If A is an invertible matrix of order 2, then det(A^-1) is equal to

Question 33 of 58intermediate🧠 RememberMCQ1 mark

If A is an invertible matrix of order 2, then det( $A^{-1}$ ) is equal to

Exercise: EXERCISE 4.4 | Q: 18 | (Page 18)

For More Understanding

Explanation

Option (B) is correct. We know that for an invertible matrix A, the product of the matrix and its inverse is the identity matrix, i.e., $A A^{-1} = I$ . Taking the determinant of both sides, we get $|A A^{-1}| = |I|$ . Using the property of determinants that the determinant of a product is the product of the determinants, this becomes $|A| \cdot |A^{-1}| = |I|$ . Since the determinant of the identity matrix of order 2 is 1, we have $|A| \cdot |A^{-1}| = 1$ . Rearranging this equation gives $|A^{-1}| = \frac{1}{|A|}$ .

Solution Steps

Step 1: Use the definition of an invertible matrix provided in the context: $A A^{-1} = I$ .
Step 2: Apply the determinant operator to both sides: $|A A^{-1}| = |I|$ .
Step 3: Apply the product rule for determinants: $|A| \cdot |A^{-1}| = |I|$ .
Step 4: Evaluate $|I|$ for a matrix of order 2. Using the determinant definition $\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} = a_{11}a_{22} - a_{21}a_{12}$ , for $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ , $|I| = 1(1) - 0(0) = 1$ .
Step 5: Solve for $|A^{-1}|$ : $|A| \cdot |A^{-1}| = 1 \implies |A^{-1}| = \frac{1}{|A|}$ .

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