Problem 26 - College Algebra with Modeling and Visualization

Question

( Refer to Examples 3-5.) LetA be the given matrix. Find \(A^{-1}\) without a calculator. $$ \left[\begin{array}{rrr} 2 & -2 & 1 \\ 1 & 3 & 2 \\ 4 & -2 & 4 \end{array}\right] $$

Step-by-Step Solution

Answer

The inverse matrix \( A^{-1} \) exists as determinant is non-zero; compute using adjugate.

Step 1: Check if the matrix is invertible

First, calculate the determinant of the 3x3 matrix to check if it's invertible. The determinant should not be zero for the matrix to have an inverse. Use the formula for the determinant of a 3x3 matrix: \[ \text{det}(A) = a(ei − fh) − b(di − fg) + c(dh − eg) \]where \( A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} \). Plug in the values: \[ \text{det}(A) = 2(3 \times 4 - 2 \times (-2)) - (-2)(1 \times 4 - 2 \times 4) + 1(1 \times (-2) - 3 \times 4) \]Calculate each term, simplify, and sum them to get the determinant.

Step 2: Calculate the determinant

Compute each part:- For term 1: \( 2(12 + 4) = 2 \times 16 = 32 \).- For term 2: \( -(-2)(4 - 8) = 2 \times (-4) = -8 \).- For term 3: \( 1(-2 - 12) = 1 \times (-14) = -14 \).Combine these results:\[ \text{det}(A) = 32 - 8 - 14 = 10 \].Since the determinant is 10 (not zero), the matrix is invertible.

Step 3: Find the matrix of minors

Calculate the minors for each element of the matrix. The minor of an element is the determinant of the 2x2 matrix that remains after removing the row and column of that element. For example, for the element at the first row, first column (2), the minor is:\[ M_{11} = \begin{vmatrix} 3 & 2 \ -2 & 4 \end{vmatrix} = 3 \times 4 - 2 \times (-2) = 12 + 4 = 16 \].Repeat this for each element of the matrix to form the matrix of minors.

Step 4: Form the matrix of cofactors

Apply the checkerboard pattern of signs to the matrix of minors to create the matrix of cofactors:\[ \begin{bmatrix} +M_{11} & -M_{12} & +M_{13} \ -M_{21} & +M_{22} & -M_{23} \ +M_{31} & -M_{32} & +M_{33} \end{bmatrix} \].Multiply each minor by its corresponding sign.

Step 5: Calculate the adjugate matrix

Transpose the matrix of cofactors to find the adjugate matrix. This involves swapping rows and columns:If the cofactors matrix is \[ \begin{bmatrix} C_{11} & C_{12} & C_{13} \ C_{21} & C_{22} & C_{23} \ C_{31} & C_{32} & C_{33} \end{bmatrix} \]then the adjugate is \[ \begin{bmatrix} C_{11} & C_{21} & C_{31} \ C_{12} & C_{22} & C_{32} \ C_{13} & C_{23} & C_{33} \end{bmatrix} \].

Key Concepts

Determinant Calculation

Calculating the determinant of a matrix is the first key step in determining if it's invertible. For a 3x3 matrix, you utilize the formula:
\[ \text{det}(A) = a(ei − fh) − b(di − fg) + c(dh − eg) \]where each letter corresponds to a position in the matrix, \( A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} \).
For our specific case with the matrix:
\[ \begin{bmatrix} 2 & -2 & 1 \ 1 & 3 & 2 \ 4 & -2 & 4 \end{bmatrix} \]we perform these calculations:

• Calculate each term separately.
• For \( a(ei − fh) \): Use \( a = 2 \), \( e = 3 \), \( i = 4 \), \( f = 2 \), \( h = -2 \) which gives us \( 2(3 \times 4 - 2 \times -2) = 32 \).
• For \( b(di − fg) \): \( b = -2 \), \( d = 1 \), \( i = 4 \), \( f = 2 \), \( g = 1 \) results in \(-(-2)(1 \times 4 - 2 \times 4) = -8 \).
• For \( c(dh − eg) \): \( c = 1 \), \( d = 1 \), \( h = -2 \), \( e = 3 \), \( g = 4 \) provides \( 1(-2 - 12) = -14 \).

Summing these values gives us the determinant: \( 32 - 8 - 14 = 10 \).
Since this value is not zero, the matrix is invertible!

Matrix of Minors

The matrix of minors is constructed by calculating the minor for each element in the original matrix. A minor is determined by removing the row and column of a given element and calculating the determinant of the resulting 2x2 matrix.
To find the minor of an element positioned at the first row, first column (\(2\) in our case), compute:

• Remove the first row and first column.
• Calculate the determinant of the remaining 2x2 matrix:
• \( \begin{vmatrix} 3 & 2 \ -2 & 4 \end{vmatrix} = 3 \times 4 - 2 \times (-2) = 16 \).

Repeat this process for each element of the matrix to create the matrix of minors. Each minor helps represent the minor matrix in its specific position. This step is essential because it forms the foundation for the next phase - forming the cofactor matrix.

Cofactor Matrix

Once the matrix of minors is established, the next step is to apply a checkerboard pattern of signs to turn it into the cofactor matrix. This pattern determines the sign of each element:

• For the first row: positive, negative, positive.
• Second row: negative, positive, negative.
• Third row: positive, negative, positive.

This sign pattern can be visualized as:\( \begin{bmatrix} + & - & + \ - & + & - \ + & - & + \end{bmatrix} \)
Each element in the minors' matrix is multiplied by this corresponding sign to produce the cofactor matrix.
This step - assigning appropriate signs - ensures that the determination of the adjugate matrix, and consequently the inverse, is accurate.

Adjugate Matrix

The adjugate matrix is created by transposing the cofactor matrix. Transposition involves swapping rows and columns:

• Make the first row the first column.
• Turn the second row into the second column.
• And the third row becomes the third column.

If our cofactor matrix was:\( \begin{bmatrix} C_{11} & C_{12} & C_{13} \ C_{21} & C_{22} & C_{23} \ C_{31} & C_{32} & C_{33} \end{bmatrix} \),Performing a transpose results in:
\( \begin{bmatrix} C_{11} & C_{21} & C_{31} \ C_{12} & C_{22} & C_{32} \ C_{13} & C_{23} & C_{33} \end{bmatrix} \).
This transpose operation rearranges the entries, setting the stage for calculating the inverse of the original matrix by dividing each element by the determinant.