Pregunta
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Find the inverse of \( \quad A=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8\end{array}\right] \)

Ask by Willis Stanley. in Saudi Arabia
Feb 26,2025

Solución de inteligencia artificial de Upstudy

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العكس للمصفوفة \( A \) هو: \[ A^{-1} = \left[\begin{array}{lll}-40 & 16 & 9 \\ 13 & -5 & -3 \\ 5 & -2 & -1\end{array}\right] \]

Solución

Matrices by following steps: - step0: Find the matrix inverse: \(\left[\begin{array}{lll}{1}&{2}&{3}\\{2}&{5}&{3}\\{1}&{0}&{8}\end{array}\right]\) - step1: Begin by adjoining the identity matrix to form the matrix: \(\left[\begin{array}{lll|lll}{1}&{2}&{3}&{1}&{0}&{0}\\{2}&{5}&{3}&{0}&{1}&{0}\\{1}&{0}&{8}&{0}&{0}&{1}\end{array}\right]\) - step2: Simplify the row: \(\left[\begin{array}{lll|lll}{1}&{2}&{3}&{1}&{0}&{0}\\{0}&{1}&{-3}&{-2}&{1}&{0}\\{1}&{0}&{8}&{0}&{0}&{1}\end{array}\right]\) - step3: Simplify the row: \(\left[\begin{array}{lll|lll}{1}&{2}&{3}&{1}&{0}&{0}\\{0}&{1}&{-3}&{-2}&{1}&{0}\\{0}&{-2}&{5}&{-1}&{0}&{1}\end{array}\right]\) - step4: Simplify the row: \(\left[\begin{array}{lll|lll}{1}&{2}&{3}&{1}&{0}&{0}\\{0}&{1}&{-3}&{-2}&{1}&{0}\\{0}&{0}&{-1}&{-5}&{2}&{1}\end{array}\right]\) - step5: Simplify the row: \(\left[\begin{array}{lll|lll}{1}&{2}&{3}&{1}&{0}&{0}\\{0}&{1}&{-3}&{-2}&{1}&{0}\\{0}&{0}&{1}&{5}&{-2}&{-1}\end{array}\right]\) - step6: Simplify the row: \(\left[\begin{array}{lll|lll}{1}&{2}&{3}&{1}&{0}&{0}\\{0}&{1}&{0}&{13}&{-5}&{-3}\\{0}&{0}&{1}&{5}&{-2}&{-1}\end{array}\right]\) - step7: Simplify the row: \(\left[\begin{array}{lll|lll}{1}&{2}&{0}&{-14}&{6}&{3}\\{0}&{1}&{0}&{13}&{-5}&{-3}\\{0}&{0}&{1}&{5}&{-2}&{-1}\end{array}\right]\) - step8: Simplify the row: \(\left[\begin{array}{lll|lll}{1}&{0}&{0}&{-40}&{16}&{9}\\{0}&{1}&{0}&{13}&{-5}&{-3}\\{0}&{0}&{1}&{5}&{-2}&{-1}\end{array}\right]\) - step9: Write the right half matrix: \(\left[\begin{array}{lll}{-40}&{16}&{9}\\{13}&{-5}&{-3}\\{5}&{-2}&{-1}\end{array}\right]\) عكس المصفوفة \( A = \left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8\end{array}\right] \) هو: \[ A^{-1} = \left[\begin{array}{lll}-40 & 16 & 9 \\ 13 & -5 & -3 \\ 5 & -2 & -1\end{array}\right] \]

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Mind Expander

To find the inverse of the matrix \( A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{bmatrix} \), we can use the formula for the inverse of a matrix, which is given by \( A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \), where \( \text{det}(A) \) is the determinant of \( A \) and \( \text{adj}(A) \) is the adjugate of \( A \). First, we calculate the determinant of \( A \): \[ \text{det}(A) = 1 \cdot (5 \cdot 8 - 3 \cdot 0) - 2 \cdot (2 \cdot 8 - 3 \cdot 1) + 3 \cdot (2 \cdot 0 - 5 \cdot 1) \] \[ = 1 \cdot (40 - 0) - 2 \cdot (16 - 3) + 3 \cdot (0 - 5) \] \[ = 40 - 2 \cdot 13 - 15 \] \[ = 40 - 26 - 15 = -1. \] Now, we know that \( \text{det}(A) = -1 \). Next, we need to find the adjugate of \( A \), which involves finding the cofactor matrix and then transposing it. The cofactor of each element is computed by taking the determinant of the 2x2 matrix that remains when the row and column of that element are removed. Calculating the cofactors: 1. For \( a_{11} = 1 \), the cofactor is \( \text{det}\begin{bmatrix} 5 & 3 \\ 0 & 8 \end{bmatrix} = 5 \cdot 8 - 3 \cdot 0 = 40 \). 2. For \( a_{12} = 2 \), the cofactor is \( -\text{det}\begin{bmatrix} 2 & 3 \\ 1 & 8 \end{bmatrix} = -(2 \cdot 8 - 3 \cdot 1) = -(16 - 3) = -13 \). 3. For \( a_{13} = 3 \), the cofactor is \( \text{det}\begin{bmatrix} 2 & 5 \\ 1 & 0 \end{bmatrix} = 2 \cdot 0 - 5 \cdot 1 = -5 \). 4. For \( a_{21} = 2 \), the cofactor is \( -\text{det}\begin{bmatrix} 2 & 3 \\ 0 & 8 \end{bmatrix} = -(2 \cdot 8 - 3 \cdot 0) = -16 \). 5. For \( a_{22} = 5 \), the cofactor is \( \text{det}\begin{bmatrix} 1 & 3 \\ 1 & 8 \end{bmatrix} = 1 \cdot 8 - 3 \cdot 1 = 5 \). 6. For \( a_{23} = 3 \), the cofactor is \( -\text{det}\begin{bmatrix} 1 & 2 \\ 1 & 0 \end{bmatrix} = -(1 \cdot 0 - 2 \cdot 1) = 2 \). 7. For \( a_{31} = 1 \), the cofactor is \( \text{det}\begin{bmatrix} 2 & 3 \\ 5 & 3 \end{bmatrix} = 2 \cdot 3 - 3 \cdot 5 = 6 - 15 = -9 \). 8. For \( a_{32} = 0 \), the cofactor is \( -\text{det}\begin{bmatrix} 1 & 3 \\ 2 & 3 \end{bmatrix} = -(1 \cdot 3 - 3 \cdot 2) = -(-3) = 3 \). 9. For \( a_{33} = 8 \), the cofactor is \( \text{det}\begin{bmatrix} 1 & 2 \\ 2 & 5 \end{bmatrix} = 1 \cdot 5 - 2 \cdot 2 = 5 -

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