Question 22 of 58intermediate🔧 ApplyShort Answer2 marks

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

[123024005]\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}
Correct Answer

Step 1: Compute determinant. Expanding along first row, |A| = 1·(2·5 - 4·0) - 2·(0·5 - 4·0) + 3·(0·0 - 2·0) = 10. Since |A| ≠\neq 0, inverse exists.

Step 2: Find cofactors. A11 = 10, A12 = 0, A13 = 0; A21 = -10, A22 = 5, A23 = 0; A31 = 2, A32 = -4, A33 = 2.

Step 3: Form adjoint. adj A = transpose of cofactor matrix = [[10, -10, 2], [0, 5, -4], [0, 0, 2]].

Step 4: Compute inverse. A−1A^{-1} = (adj A)/|A| = (1/10) adj A = [[1, -1, 1/5], [0, 1/2, -2/5], [0, 0, 1/5]].

Hence, the inverse is [1−115012−250015]\begin{bmatrix} 1 & -1 & \frac{1}{5} \\ 0 & \frac{1}{2} & -\frac{2}{5} \\ 0 & 0 & \frac{1}{5} \end{bmatrix}.

Exercise: EXERCISE 4.4 | Q: 7 | (Page 17)
For More Understanding

Explanation

The solution follows the textbook method: first verify the determinant is non-zero to ensure invertibility. Then compute each cofactor using minors, form the adjoint as the transpose of the cofactor matrix, and finally divide by the determinant to obtain the inverse. The matrix is upper triangular, so the determinant is simply the product of diagonal entries, but expansion along the first row is shown to align with the example in the context.

Solution Steps

  1. Compute determinant |A| = 10, non-zero, so inverse exists.

  2. Find cofactors: A11=10, A12=0, A13=0; A21=-10, A22=5, A23=0; A31=2, A32=-4, A33=2.

  3. Form adjoint as transpose of cofactor matrix.

  4. Divide adjoint by determinant to get A−1A^{-1}.

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