Question 20 of 58beginner🔧 ApplyShort Answer2 marks

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

[2−243]\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}
Correct Answer

A−1A^{-1} = 114[32−42]\frac{1}{14} \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}

Exercise: EXERCISE 4.4 | Q: 5 | (Page 17)
For More Understanding

Explanation

To find the inverse of a 2×22 \times 2 matrix, we first verify that the determinant is non-zero, then use the formula A−1A^{-1} = (1/|A|) × adj(A). The adjoint of a 2×22 \times 2 matrix is obtained by interchanging diagonal elements and changing signs of off-diagonal elements.

Solution Steps

  1. Step 1: Find the determinant |A| = (2)(3) - (-2)(4) = 6 + 8 = 14. Since |A| ≠\neq 0, inverse exists.

  2. Step 2: Find adj(A) by interchanging diagonal elements and changing signs of off-diagonal elements: adj(A) = [32−42]\begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}

  3. Step 3: Apply formula A−1A^{-1} = (1/|A|) × adj(A) = (1/14) × [32−42]=114[32−42]\begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix} = \frac{1}{14} \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}

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