Question 25 of 58intermediate🔧 ApplyShort Answer2 marks

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

[1−1202−33−24]\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}
Correct Answer

The inverse of the matrix is [−20192−361−2]\begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}.

Exercise: EXERCISE 4.4 | Q: 10 | (Page 18)
For More Understanding

Explanation

The textbook context directly provides the solution to this question. It shows that when the given matrix is multiplied by the matrix [−20192−361−2]\begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}, the product equals the identity matrix II. This confirms that the second matrix is indeed the inverse of the given matrix. The calculation shows each element of the product matrix simplifies to give [100010001]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, verifying the inverse relationship.

Solution Steps

  1. Step 1: Let A=[1−1202−33−24]A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} and verify that ∣A∣≠0|A| \neq 0 (matrix is nonsingular).

  2. Step 2: Find the inverse by computing A−1A^{-1} such that A⋅A−1=IA \cdot A^{-1} = I.

  3. Step 3: The inverse is given by A−1=[−20192−361−2]A^{-1} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}, which can be verified by multiplication: A⋅A−1=[100010001]=IA \cdot A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I.

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