Problem 21 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The inverse of the matrix is \( \begin{bmatrix} \frac{9}{4} & 0 & -2 \\ -\frac{3}{2} & -\frac{1}{2} & \frac{3}{2} \\ -\frac{5}{4} & \frac{1}{2} & -\frac{3}{2} \end{bmatrix} \)

Step 1: Check if the matrix is invertible

To find the inverse of a matrix, first, we need to check if the matrix is invertible. A matrix is invertible if its determinant is non-zero. Calculate the determinant of the matrix:\[\text{det}(A) = \begin{vmatrix} 0 & -2 & 2 \ 3 & 1 & 3 \ 1 & -2 & 3 \end{vmatrix}\]Use the expansion of the first row to calculate the determinant:\[\text{det}(A) = 0(\text{minor}) - (-2)(3(-2) - 1 \cdot 3) + 2(3(-2) - 1 \cdot 1)\]\[= -2(-6 - 3) + 2(-6 - 1)\]\[= -2 (-9) + 2(-7)\]\[= 18 - 14 = 4\]Since the determinant is 4, which is not zero, the matrix is invertible.

Step 2: Find the adjugate matrix

The next step is to find the adjugate (adjoint) of the matrix, which involves calculating the cofactor matrix and then transposing it. The cofactor of each element is found by finding the determinant of the 2x2 matrix that remains after removing the row and column of each element. Compute the cofactor matrix:\[\begin{bmatrix} \begin{vmatrix} 1 & 3 \ -2 & 3 \end{vmatrix} & -\begin{vmatrix} 3 & 3 \ 1 & 3 \end{vmatrix} & \begin{vmatrix} 3 & 1 \ 1 & -2 \end{vmatrix} \ -\begin{vmatrix} -2 & 2 \ -2 & 3 \end{vmatrix} & \begin{vmatrix} 0 & 2 \ 1 & 3 \end{vmatrix} & -\begin{vmatrix} 0 & -2 \ 1 & -2 \end{vmatrix} \ \begin{vmatrix} -2 & 2 \ 1 & 3 \end{vmatrix} & -\begin{vmatrix} 0 & 2 \ 3 & 3 \end{vmatrix} & \begin{vmatrix} 0 & -2 \ 3 & 1 \end{vmatrix}\end{bmatrix}\]Compute each minor and then the corresponding cofactor matrix:\[\begin{bmatrix} 1(3) - 3(-2) & -[3(3) - 3(1)] & 3(-2) - 1(1) \ -[-2(3) - (-2)(2)] & 0(3) - 2(1) & -[0(-2) - 2(1)] \ -2(3) - 2(1) & -[0(3) - 2(3)] & 0(1) - 2(3)\end{bmatrix}\]\[\begin{bmatrix} 3 + 6 & -[9 - 3] & -5 \ 0 & -2 & 2 \ -8 & 6 & -6\end{bmatrix}\]Transpose this cofactor matrix to find the adjugate:\[\text{adj}(A) =\begin{bmatrix} 9 & 0 & -8 \ -6 & -2 & 6 \ -5 & 2 & -6\end{bmatrix}\]

Step 3: Compute the inverse matrix

The inverse of a matrix is given by the formula:\[A^{-1} = \frac{1}{\text{det}(A)} \times \text{adj}(A)\]Given the determinant found earlier is \(4\), and the adjugate matrix is:\[\begin{bmatrix} 9 & 0 & -8 \ -6 & -2 & 6 \ -5 & 2 & -6\end{bmatrix}\]The inverse matrix is:\[A^{-1} = \frac{1}{4} \times \begin{bmatrix} 9 & 0 & -8 \ -6 & -2 & 6 \ -5 & 2 & -6\end{bmatrix} = \begin{bmatrix} \frac{9}{4} & 0 & -2 \ -\frac{3}{2} & -\frac{1}{2} & \frac{3}{2} \ -\frac{5}{4} & \frac{1}{2} & -\frac{3}{2}\end{bmatrix}\]

Step 4: Verify the inverse matrix

To ensure the accuracy of our inverse, we multiply the original matrix by its calculated inverse to see if we get the identity matrix.Multiply:\[\begin{bmatrix}0 & -2 & 2 \3 & 1 & 3 \1 & -2 & 3\end{bmatrix} \times \begin{bmatrix}\frac{9}{4} & 0 & -2 \-\frac{3}{2} & -\frac{1}{2} & \frac{3}{2} \-\frac{5}{4} & \frac{1}{2} & -\frac{3}{2}\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \0 & 1 & 0 \0 & 0 & 1\end{bmatrix}\]If the result is the identity matrix, the inverse is correct. After performing the multiplication, confirm the answer matches the identity matrix.

Key Concepts

Determinant Calculation

The determinant plays a vital role when it comes to finding the inverse of a matrix. It helps us determine whether a matrix is invertible. If the determinant of a matrix is zero, the matrix does not have an inverse. In our example, we began by calculating the determinant of the given matrix. To find the determinant of a 3x3 matrix, we use a method called cofactor expansion. This involves expanding along any row or column. In our solution, we used the first row. Start by identifying the minors for each element in the row and calculate them. Then, multiply each minor by its corresponding element from the row, applying the correct sign according to the position using the checkerboard pattern of signs (positive, negative, positive, etc.). Once all these calculations are done, sum them up to get the determinant. In this case, the detailed calculation yielded a determinant of 4, which is non-zero. Therefore, the matrix is invertible, and we could proceed to find its inverse.

Adjugate Matrix

The adjugate or adjoint of a matrix is essential for computing the inverse. This is because the inverse of a matrix involves dividing the adjugate by the determinant. The adjugate matrix is simply the transpose of the cofactor matrix. Finding cofactors requires us to compute the determinant of smaller 2x2 matrices by removing the row and column of each element in the original matrix. Remember that each position in the cofactor matrix needs to be adjusted using the checkerboard pattern of signs. After calculating the cofactors, organize them into a matrix - this is your cofactor matrix. Then, transpose it, which means swapping its rows and columns, to get the adjugate matrix. In our example, this involved numerous calculations of 2x2 determinants and the resulting adjugate was prepared by transposing the derived cofactor matrix.

Matrix Multiplication Verification

Once we obtained the inverse, it's crucial to verify its correctness. This is where matrix multiplication comes in handy. The original matrix multiplied by its inverse should yield an identity matrix.The identity matrix is special because when you multiply a matrix by it, you get the original matrix back. For a 3x3 matrix, the identity matrix is:\[\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}\]To verify our solution, we multiplied the original matrix with the inverse we calculated. Each element in the resulting product matrix should match the corresponding element in the identity matrix.If both matrices match perfectly, it confirms that our calculations for the inverse are correct. For our example, the multiplication indeed resulted in the identity matrix, validating the accuracy of our inverse.