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Math  /  Algebra

QuestionFind the inverse of the 3×33 \times 3 matrix AA A=[121910301]A=\left[\begin{array}{lll} 1 & 2 & 1 \\ 9 & 1 & 0 \\ 3 & 0 & 1 \end{array}\right]

Studdy Solution

STEP 1

What is this asking? We need to find a matrix, let's call it A−1A^{-1}, that when multiplied by the given matrix AA results in the identity matrix. Watch out! Not all matrices have inverses!
If the determinant of AA is zero, then the inverse doesn't exist.

STEP 2

1. Calculate the Determinant
2. Find the Matrix of Minors
3. Create the Matrix of Cofactors
4. Find the Adjugate
5. Calculate the Inverse

STEP 3

Let's **calculate the determinant** of matrix AA!
This is super important because if the determinant is zero, the inverse just doesn't exist.
We'll use the formula for the determinant of a 3x3 matrix.

STEP 4

det⁡(A)=1⋅(1⋅1−0⋅0)−2⋅(9⋅1−0⋅3)+1⋅(9⋅0−1⋅3) \det(A) = 1 \cdot (1 \cdot 1 - 0 \cdot 0) - 2 \cdot (9 \cdot 1 - 0 \cdot 3) + 1 \cdot (9 \cdot 0 - 1 \cdot 3)

STEP 5

det⁡(A)=1−18−3=−20 \det(A) = 1 - 18 - 3 = \mathbf{-20}

STEP 6

Great! The determinant is **-20**, which isn't zero, so the inverse exists!
Let's proceed!

STEP 7

Now, let's **build the matrix of minors**.
Each element of this matrix is the determinant of a smaller 2×22 \times 2 matrix, which we get by ignoring the row and column of the corresponding element in AA.

STEP 8

Minors(A)=[19−32−2−6−1−9−17] \text{Minors}(A) = \begin{bmatrix} 1 & 9 & -3 \\ 2 & -2 & -6 \\ -1 & -9 & -17 \end{bmatrix}

STEP 9

Minors(A)=[19−32−2−6−1−9−17] \text{Minors}(A) = \begin{bmatrix} 1 & 9 & -3 \\ 2 & -2 & -6 \\ -1 & -9 & -17 \end{bmatrix}

STEP 10

To get the **matrix of cofactors**, we just need to change the signs of some elements in the matrix of minors.
We multiply each element by (−1)i+j(-1)^{i+j}, where ii is the row number and jj is the column number.

STEP 11

[1−9−3−2−26−19−17]\begin{bmatrix} 1 & -9 & -3 \\ -2 & -2 & 6 \\ -1 & 9 & -17 \end{bmatrix}

STEP 12

The **adjugate** is simply the transpose of the matrix of cofactors.
We swap rows and columns!

STEP 13

Adj(A)=[1−2−1−9−29−36−17] \text{Adj}(A) = \begin{bmatrix} 1 & -2 & -1 \\ -9 & -2 & 9 \\ -3 & 6 & -17 \end{bmatrix}

STEP 14

Finally, we can **calculate the inverse**!
We divide the adjugate by the determinant.

STEP 15

A−1=1−20[1−2−1−9−29−36−17] A^{-1} = \frac{1}{-20} \begin{bmatrix} 1 & -2 & -1 \\ -9 & -2 & 9 \\ -3 & 6 & -17 \end{bmatrix}

STEP 16

A−1=[−120220120920220−920320−6201720] A^{-1} = \begin{bmatrix} -\frac{1}{20} & \frac{2}{20} & \frac{1}{20} \\ \frac{9}{20} & \frac{2}{20} & -\frac{9}{20} \\ \frac{3}{20} & -\frac{6}{20} & \frac{17}{20} \end{bmatrix}

STEP 17

The inverse of matrix AA is: A−1=[−120110120920110−920320−3101720] A^{-1} = \begin{bmatrix} -\frac{1}{20} & \frac{1}{10} & \frac{1}{20} \\ \frac{9}{20} & \frac{1}{10} & -\frac{9}{20} \\ \frac{3}{20} & -\frac{3}{10} & \frac{17}{20} \end{bmatrix}

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