Question 17 of 58intermediate🔧 ApplyShort Answer2 marks

Find adjoint of each of the matrices:

[1−12235−201]\begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}
Correct Answer

adj A = [31−11−125−1625]\begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}

Exercise: EXERCISE 4.4 | Q: 2 | (Page 17)
For More Understanding

Explanation

According to Definition 3 in the textbook, the adjoint of a square matrix A = [aij]n×n is the transpose of the matrix [Aij]n×n, where Aij is the cofactor of element aij. For a 3×33 \times 3 matrix, we first calculate all nine cofactors, arrange them in a matrix, and then take its transpose.

Solution Steps

  1. Step 1: Calculate cofactors for Row 1: A₁₁ = |3 5; 0 1| = 3, A₁₂ = -|2 5; -2 1| = -12, A₁₃ = |2 3; -2 0| = 6

  2. Step 2: Calculate cofactors for Row 2: A₂₁ = -|-1 2; 0 1| = 1, A₂₂ = |1 2; -2 1| = 5, A₂₃ = -|1 -1; -2 0| = 2

  3. Step 3: Calculate cofactors for Row 3: A₃₁ = |-1 2; 3 5| = -11, A₃₂ = -|1 2; 2 5| = -1, A₃₃ = |1 -1; 2 3| = 5

  4. Step 4: Form matrix of cofactors: [3 -12 6; 1 5 2; -11 -1 5]

  5. Step 5: Take transpose to get adj A = [3 1 -11; -12 5 -1; 6 2 5]

PreviousNext