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Math  /  Algebra

QuestionFind the inverse of each matrix. 17) [−25−2−220−3−22]\left[\begin{array}{ccc}-2 & 5 & -2 \\ -2 & 2 & 0 \\ -3 & -2 & 2\end{array}\right]

Studdy Solution

STEP 1

1. The problem requires finding the inverse of a given 3×33 \times 3 matrix.
2. The matrix is invertible, meaning its determinant is non-zero.
3. The inverse of a matrix AA is denoted A−1A^{-1} and satisfies AA−1=IA A^{-1} = I, where II is the identity matrix.
4. The inverse of a 3×33 \times 3 matrix can be found using the formula involving the adjugate matrix and the determinant.

STEP 2

1. Calculate the determinant of the given matrix.
2. Find the matrix of minors.
3. Find the matrix of cofactors.
4. Transpose the matrix of cofactors to get the adjugate matrix.
5. Divide the adjugate matrix by the determinant to find the inverse matrix.

STEP 3

Calculate the determinant of the given matrix A=[−25−2−220−3−22]A = \left[\begin{array}{ccc}-2 & 5 & -2 \\ -2 & 2 & 0 \\ -3 & -2 & 2\end{array}\right] using the formula for the determinant of a 3×33 \times 3 matrix. det(A)=−2(∣20−22∣)−5(∣−20−32∣)−2(∣−22−3−2∣) \text{det}(A) = -2 \left( \begin{vmatrix} 2 & 0 \\ -2 & 2 \end{vmatrix} \right) - 5 \left( \begin{vmatrix} -2 & 0 \\ -3 & 2 \end{vmatrix} \right) - 2 \left( \begin{vmatrix} -2 & 2 \\ -3 & -2 \end{vmatrix} \right)

STEP 4

Calculate the 2×22 \times 2 determinants involved: ∣20−22∣=2⋅2−0⋅(−2)=4 \begin{vmatrix} 2 & 0 \\ -2 & 2 \end{vmatrix} = 2 \cdot 2 - 0 \cdot (-2) = 4 ∣−20−32∣=−2⋅2−0⋅(−3)=−4 \begin{vmatrix} -2 & 0 \\ -3 & 2 \end{vmatrix} = -2 \cdot 2 - 0 \cdot (-3) = -4 ∣−22−3−2∣=−2⋅(−2)−2⋅(−3)=4+6=10 \begin{vmatrix} -2 & 2 \\ -3 & -2 \end{vmatrix} = -2 \cdot (-2) - 2 \cdot (-3) = 4 + 6 = 10

STEP 5

Substitute these values back into the determinant formula: det(A)=−2(4)−5(−4)−2(10)=−8+20−20=−8+20−20=−8 \text{det}(A) = -2 (4) - 5 (-4) - 2 (10) = -8 + 20 - 20 = -8 + 20 - 20 = -8

STEP 6

Find the matrix of minors. For each element in AA, find the determinant of the submatrix formed by deleting the row and column of that element.
Minor of a11a_{11}: ∣20−22∣=4 \begin{vmatrix} 2 & 0 \\ -2 & 2 \end{vmatrix} = 4 Minor of a12a_{12}: ∣−20−32∣=−4 \begin{vmatrix} -2 & 0 \\ -3 & 2 \end{vmatrix} = -4 Minor of a13a_{13}: ∣−22−3−2∣=10 \begin{vmatrix} -2 & 2 \\ -3 & -2 \end{vmatrix} = 10 Minor of a21a_{21}: ∣5−2−22∣=6 \begin{vmatrix} 5 & -2 \\ -2 & 2 \end{vmatrix} = 6 Minor of a22a_{22}: ∣−2−2−32∣=−10 \begin{vmatrix} -2 & -2 \\ -3 & 2 \end{vmatrix} = -10 Minor of a23a_{23}: ∣−25−3−2∣=19 \begin{vmatrix} -2 & 5 \\ -3 & -2 \end{vmatrix} = 19 Minor of a31a_{31}: ∣5−220∣=−4 \begin{vmatrix} 5 & -2 \\ 2 & 0 \end{vmatrix} = -4 Minor of a32a_{32}: ∣−2−2−20∣=−4 \begin{vmatrix} -2 & -2 \\ -2 & 0 \end{vmatrix} = -4 Minor of a33a_{33}: ∣−25−22∣=−6 \begin{vmatrix} -2 & 5 \\ -2 & 2 \end{vmatrix} = -6

STEP 7

Form the matrix of minors: M=[4−4106−1019−4−4−6] M = \left[\begin{array}{ccc} 4 & -4 & 10 \\ 6 & -10 & 19 \\ -4 & -4 & -6 \end{array}\right]

STEP 8

Form the matrix of cofactors by applying the checkerboard pattern of signs: C=[4410−6−10−19−44−6] C = \left[\begin{array}{ccc} 4 & 4 & 10 \\ -6 & -10 & -19 \\ -4 & 4 & -6 \end{array}\right]

STEP 9

Transpose the matrix of cofactors to get the adjugate matrix: adj(A)=[4−6−44−10410−19−6] \text{adj}(A) = \left[\begin{array}{ccc} 4 & -6 & -4 \\ 4 & -10 & 4 \\ 10 & -19 & -6 \end{array}\right]

STEP 10

Divide the adjugate matrix by the determinant to get the inverse A−1A^{-1}: A−1=1det(A)adj(A)=1−8[4−6−44−10410−19−6]=[−123412−1254−12−5419834] A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) = \frac{1}{-8} \left[\begin{array}{ccc} 4 & -6 & -4 \\ 4 & -10 & 4 \\ 10 & -19 & -6 \end{array}\right] = \left[\begin{array}{ccc} -\frac{1}{2} & \frac{3}{4} & \frac{1}{2} \\ -\frac{1}{2} & \frac{5}{4} & -\frac{1}{2} \\ -\frac{5}{4} & \frac{19}{8} & \frac{3}{4} \end{array}\right]
The inverse of the given matrix is: A−1=[−123412−1254−12−5419834] A^{-1} = \left[\begin{array}{ccc} -\frac{1}{2} & \frac{3}{4} & \frac{1}{2} \\ -\frac{1}{2} & \frac{5}{4} & -\frac{1}{2} \\ -\frac{5}{4} & \frac{19}{8} & \frac{3}{4} \end{array}\right]

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