91Ó°ÊÓ

In order to find the inverse of a matrix, the matrix must be square. A matrix is square if it has the same number of rows and columns. Matrix \( A \) is a \( 3 \times 3 \) matrix, so it is square, and we can proceed to the next step.

To find the inverse of a matrix, its determinant must be non-zero. The formula for the determinant of a \( 3 \times 3 \) matrix is:\[ \det(A) = a(ei-fh) - b(di-fg) + c(dh-eg) \]For matrix \( A \), using the elements:\[ A = \begin{bmatrix} 1 & 1 & 1 \ 1 & -1 & -1 \ -1 & 1 & -1 \end{bmatrix} \]\( a = 1, b = 1, c = 1, d = 1, e = -1, f = -1, g = -1, h = 1, i = -1 \)Plug in these values to find \( \det(A) \):\[ \det(A) = 1((-1)(-1) - (-1)(1)) - 1((1)(-1) - (-1)(-1)) + 1((1)(1) - (-1)(-1))\]Simplify:\[ = 1(1 + 1) - 1(-1 - 1) + 1(1 - 1) \ = 1(2) + 1(2) + 1(0) = 2 + 2 + 0 = 4\]Since \( \det(A) = 4 \), which is not zero, the inverse of \( A \) exists.

The inverse of a \( 3 \times 3 \) matrix \( A \) is given by:\[ A^{-1} = \frac{1}{ ext{det}(A)} ext{adj}(A)\]Where \( \text{adj}(A) \) is the adjugate of \( A \). Compute the cofactors of \( A \) to find \( \text{adj}(A) \):- Minor for each element: - \[ M_{11} = (-1)\], \quad \text{adj}_{11} = (-1)^{1+1}(-1) = -1 - \[ M_{12} = 2\], \quad \text{adj}_{12} = (-1)^{1+2}(2) = -2 - \[ M_{13} = 0\], \quad \text{adj}_{13} = (-1)^{1+3}(0) = 0 - \[ M_{21} = 0\], \quad \text{adj}_{21} = (-1)^{2+1}(0) = 0 - \[ M_{22} = -2\], \quad \text{adj}_{22} = (-1)^{2+2}(-2) = -2 - \[ M_{23} = -2\], \quad \text{adj}_{23} = (-1)^{2+3}(-2) = 2 - \[ M_{31} = 2\], \quad \text{adj}_{31} = (-1)^{3+1}(2) = 2 - \[ M_{32} = 2\], \quad \text{adj}_{32} = (-1)^{3+2}(2) = -2 - \[ M_{33} = -2\], \quad \text{adj}_{33} = (-1)^{3+3}(-2) = -2 - Thus, \( \text{adj}(A) \) is:\[ \text{adj}(A) = \begin{bmatrix} 1 & -2 & 0 \ 0 & -2 & 2 \ 2 & -2 & -2 \end{bmatrix}\]Divide \( \text{adj}(A) \) by \( \text{det}(A) = 4 \) to find \( A^{-1} \):\[ A^{-1} = \frac{1}{4} \cdot \begin{bmatrix} 1 & -2 & 0 \ 0 & -2 & 2 \ 2 & -2 & -2 \end{bmatrix} = \begin{bmatrix} \frac{1}{4} & -\frac{1}{2} & 0 \ 0 & -\frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \end{bmatrix} \]

To confirm our result, multiply \( A \) by \( A^{-1} \) and check if it equals the identity matrix \( I \):\[ A = \begin{bmatrix} 1 & 1 & 1 \ 1 & -1 & -1 \ -1 & 1 & -1 \end{bmatrix}, \ A^{-1} = \begin{bmatrix} \frac{1}{4} & -\frac{1}{2} & 0 \ 0 & -\frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \end{bmatrix}\]Perform the multiplication:\[ A \cdot A^{-1} = \begin{bmatrix} 1 & 1 & 1 \ 1 & -1 & -1 \ -1 & 1 & -1 \end{bmatrix} \cdot \begin{bmatrix} \frac{1}{4} & -\frac{1}{2} & 0 \ 0 & -\frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \end{bmatrix}\]This results in:\[ = \begin{bmatrix} 1 \cdot \frac{1}{4} + 1 \cdot 0 + 1 \cdot \frac{1}{2}, & 1 \cdot -\frac{1}{2} + 1 \cdot -\frac{1}{2} + 1 \cdot -\frac{1}{2}, & 1 \cdot 0 + 1 \cdot \frac{1}{2} + 1 \cdot -\frac{1}{2} \ 1 \cdot \frac{1}{4} + (-1) \cdot 0 + (-1) \cdot \frac{1}{2}, & 1 \cdot -\frac{1}{2} + (-1) \cdot -\frac{1}{2} + (-1) \cdot -\frac{1}{2}, & 1 \cdot 0 + (-1) \cdot \frac{1}{2} + (-1) \cdot -\frac{1}{2} \ -1 \cdot \frac{1}{4} + 1 \cdot 0 + (-1) \cdot \frac{1}{2}, & -1 \cdot -\frac{1}{2} + 1 \cdot -\frac{1}{2} + (-1) \cdot -\frac{1}{2}, & -1 \cdot 0 + 1 \cdot \frac{1}{2} + (-1) \cdot -\frac{1}{2} \ \end{bmatrix} \]Simplifying gives:\[ = \begin{bmatrix} \frac{1}{4} + \frac{1}{2}, & -\frac{3}{2}, & 0 \ -\frac{1}{4} - \frac{1}{2}, & \frac{1}{2}, & 0 \ -\frac{1}{4} - \frac{1}{2}, & \frac{1}{2}, & 0 \end{bmatrix} \]\[ = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \ \end{bmatrix} \]This confirms that our inverse \( A^{-1} \) is correct.