Question

question 12
find the inverse of the following matrix, if possible. reduce all fractions to lowest terms.
$a = \

$$\begin{bmatrix}-2 & 0 & -4\\\\ -3 & -5 & 2\\\\ 1 & 2 & 1\\end{bmatrix}$$

$
answer 2 points
selecting an option will display any text boxes needed to complete your answer.
does this matrix have an inverse?
yes no

Explanation:

Step1: Calculate the determinant of matrix \( A \)

The matrix \( A =

$$\begin{bmatrix}-2&0&-4\\-3&-5&2\\1&2&1\end{bmatrix}$$

\). The determinant of a \( 3 \times 3 \) matrix \(

$$\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}$$

\) is given by \( \det(A)=a(ei - fh)-b(di - fg)+c(dh - eg) \).

Substituting the values:

$$
LATEXBLOCK2
$$

Since the determinant \( \det(A) = 22
eq0 \), the matrix is invertible (so the answer to "Does this matrix have an inverse?" is Yes).

Step2: Find the matrix of minors

The minor of an element \( a_{ij} \) is the determinant of the submatrix obtained by removing row \( i \) and column \( j \).

• Minor of \( -2 \) (row 1, column 1): \( \det

$$\begin{bmatrix}-5&2\\2&1\end{bmatrix}$$

=(-5)(1)-(2)(2)=-5 - 4=-9 \)

• Minor of \( 0 \) (row 1, column 2): \( \det

$$\begin{bmatrix}-3&2\\1&1\end{bmatrix}$$

=(-3)(1)-(2)(1)=-3 - 2=-5 \)

• Minor of \( -4 \) (row 1, column 3): \( \det

$$\begin{bmatrix}-3&-5\\1&2\end{bmatrix}$$

=(-3)(2)-(-5)(1)=-6 + 5=-1 \)

• Minor of \( -3 \) (row 2, column 1): \( \det

$$\begin{bmatrix}0&-4\\2&1\end{bmatrix}$$

=(0)(1)-(-4)(2)=0 + 8 = 8 \)

• Minor of \( -5 \) (row 2, column 2): \( \det

$$\begin{bmatrix}-2&-4\\1&1\end{bmatrix}$$

=(-2)(1)-(-4)(1)=-2 + 4 = 2 \)

• Minor of \( 2 \) (row 2, column 3): \( \det

$$\begin{bmatrix}-2&0\\1&2\end{bmatrix}$$

=(-2)(2)-(0)(1)=-4 - 0=-4 \)

• Minor of \( 1 \) (row 3, column 1): \( \det

$$\begin{bmatrix}0&-4\\-5&2\end{bmatrix}$$

=(0)(2)-(-4)(-5)=0 - 20=-20 \)

• Minor of \( 2 \) (row 3, column 2): \( \det

$$\begin{bmatrix}-2&-4\\-3&2\end{bmatrix}$$

=(-2)(2)-(-4)(-3)=-4 - 12=-16 \)

• Minor of \( 1 \) (row 3, column 3): \( \det

$$\begin{bmatrix}-2&0\\-3&-5\end{bmatrix}$$

=(-2)(-5)-(0)(-3)=10 - 0 = 10 \)

Matrix of minors: \(

$$\begin{bmatrix}-9&-5&-1\\8&2&-4\\-20&-16&10\end{bmatrix}$$

\)

Step3: Find the matrix of cofactors

The cofactor matrix is obtained by multiplying the minors by \( (-1)^{i + j} \), where \( i \) is the row number and \( j \) is the column number.

• Cofactor of \( -9 \) (row 1, column 1): \( (-1)^{1+1}(-9)=-9 \)
• Cofactor of \( -5 \) (row 1, column 2): \( (-1)^{1+2}(-5)=5 \)
• Cofactor of \( -1 \) (row 1, column 3): \( (-1)^{1+3}(-1)=-1 \)
• Cofactor of \( 8 \) (row 2, column 1): \( (-1)^{2+1}(8)=-8 \)
• Cofactor of \( 2 \) (row 2, column 2): \( (-1)^{2+2}(2)=2 \)
• Cofactor of \( -4 \) (row 2, column 3): \( (-1)^{2+3}(-4)=4 \)
• Cofactor of \( -20 \) (row 3, column 1): \( (-1)^{3+1}(-20)=-20 \)
• Cofactor of \( -16 \) (row 3, column 2): \( (-1)^{3+2}(-16)=16 \)
• Cofactor of \( 10 \) (row 3, column 3): \( (-1)^{3+3}(10)=10 \)

Cofactor matrix: \(

$$\begin{bmatrix}-9&5&-1\\-8&2&4\\-20&16&10\end{bmatrix}$$

\)

Step4: Transpose the cofactor matrix (adjugate matrix)

The adjugate matrix is the transpose of the cofactor matrix. Transposing means swapping rows and columns.

Adjugate matrix \( \text{adj}(A)=

$$\begin{bmatrix}-9&-8&-20\\5&2&16\\-1&4&10\end{bmatrix}$$

\)

Step5: Find the inverse matrix

The inverse of a matrix \( A \) is given by \( A^{-1}=\frac{1}{\det(A)}\text{adj}(A) \). Since \( \det(A) = 22 \), we have:

$$ A^{-1}=\frac{1}{22}
LATEXBLOCK15
=
LATEXBLOCK16
$$

Answer:

Does this matrix have an inverse? Yes

The inverse matrix is \(

$$\begin{bmatrix}-\frac{9}{22}&-\frac{4}{11}&-\frac{10}{11}\\\frac{5}{22}&\frac{1}{11}&\frac{8}{11}\\-\frac{1}{22}&\frac{2}{11}&\frac{5}{11}\end{bmatrix}$$

\)