How to find the inverse of this matrix? (in order to solve the question).

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.

• Well, $(A)_{11}=a$ is obvious. $(A^{-1})_{11}$ is clearly $(a^2+1)/\det A$ by cofactors, so it remains to evaluate $\det A$. – user10354138 Mar 4 at 11:16
• @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. – Pwaol Mar 4 at 11:20
• It amounts to this. – user10354138 Mar 4 at 11:23

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 $$x = z - ay = \frac{1}{a(1-a)} - \frac{1}{1-a} = \frac{1}{a}$$.

• 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? – Pwaol Mar 4 at 11:24
• @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. – levap Mar 4 at 11:26
• Thanks alot! I appreciate the help. – Pwaol Mar 4 at 11:27