Problem 18
Question
Find the inverse of the matrix if it exists. $$ \left[\begin{array}{rrr}{5} & {7} & {4} \\ {3} & {-1} & {3} \\ {6} & {7} & {5}\end{array}\right] $$
Step-by-Step Solution
Verified Answer
The inverse exists and is \( \begin{bmatrix} 26 & 7 & -25 \\ -3 & -5 & 3 \\ -27 & -7 & 26 \end{bmatrix} \).
1Step 1: Calculate the Determinant
First, we need to check if the given matrix has an inverse, which only exists if the determinant of the matrix is non-zero. The determinant of a 3x3 matrix: \[ \begin{vmatrix} a & b & c \ d & e & f \ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg) \]For our matrix \( A = \begin{bmatrix} 5 & 7 & 4 \ 3 & -1 & 3 \ 6 & 7 & 5 \end{bmatrix} \), substitute the values:\[ \text{det}(A) = 5((-1)(5) - (3)(7)) - 7((3)(5) - (3)(6)) + 4((3)(7) - (-1)(6)) \] Calculating: \[ = 5(-5 - 21) - 7(15 - 18) + 4(21 + 6) \] \[ = 5(-26) + 7(3) + 4(27) \] \[ = -130 + 21 + 108 \] \[ = -1 \].Since the determinant is not zero, the matrix has an inverse.
2Step 2: Calculate the Matrix of Minors
Calculate each minor of the matrix, which involves the determinant of each 2x2 submatrix obtained by removing one row and one column. For example, the minor of element at (1,1) is the determinant of its complementary 2x2 matrix:For \( M_{11} \):\[ \begin{vmatrix} -1 & 3 \ 7 & 5 \end{vmatrix} = (-1)(5) - (3)(7) = -5 - 21 = -26 \].Proceed similarly for all minors:\( M_{12} = \begin{vmatrix} 3 & 3 \ 6 & 5 \end{vmatrix} = -3 \)\( M_{13} = \begin{vmatrix} 3 & -1 \ 6 & 7 \end{vmatrix} = 27 \)\( M_{21} = \begin{vmatrix} 7 & 4 \ 7 & 5 \end{vmatrix} = 7 \)\( M_{22} = \begin{vmatrix} 5 & 4 \ 6 & 5 \end{vmatrix} = 5 \)\( M_{23} = \begin{vmatrix} 5 & 7 \ 6 & 7 \end{vmatrix} = -7 \)\( M_{31} = \begin{vmatrix} 7 & 4 \ -1 & 3 \end{vmatrix} = 25 \)\( M_{32} = \begin{vmatrix} 5 & 4 \ 3 & 3 \end{vmatrix} = 3 \)\( M_{33} = \begin{vmatrix} 5 & 7 \ 3 & -1 \end{vmatrix} = -26 \).
3Step 3: Calculate the Cofactor Matrix
The cofactor matrix is obtained by applying a sign change according to the positions to the matrix of minors:\[\begin{bmatrix}+M_{11} & -M_{12} & +M_{13} \-M_{21} & +M_{22} & -M_{23} \+M_{31} & -M_{32} & +M_{33}\end{bmatrix} = \begin{bmatrix}-26 & 3 & 27 \-7 & 5 & 7 \25 & -3 & -26\end{bmatrix}\].
4Step 4: Transpose the Cofactor Matrix
Transpose the cofactor matrix to obtain the adjugate matrix. Transposing means swapping the rows and columns:\[\begin{bmatrix}-26 & -7 & 25 \3 & 5 & -3 \27 & 7 & -26\end{bmatrix}\].
5Step 5: Multiply by the Reciprocal of the Determinant
Finally, the inverse of the matrix is obtained by multiplying the adjugate matrix by the reciprocal of the determinant (which was found to be \(-1\)). The inverse \( A^{-1} \) is:\[A^{-1} = -1 \times \begin{bmatrix}-26 & -7 & 25 \3 & 5 & -3 \27 & 7 & -26\end{bmatrix} = \begin{bmatrix}26 & 7 & -25 \-3 & -5 & 3 \-27 & -7 & 26\end{bmatrix}\].
Key Concepts
DeterminantMatrix of MinorsCofactor MatrixAdjugate Matrix
Determinant
The determinant is a special number that can be calculated from a square matrix. Imagine the determinant as a measure of a matrix's "volume" or how much space it fills. For a given 3x3 matrix, the formula involves the elements of the matrix:\[\begin{vmatrix} a & b & c \ d & e & f \ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg)\]The determinant helps us understand whether a matrix has an inverse. If the determinant is zero, the matrix does not have an inverse. When calculating for our specific matrix, we found the determinant to be -1, which is non-zero. Hence, the matrix has an inverse.
This makes the determinant a crucial step in finding the inverse.
This makes the determinant a crucial step in finding the inverse.
Matrix of Minors
A matrix of minors involves calculating a smaller matrix's determinant within the larger matrix. To find a minor for an element, remove its row and column, then calculate the determinant of the resulting 2x2 matrix. For instance, if you take the element in the first row and first column of our original matrix, the corresponding minor is computed from:\[\begin{vmatrix} -1 & 3 \ 7 & 5 \end{vmatrix}\]The operation here results in \(-26\) for the minor. Repeat this process for every element in the matrix to form the complete matrix of minors.
This step helps in preparing the data necessary to form the cofactor matrix.
This step helps in preparing the data necessary to form the cofactor matrix.
Cofactor Matrix
The cofactor matrix is derived from the matrix of minors by applying alternations of signs. This is often referred to as the checkerboard pattern of pluses and minuses. For the same position of each minor matrix element, multiply the minor by \( (-1)^{i+j} \), where i and j represent the row and column indices respectively. This turns the matrix of minors into:\[\begin{bmatrix}+M_{11} & -M_{12} & +M_{13} \-M_{21} & +M_{22} & -M_{23} \+M_{31} & -M_{32} & +M_{33}\end{bmatrix} =\begin{bmatrix}-26 & 3 & 27 \-7 & 5 & 7 \25 & -3 & -26\end{bmatrix}\]This process is essential to obtaining the adjugate matrix, a step closer to finding the inverse.
Adjugate Matrix
The adjugate matrix is essentially the transpose of the cofactor matrix. Transposing a matrix means converting its rows into columns and vice versa. This transformation shifts our understanding of the matrix to be prepared for calculating the inverse. For our example, the adjugate matrix becomes:\[\begin{bmatrix}-26 & -7 & 25 \3 & 5 & -3 \27 & 7 & -26\end{bmatrix}\]From here, the final step towards the matrix inverse involves multiplying the adjugate matrix by the reciprocal of the determinant of the original matrix. This yields the inverse matrix that allows us to compute different operations, like solving systems of equations with this matrix.
Other exercises in this chapter
Problem 18
A matrix is given. (a) Determine whether the matrix is in row-echelon form. (b) Determine whether the matrix is in reduced row-echelon form. (c) Write the syste
View solution Problem 18
\(13-44=\) Find the partial fraction decomposition of the rational function. $$ \frac{x-12}{x^{2}-4 x} $$
View solution Problem 18
Solve the matrix equation for the unknown matrix \(X,\) or explain why no solution exists. $$ \begin{array}{l}{A=\left[\begin{array}{ll}{4} & {6} \\ {1} & {3}\e
View solution Problem 18
\(17-36\) . Find the complete solution of the linear system, or show that it is inconsistent. $$ \left\\{\begin{aligned} x-y+z &=0 \\ y+2 z &=-2 \\ x+y-z &=2 \e
View solution