2 + (1 - a) R_3} \left( \begin{array}{ccc|c} 1 & a & -1 & 0 \\ 0 & 0 & a(1-a) & 1 \\ 0 & -1 - a & a + 1 & 0 \end{array} \right) $$ which translates into the system $$ x + ay - z = 0, \\ a(1-a)z = 1, \\ (a+1)z = (a+1)y. $$

Hence, $z = y = \frac{1}{a(1-a)}$ and $x = z - ay = \frac{1}{a(1-a)} - \frac{1}{1-a} = \frac{1}{a}$.

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  • $\begingroup$ Thanks, I appreciate the help, seems like I'm solving some unrelevant question since I didn't learn about the adjugate matrix, do you think there's anyway to find $(A^{-1})_{11}$ without using the formula for the adjugate matrix? $\endgroup$ – Pwaol Mar 4 at 11:24
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    $\begingroup$ @Pwaol: There are many ways to solve this question. You can perform Gaussian elimination or even work up directly with the definition. I'll add another solution in a couple of minutes. $\endgroup$ – levap Mar 4 at 11:26
  • $\begingroup$ Thanks alot! I appreciate the help. $\endgroup$ – Pwaol Mar 4 at 11:27

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