Problem 16 - Precalculus Mathematics for Calculus

Question

Find the inverse of the matrix if it exists. $$\left[\begin{array}{lll} 4 & 2 & 3 \\ 3 & 3 & 2 \\ 1 & 0 & 1 \end{array}\right]$$

Step-by-Step Solution

Answer

The inverse is \(\begin{bmatrix} 3 & 2 & -5 \\ -1 & 1 & 1 \\ -3 & -3 & 6 \end{bmatrix}\).

Step 1: Check if the Matrix is Invertible

To find the inverse of a matrix, we must first determine if it is invertible by checking if its determinant is non-zero. The matrix is invertible only if its determinant is not equal to zero. We calculate the determinant for the given matrix: \[\text{det}(A) = \begin{vmatrix} 4 & 2 & 3 \ 3 & 3 & 2 \ 1 & 0 & 1 \end{vmatrix}\] Using the formula for a 3x3 determinant:\[\text{det}(A) = 4(3 \cdot 1 - 2 \cdot 0) - 2(3 \cdot 1 - 2 \cdot 1) + 3(3 \cdot 0 - 3 \cdot 1)\]\[= 4(3) - 2(1) - 3(3)\]\[= 12 - 2 - 9 = 1\]Since the determinant is 1, the matrix is invertible.

Step 2: Find the Adjoint of the Matrix

The adjoint of a matrix is the transpose of its cofactor matrix. We first find the cofactors of the original matrix:1. Minor of 1st element (4): \[\begin{vmatrix} 3 & 2 \ 0 & 1 \end{vmatrix} = 3(1) - 2(0) = 3\]2. Minor of 2nd element (2): \[\begin{vmatrix} 3 & 2 \ 1 & 1 \end{vmatrix} = 3(1) - 2(1) = 1\] (change sign due to position)3. Minor of 3rd element (3): \[\begin{vmatrix} 3 & 3 \ 1 & 0 \end{vmatrix} = 3(0) - 3(1) = -3\]4. Minor of 4th element (3): \[\begin{vmatrix} 2 & 3 \ 0 & 1 \end{vmatrix} = 2(1) - 3(0) = 2\] (change sign due to position)5. Minor of 5th element (3): \[\begin{vmatrix} 4 & 3 \ 1 & 1 \end{vmatrix} = 4(1) - 3(1) = 1\]6. Minor of 6th element (2): \[\begin{vmatrix} 4 & 3 \ 1 & 0 \end{vmatrix} = 4(0) - 3(1) = -3\] (change sign due to position)7. Minor of 7th element (1): \[\begin{vmatrix} 2 & 3 \ 3 & 2 \end{vmatrix} = 2(2) - 3(3) = -5\]8. Minor of 8th element (0): \[\begin{vmatrix} 4 & 3 \ 3 & 2 \end{vmatrix} = 4(2) - 3(3) = -1\] (change sign due to position)9. Minor of 9th element (1): \[\begin{vmatrix} 4 & 2 \ 3 & 3 \end{vmatrix} = 4(3) - 2(3) = 6\]Now, we'll write out the adjugate matrix using these cofactors:\[\text{Cofactor Matrix} = \begin{bmatrix} 3 & -1 & -3 \ 2 & 1 & -3 \ -5 & 1 & 6 \end{bmatrix}\]\[\text{Adjugate (Transpose of Cofactor Matrix)}= \begin{bmatrix} 3 & 2 & -5 \ -1 & 1 & 1 \ -3 & -3 & 6 \end{bmatrix}\]

Step 3: Calculate the Inverse Using the Determinant and Adjugate

The inverse of a matrix is given by the formula:\[A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{Adjugate}(A)\]Since the determinant is 1, the inverse is simply the adjugate matrix itself:\[A^{-1} = \begin{bmatrix} 3 & 2 & -5 \ -1 & 1 & 1 \ -3 & -3 & 6 \end{bmatrix}\]

Key Concepts

Determinant Calculation

The determinant calculation is a crucial first step in determining if a matrix is invertible. For a 3x3 matrix, like the one in this exercise, the determinant helps us understand whether an inverse exists. If the determinant is non-zero, the matrix is invertible.

The determinant of a 3x3 matrix is calculated using a specific pattern. The given matrix:

• \[\begin{bmatrix} 4 & 2 & 3 \ 3 & 3 & 2 \ 1 & 0 & 1 \end{bmatrix}\]

requires multiplication and subtraction steps that are dictated by its structure. We start by multiplying across three diagonals and adjusting their result by the signs of the diagonals.
The formula used is:

• \[\text{det}(A) = 4(3 \times 1 - 2 \times 0) - 2(3 \times 1 - 2 \times 1) + 3(3 \times 0 - 3 \times 1) \]

The calculations step-by-step were:

• \[= 4(3) - 2(1) - 3(3)\]
• \[= 12 - 2 - 9 = 1\]

With a non-zero determinant of 1, we confirm that this matrix indeed has an inverse.

Cofactor Matrix

Constructing the cofactor matrix involves finding the minors of each element in the original matrix, then adjusting the sign based on the element's position. This matrix will be crucial for calculating the adjugate matrix.

For each element in the matrix:

• Identify the minor by removing the row and column of the current element and calculating the determinant of the remaining 2x2 matrix.
• Apply the sign change based on the position using the pattern of \[ +, -, + \] within the matrix.

For the given matrix, the minors and sign-adjusted results were:

• Minor(1, 1): \[ \begin{vmatrix} 3 & 2 \ 0 & 1 \end{vmatrix} = 3 \]
• Minor(1, 2): \[ \begin{vmatrix} 3 & 2 \ 1 & 1 \end{vmatrix} = -1 \] (Change sign due to position)
• And so on through all elements...

The ensuing cofactor matrix became:

• \[\begin{bmatrix} 3 & -1 & -3 \ 2 & 1 & -3 \ -5 & -1 & 6 \end{bmatrix}\]

This cofactor matrix is key to obtaining the adjugate matrix.

Adjugate Matrix

The adjugate matrix is used in finding the inverse of an invertible matrix. It is derived by transposing the cofactor matrix, an operation that switches its rows with columns.

The cofactor matrix found earlier:

• \[\begin{bmatrix}3 & -1 & -3 \2 & 1 & -3 \-5 & 1 & 6\end{bmatrix} \]

By transposing, we form the adjugate matrix:

• \[\begin{bmatrix}3 & 2 & -5 \-1 & 1 & 1 \-3 & -3 & 6\end{bmatrix} \]

Transposing means simply flip rows to columns:

• Original row 1 becomes column 1
• Original row 2 becomes column 2
• Original row 3 becomes column 3

Inverting the matrix uses this adjugate matrix. Since the determinant was 1, the inverse is directly this adjugate matrix itself; therefore:

• \[A^{-1} = \begin{bmatrix} 3 & 2 & -5 \ -1 & 1 & 1 \ -3 & -3 & 6 \end{bmatrix} \]

Understanding this conclusion is key for operations that involve inverses.