Problem 15 - Algebra and Trigonometry

Question

Find the inverse of the matrix if it exists. \(\left[\begin{array}{rrr}{2} & {4} & {1} \\ {-1} & {1} & {-1} \\ {1} & {4} & {0}\end{array}\right]\)

Step-by-Step Solution

Answer

Inverse found with \(\frac{1}{7}\) times the adjugate.

Step 1: Determine the Determinant

To find the inverse of a matrix, first check if it is invertible by calculating the determinant. For a 3x3 matrix \(A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix}\), the determinant \( \text{det}(A) \) is given by: \[ \text{det}(A) = a(ei − fh) − b(di − fg) + c(dh − eg)\]Substituting the given matrix values:\[\text{det}(A) = 2(1\cdot 0 - (-1)\cdot 4) - 4((-1)\cdot 0 - (-1)\cdot 1) + 1((-1)\cdot 4 - 1\cdot 1) = 2\cdot 4 + 4\cdot 1 + 1(-5) = 8 + 4 - 5 = 7.\]The determinant is 7, so the matrix is invertible.

Step 2: Calculate the Matrix of Minors

The matrix of minors is obtained by calculating the determinant of each 2x2 minor of the original matrix. Each element in the matrix of minors is formed by removing the row and column of the element and calculating the determinant of the remaining 2x2 matrix.For example, the minor of the first element \(a = 2\) is determined by removing the first row and column:\[M_{11} = \begin{vmatrix}1 & -1 \ 4 & 0 \end{vmatrix} = (1)(0) - (-1)(4) = 4\]Calculating the rest:\[M_{12} = \begin{vmatrix} -1 & -1 \ 1 & 0 \end{vmatrix} = 0 - (-1) = 1\]\[M_{13} = \begin{vmatrix} -1 & 1 \ 1 & 4 \end{vmatrix} = (-1)(4) - (1)(1) = -4 - 1 = -5\]Repeat this process for each element in the matrix. The resulting matrix of minors is:\[\begin{bmatrix} 4 & 1 & -5 \ -4 & 2 & -5 \ 5 & 0 & 2 \end{bmatrix}\]

Step 3: Form the Cofactor Matrix

The cofactor matrix is obtained by applying a checkerboard pattern of signs (+/-) to the matrix of minors. This pattern begins with a '+' sign for the top-left element. The signs alternate as follows:\[\text{Cofactor Matrix} = \begin{bmatrix} +4 & -1 & +(-5) \ -(-4) & +2 & -(-5) \ +5 & -0 & +2 \end{bmatrix} = \begin{bmatrix} 4 & -1 & -5 \ 4 & 2 & 5 \ 5 & 0 & 2 \end{bmatrix}\]

Step 4: Calculate the Adjugate (Transpose of the Cofactor Matrix)

The adjugate (or adjoint) of a matrix is the transpose of its cofactor matrix. Transpose the cofactor matrix by swapping rows and columns:\[\text{Adjugate Matrix} = \begin{bmatrix} 4 & 4 & 5 \ -1 & 2 & 0 \ -5 & 5 & 2 \end{bmatrix}\]

Step 5: Compute the Inverse of the Matrix

To find the inverse of the original matrix \(A\), use the formula:\[A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{Adjugate Matrix}\]We found earlier that \( \text{det}(A) = 7 \). Therefore, the inverse matrix \( A^{-1} \) is:\[A^{-1} = \frac{1}{7} \cdot \begin{bmatrix} 4 & 4 & 5 \ -1 & 2 & 0 \ -5 & 5 & 2 \end{bmatrix} = \begin{bmatrix} \frac{4}{7} & \frac{4}{7} & \frac{5}{7} \ -\frac{1}{7} & \frac{2}{7} & 0 \ -\frac{5}{7} & \frac{5}{7} & \frac{2}{7} \end{bmatrix}\]

Key Concepts

Determinant Calculation

The determinant of a matrix is a crucial step in determining whether a matrix is invertible. For a 3x3 matrix, we can calculate the determinant using the formula:

• For matrix \[A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix}\]
• The determinant, denoted as \( \text{det}(A) \), is computed as\[a(ei - fh) - b(di - fg) + c(dh - eg)\]

Substitute the specific values of the given matrix into this formula.
For instance, using the matrix:\[\begin{bmatrix}2 & 4 & 1 \-1 & 1 & -1 \1 & 4 & 0\end{bmatrix}\]Plugging in the values:\[\text{det}(A) = 2(1\cdot0 - (-1)\cdot4) - 4((-1)\cdot0 - (-1)\cdot1) + 1((-1)\cdot4 - 1\cdot1)\]After calculation, the determinant is 7, which shows the matrix is invertible because the determinant is not zero.
Understanding how to compute and interpret determinants is fundamental when working with matrices.

Matrix of Minors

The matrix of minors is the foundation for further steps in finding the inverse. Each element in the matrix of minors corresponds to the determinant of a 2x2 submatrix created by removing the row and column of the respective element from the original matrix.

• To find a minor, exclude the row and column where the element is located.
• Calculate the determinant of the remaining 2x2 matrix.For example, to find the minor of the first element \((a = 2)\), consider:\[M_{11} = \begin{vmatrix} 1 & -1 \ 4 & 0 \end{vmatrix} = (1)(0) - (-1)(4) = 4\]

Repeat this process to find all the minors:\[\begin{bmatrix}4 & 1 & -5 \-4 & 2 & -5 \5 & 0 & 2\end{bmatrix}\]Understanding how to form the matrix of minors lays the groundwork for finding the cofactor matrix.

Cofactor Matrix

The cofactor matrix introduces a pattern of alternating signs to the matrix of minors. This step adjusts the signs of the elements of the matrix of minors using a checkerboard pattern of plus and minus signs.

• Start with a '+' sign at the top-left and alternate as you move along rows and columns.
• Apply this pattern: + - +, - + -, + - +.For the minor calculated above, the cofactor is given by:\[\begin{bmatrix}+4 & -1 & -5 \4 & +2 & -5 \5 & -0 & +2\end{bmatrix}\]
• Sign switching is done as follows:\[\begin{bmatrix}4 & -1 & -5 \4 & 2 & 5 \5 & 0 & 2\end{bmatrix}\]

The cofactor matrix is a vital player in transforming the matrix of minors to prepare for the adjugate matrix.

Adjugate Matrix

The adjugate, or adjoint, of a matrix is the transpose of its cofactor matrix. This step sets the stage for calculating the matrix inverse.

• To find the adjugate, swap the rows and columns of the cofactor matrix.
• Transposing the cofactor previously calculated gives:\[\begin{bmatrix}4 & 4 & 5 \-1 & 2 & 0 \-5 & 5 & 2\end{bmatrix}\]

The adjugate matrix retains all the properties necessary for making the upcoming inverse calculation possible.

Inverse Matrix Computation

Finally, computing the inverse matrix utilizes both the determinant and the adjugate. The inverse of matrix \(A\) is calculated using the formula:

• \[A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{Adjugate Matrix}\]
• From our calculations, \(\text{det}(A) = 7\).

Inserting the adjugate matrix result:\[A^{-1} = \frac{1}{7} \cdot \begin{bmatrix}4 & 4 & 5 \-1 & 2 & 0 \-5 & 5 & 2\end{bmatrix} = \begin{bmatrix}\frac{4}{7} & \frac{4}{7} & \frac{5}{7} \-\frac{1}{7} & \frac{2}{7} & 0 \-\frac{5}{7} & \frac{5}{7} & \frac{2}{7}\end{bmatrix}\]This inverse matrix can now be used to solve systems of equations or find transformations, provided the original determinant isn't zero.