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Math  /  Algebra

QuestionFind the inverse of the matrix. [−1−1232−4−0.500.5]\left[\begin{array}{ccc} -1 & -1 & 2 \\ 3 & 2 & -4 \\ -0.5 & 0 & 0.5 \end{array}\right] - [210114212]\left[\begin{array}{lll}2 & 1 & 0 \\ 1 & 1 & 4 \\ 2 & 1 & 2\end{array}\right] [204101222]\left[\begin{array}{lll}2 & 0 & 4 \\ 1 & 0 & 1 \\ 2 & 2 & 2\end{array}\right] [212114211]\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 1 & 4 \\ 2 & 1 & 1\end{array}\right] [212142112]\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 4 & 2 \\ 1 & 1 & 2\end{array}\right]

Studdy Solution

STEP 1

1. We are given a 3×33 \times 3 matrix and need to find its inverse.
2. The matrix is invertible, meaning it has a non-zero determinant.
3. The inverse of a matrix A A is denoted as A−1 A^{-1} and satisfies A⋅A−1=I A \cdot A^{-1} = I , where I I is the identity matrix.

STEP 2

1. Calculate the determinant of the matrix.
2. Find the matrix of minors.
3. Find the cofactor matrix.
4. Find the adjugate (transpose of the cofactor matrix).
5. Calculate the inverse using the formula A−1=1det(A)⋅adj(A) A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) .

STEP 3

Calculate the determinant of the matrix:
Given matrix: A=[−1−1232−4−0.500.5] A = \begin{bmatrix} -1 & -1 & 2 \\ 3 & 2 & -4 \\ -0.5 & 0 & 0.5 \end{bmatrix}
The determinant of a 3×33 \times 3 matrix A=[abcdefghi] A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} is calculated as:
det(A)=a(ei−fh)−b(di−fg)+c(dh−eg)\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)
Substitute the values:
det(A)=(−1)((2)(0.5)−(−4)(0))−(−1)((3)(0.5)−(−4)(−0.5))+(2)((3)(0)−(2)(−0.5))\text{det}(A) = (-1)((2)(0.5) - (-4)(0)) - (-1)((3)(0.5) - (-4)(-0.5)) + (2)((3)(0) - (2)(-0.5))
Calculate each term:
=(−1)(1)−(−1)(1.5−2)+(2)(1)= (-1)(1) - (-1)(1.5 - 2) + (2)(1)
=−1+0.5+2= -1 + 0.5 + 2
=1.5= 1.5

STEP 4

Find the matrix of minors for matrix A A .
Each element of the matrix of minors is the determinant of the 2×22 \times 2 matrix formed by deleting the row and column of that element.
Calculate the minors:
Minor11=∣2−400.5∣=(2)(0.5)−(−4)(0)=1\text{Minor}_{11} = \begin{vmatrix} 2 & -4 \\ 0 & 0.5 \end{vmatrix} = (2)(0.5) - (-4)(0) = 1
Minor12=∣3−4−0.50.5∣=(3)(0.5)−(−4)(−0.5)=1.5−2=−0.5\text{Minor}_{12} = \begin{vmatrix} 3 & -4 \\ -0.5 & 0.5 \end{vmatrix} = (3)(0.5) - (-4)(-0.5) = 1.5 - 2 = -0.5
Minor13=∣32−0.50∣=(3)(0)−(2)(−0.5)=1\text{Minor}_{13} = \begin{vmatrix} 3 & 2 \\ -0.5 & 0 \end{vmatrix} = (3)(0) - (2)(-0.5) = 1
Minor21=∣−12−0.50.5∣=(−1)(0.5)−(2)(−0.5)=0.5\text{Minor}_{21} = \begin{vmatrix} -1 & 2 \\ -0.5 & 0.5 \end{vmatrix} = (-1)(0.5) - (2)(-0.5) = 0.5
Minor22=∣−12−0.50.5∣=(−1)(0.5)−(2)(−0.5)=0.5\text{Minor}_{22} = \begin{vmatrix} -1 & 2 \\ -0.5 & 0.5 \end{vmatrix} = (-1)(0.5) - (2)(-0.5) = 0.5
Minor23=∣−1−1−0.50∣=(−1)(0)−(−1)(−0.5)=−0.5\text{Minor}_{23} = \begin{vmatrix} -1 & -1 \\ -0.5 & 0 \end{vmatrix} = (-1)(0) - (-1)(-0.5) = -0.5
Minor31=∣−123−4∣=(−1)(−4)−(2)(3)=−2\text{Minor}_{31} = \begin{vmatrix} -1 & 2 \\ 3 & -4 \end{vmatrix} = (-1)(-4) - (2)(3) = -2
Minor32=∣−123−4∣=(−1)(−4)−(2)(3)=−2\text{Minor}_{32} = \begin{vmatrix} -1 & 2 \\ 3 & -4 \end{vmatrix} = (-1)(-4) - (2)(3) = -2
Minor33=∣−1−132∣=(−1)(2)−(−1)(3)=1\text{Minor}_{33} = \begin{vmatrix} -1 & -1 \\ 3 & 2 \end{vmatrix} = (-1)(2) - (-1)(3) = 1

STEP 5

Find the cofactor matrix by applying the checkerboard pattern of signs to the matrix of minors:
Cofactor matrix=[10.51−0.50.50.5−2−21]\text{Cofactor matrix} = \begin{bmatrix} 1 & 0.5 & 1 \\ -0.5 & 0.5 & 0.5 \\ -2 & -2 & 1 \end{bmatrix}

STEP 6

Find the adjugate by transposing the cofactor matrix:
Adjugate=[1−0.5−20.50.5−210.51]\text{Adjugate} = \begin{bmatrix} 1 & -0.5 & -2 \\ 0.5 & 0.5 & -2 \\ 1 & 0.5 & 1 \end{bmatrix}

STEP 7

Calculate the inverse using the formula A−1=1det(A)⋅adj(A) A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) :
A−1=11.5[1−0.5−20.50.5−210.51]A^{-1} = \frac{1}{1.5} \begin{bmatrix} 1 & -0.5 & -2 \\ 0.5 & 0.5 & -2 \\ 1 & 0.5 & 1 \end{bmatrix}
=[11.5−0.51.5−21.50.51.50.51.5−21.511.50.51.511.5]= \begin{bmatrix} \frac{1}{1.5} & \frac{-0.5}{1.5} & \frac{-2}{1.5} \\ \frac{0.5}{1.5} & \frac{0.5}{1.5} & \frac{-2}{1.5} \\ \frac{1}{1.5} & \frac{0.5}{1.5} & \frac{1}{1.5} \end{bmatrix}
=[23−13−431313−43231323]= \begin{bmatrix} \frac{2}{3} & \frac{-1}{3} & \frac{-4}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{-4}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \end{bmatrix}
The inverse of the matrix is:
[23−13−431313−43231323]\boxed{\begin{bmatrix} \frac{2}{3} & \frac{-1}{3} & \frac{-4}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{-4}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \end{bmatrix}}

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