Studdy AI Logo
Math  /  Algebra

QuestionFind the inverse of the matrix A=[815−17−14234−2] A = \left[\begin{array}{rrr}8 & 1 & 5 \\ -17 & -14 & 2 \\ 3 & 4 & -2\end{array}\right] .

Studdy Solution

STEP 1

What is this asking? We need to find the *magical mirror matrix* that *undoes* the action of matrix AA, turning it back into the identity matrix. Watch out! Not every matrix has an inverse!
If the determinant is zero, then *poof*, no inverse exists!

STEP 2

1. Calculate the determinant.
2. Find the adjugate matrix.
3. Calculate the inverse matrix.

STEP 3

Let's **kick things off** by calculating the determinant of AA!
This tells us if our matrix even *has* an inverse.
If the determinant is zero, then *no inverse for you*!

STEP 4

We'll use the **cofactor expansion** along the first row because, why not?
It's as good a row as any!

STEP 5

det⁡(A)=8∣−1424−2∣−1∣−1723−2∣+5∣−17−1434∣ \det(A) = 8 \begin{vmatrix} -14 & 2 \\ 4 & -2 \end{vmatrix} - 1 \begin{vmatrix} -17 & 2 \\ 3 & -2 \end{vmatrix} + 5 \begin{vmatrix} -17 & -14 \\ 3 & 4 \end{vmatrix}

STEP 6

det⁡(A)=8((−14)⋅(−2)−2⋅4)−1((−17)⋅(−2)−2⋅3)+5((−17)⋅4−(−14)⋅3) \det(A) = 8((-14) \cdot (-2) - 2 \cdot 4) - 1((-17) \cdot (-2) - 2 \cdot 3) + 5((-17) \cdot 4 - (-14) \cdot 3)

STEP 7

det⁡(A)=8(28−8)−(34−6)+5(−68+42) \det(A) = 8(28 - 8) - (34 - 6) + 5(-68 + 42)

STEP 8

det⁡(A)=8(20)−(28)+5(−26) \det(A) = 8(20) - (28) + 5(-26)

STEP 9

det⁡(A)=160−28−130 \det(A) = 160 - 28 - 130

STEP 10

det⁡(A)=2 \det(A) = \mathbf{2}

STEP 11

*Phew*, our determinant is **2**, which isn't zero!
That means an inverse *does* exist! *Onwards to victory*!

STEP 12

Now, let's find the **adjugate matrix**, which is the *transpose of the cofactor matrix*.
It's like finding the *secret code* to unlock the inverse!

STEP 13

The cofactor matrix is found by calculating the determinant of each *minor matrix* (the matrix you get when you delete a row and column) and multiplying by (−1)i+j(-1)^{i+j}, where ii and jj are the row and column numbers.

STEP 14

After some serious number crunching (which I know you can handle!), the cofactor matrix of AA is: C=[−2028−26−22−41−29−115−69−97]C = \begin{bmatrix} -20 & 28 & -26 \\ -22 & -41 & -29 \\ -115 & -69 & -97 \end{bmatrix}

STEP 15

Now, *flip it*!
The adjugate is the transpose of the cofactor matrix: adj(A)=[−20−22−11528−41−69−26−29−97]\text{adj}(A) = \begin{bmatrix} -20 & -22 & -115 \\ 28 & -41 & -69 \\ -26 & -29 & -97 \end{bmatrix}

STEP 16

*Almost there*!
Now we just need to **divide** the adjugate matrix by the determinant we found earlier, which was **2**.

STEP 17

A−1=1det⁡(A)⋅adj(A)A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A)

STEP 18

A−1=12[−20−22−11528−41−69−26−29−97]A^{-1} = \frac{1}{2} \begin{bmatrix} -20 & -22 & -115 \\ 28 & -41 & -69 \\ -26 & -29 & -97 \end{bmatrix}

STEP 19

A−1=[−10−11−57.514−20.5−34.5−13−14.5−48.5]A^{-1} = \begin{bmatrix} -10 & -11 & -57.5 \\ 14 & -20.5 & -34.5 \\ -13 & -14.5 & -48.5 \end{bmatrix}

STEP 20

The inverse of matrix AA is: A−1=[−10−11−57.514−20.5−34.5−13−14.5−48.5]A^{-1} = \begin{bmatrix} -10 & -11 & -57.5 \\ 14 & -20.5 & -34.5 \\ -13 & -14.5 & -48.5 \end{bmatrix}

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord