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Question: Find inverse of matrix whose rows are 3,0,3 and 4,2,5 and 1,1,1...

Find inverse of matrix whose rows are 3,0,3 and 4,2,5 and 1,1,1

Answer

[1−12−1301−231−2]\begin{bmatrix} 1 & -1 & 2 \\ -\frac{1}{3} & 0 & 1 \\ -\frac{2}{3} & 1 & -2 \end{bmatrix}

Explanation

Solution

To find the inverse of the given matrix AA, we will use the formula A−1=1det⁡(A)adj(A)A^{-1} = \frac{1}{\det(A)} \text{adj}(A).

Given matrix: A=[303425111]A = \begin{bmatrix} 3 & 0 & 3 \\ 4 & 2 & 5 \\ 1 & 1 & 1 \end{bmatrix}

1. Calculate the determinant of A (det⁡(A)\det(A)): det⁡(A)=3(2×1−5×1)−0(4×1−5×1)+3(4×1−2×1)\det(A) = 3(2 \times 1 - 5 \times 1) - 0(4 \times 1 - 5 \times 1) + 3(4 \times 1 - 2 \times 1) det⁡(A)=3(2−5)−0+3(4−2)\det(A) = 3(2 - 5) - 0 + 3(4 - 2) det⁡(A)=3(−3)+3(2)\det(A) = 3(-3) + 3(2) det⁡(A)=−9+6\det(A) = -9 + 6 det⁡(A)=−3\det(A) = -3

Since det⁡(A)≠0\det(A) \neq 0, the inverse exists.

2. Calculate the cofactor matrix (C): The cofactor CijC_{ij} is (−1)i+j(-1)^{i+j} times the determinant of the submatrix obtained by deleting the ii-th row and jj-th column.

C11=+(2×1−5×1)=2−5=−3C_{11} = +(2 \times 1 - 5 \times 1) = 2 - 5 = -3 C12=−(4×1−5×1)=−(4−5)=1C_{12} = -(4 \times 1 - 5 \times 1) = -(4 - 5) = 1 C13=+(4×1−2×1)=4−2=2C_{13} = +(4 \times 1 - 2 \times 1) = 4 - 2 = 2

C21=−(0×1−3×1)=−(0−3)=3C_{21} = -(0 \times 1 - 3 \times 1) = -(0 - 3) = 3 C22=+(3×1−3×1)=3−3=0C_{22} = +(3 \times 1 - 3 \times 1) = 3 - 3 = 0 C23=−(3×1−0×1)=−(3−0)=−3C_{23} = -(3 \times 1 - 0 \times 1) = -(3 - 0) = -3

C31=+(0×5−3×2)=0−6=−6C_{31} = +(0 \times 5 - 3 \times 2) = 0 - 6 = -6 C32=−(3×5−3×4)=−(15−12)=−3C_{32} = -(3 \times 5 - 3 \times 4) = -(15 - 12) = -3 C33=+(3×2−0×4)=6−0=6C_{33} = +(3 \times 2 - 0 \times 4) = 6 - 0 = 6

The cofactor matrix CC is: C=[−31230−3−6−36]C = \begin{bmatrix} -3 & 1 & 2 \\ 3 & 0 & -3 \\ -6 & -3 & 6 \end{bmatrix}

3. Calculate the adjoint of A (adj(A)\text{adj}(A)): The adjoint of A is the transpose of the cofactor matrix. adj(A)=CT=[−33−610−32−36]\text{adj}(A) = C^T = \begin{bmatrix} -3 & 3 & -6 \\ 1 & 0 & -3 \\ 2 & -3 & 6 \end{bmatrix}

4. Calculate the inverse of A (A−1A^{-1}): A−1=1det⁡(A)adj(A)A^{-1} = \frac{1}{\det(A)} \text{adj}(A) A−1=1−3[−33−610−32−36]A^{-1} = \frac{1}{-3} \begin{bmatrix} -3 & 3 & -6 \\ 1 & 0 & -3 \\ 2 & -3 & 6 \end{bmatrix} A−1=[−3−33−3−6−31−30−3−3−32−3−3−36−3]A^{-1} = \begin{bmatrix} \frac{-3}{-3} & \frac{3}{-3} & \frac{-6}{-3} \\ \frac{1}{-3} & \frac{0}{-3} & \frac{-3}{-3} \\ \frac{2}{-3} & \frac{-3}{-3} & \frac{6}{-3} \end{bmatrix} A−1=[1−12−1301−231−2]A^{-1} = \begin{bmatrix} 1 & -1 & 2 \\ -\frac{1}{3} & 0 & 1 \\ -\frac{2}{3} & 1 & -2 \end{bmatrix}