Question 24 of 58intermediate🔧 ApplyShort Answer2 marks

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

[2134−10−721]\begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}
Correct Answer

Let A = [2134−10−721]\begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}

Step 1: Find |A|

|A| = 2[(-1)(1) - (0)(2)] - 1[(4)(1) - (0)(-7)] + 3[(4)(2) - (-1)(-7)] = 2(-1) - 1(4) + 3(8-7) = -2 - 4 + 3 = -3 ≠\neq 0

Since |A| ≠\neq 0, inverse exists.

Step 2: Find cofactors

A₁₁ = -1, A₁₂ = -4, A₁₃ = 1 A₂₁ = 5, A₂₂ = 23, A₂₃ = -11 A₃₁ = 3, A₃₂ = 12, A₃₃ = -6

Step 3: Find adj(A)

adj(A) = [−153−423121−11−6]\begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix}

Step 4: Find A⁻¹

A⁻¹ = 1∣A∣\frac{1}{|A|} × adj(A) = 1−3[−153−423121−11−6]\frac{1}{-3} \begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix}

A⁻¹ = [13−53−143−233−4−131132]\begin{bmatrix} \frac{1}{3} & -\frac{5}{3} & -1 \\ \frac{4}{3} & -\frac{23}{3} & -4 \\ -\frac{1}{3} & \frac{11}{3} & 2 \end{bmatrix}

Exercise: EXERCISE 4.4 | Q: 9 | (Page 18)
For More Understanding

Explanation

This question is from Exercise 4.4, Question 9. The solution follows the standard method taught in NCERT: first verify the determinant is non-zero (to confirm inverse exists), then calculate all 9 cofactors using the minor method, transpose to get the adjoint, and finally apply the formula A⁻¹ = (1/|A|) × adj(A). The context shows similar calculations in Example 17 where cofactors are computed element-wise for a 3×33 \times 3 matrix.

Solution Steps

  1. Step 1: Calculate determinant |A| = -3 ≠\neq 0, so inverse exists

  2. Step 2: Calculate all 9 cofactors Aᵢⱼ using minor determinants

  3. Step 3: Form adj(A) by transposing the cofactor matrix

  4. Step 4: Apply formula A⁻¹ = (1/|A|) × adj(A)

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