Question

question 12 of 34 step 1 of 1
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?
\boxed{yes} \boxed{no}

Explanation:

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

The formula for the determinant of a \( 3 \times 3 \) matrix \(

$$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$

\) is \( \det(A)=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31}) \).

For \( A =

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

\), we have:

\( \det(A)= -2((-5)(1)- (2)(2)) - 0((-3)(1)- (2)(1)) + (-4)((-3)(2)- (-5)(1)) \)

Step2: Simplify each term

First term: \( -2((-5) - 4)= -2(-9) = 18 \)

Second term: \( -0((-3) - 2)= 0 \)

Third term: \( -4((-6) + 5)= -4(-1) = 4 \)

Now, sum the terms: \( \det(A)=18 + 0 + 4 = 22 \)

Since the determinant of \( A \) is not zero (\( \det(A) = 22
eq 0 \)), the matrix is invertible.

Answer:

Yes