Question 26 of 58intermediate🔧 ApplyShort Answer2 marks

Find the inverse of each of the matrices (if it exists):

[1000cos⁡αsin⁡α0sin⁡α−cos⁡α]\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{bmatrix}
Correct Answer

First, compute the determinant |A|. Expanding along the first row: |A| = 1·det([[cosα, sinα],[sinα, -cosα]]) = -cos²α - sin²α = -1. Since |A| ≠\neq 0, the inverse exists.

Next, find the cofactor matrix. The cofactors are: C11 = -1, C12 = 0, C13 = 0; C21 = 0, C22 = -cosα, C23 = -sinα; C31 = 0, C32 = -sinα, C33 = cosα. The adjoint is the transpose of the cofactor matrix, which is the same matrix: adj(A) = [[-1,0,0],[0,-cosα,-sinα],[0,-sinα,cosα]].

Finally, A−1A^{-1} = adj(A)/|A| = adj(A)/(-1) = -adj(A) = [[1,0,0],[0,cosα,sinα],[0,sinα,-cosα]]. Hence, the inverse is the matrix itself.

Exercise: EXERCISE 4.4 | Q: 11 | (Page 18)
For More Understanding

Explanation

The student correctly applied the method: compute determinant to check invertibility, then find cofactors, adjoint, and finally use A−1A^{-1} = adj(A)/|A|. The matrix turns out to be its own inverse, which is consistent with the property that A·A = I.

Solution Steps

  1. Step 1: Compute determinant |A| = -1 (non-zero, so inverse exists).

  2. Step 2: Find cofactor matrix: C11 = -1, C12 = 0, C13 = 0; C21 = 0, C22 = -cosα, C23 = -sinα; C31 = 0, C32 = -sinα, C33 = cosα.

  3. Step 3: Adjoint is transpose of cofactor matrix, which is the same matrix.

  4. Step 4: Inverse = adj(A)/|A| = -adj(A) = [[1,0,0],[0,cosα,sinα],[0,sinα,-cosα]].

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