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Math  /  Algebra

QuestionFind the inverse of [1−120−1−1102]\left[\begin{array}{ccc} 1 & -1 & 2 \\ 0 & -1 & -1 \\ 1 & 0 & 2 \end{array}\right]

Studdy Solution

STEP 1

1. The given matrix is a 3×33 \times 3 matrix.
2. The matrix is invertible, meaning its determinant is non-zero.
3. We will use the formula for the inverse of a 3×33 \times 3 matrix involving the adjugate and determinant.

STEP 2

1. Calculate the determinant of the matrix.
2. Find the matrix of minors.
3. Find the matrix of cofactors.
4. Find the adjugate (transpose of the cofactor matrix).
5. Calculate the inverse using the determinant and adjugate.

STEP 3

Calculate the determinant of the matrix. The determinant of a 3×33 \times 3 matrix
[abcdefghi]\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)\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)
For the given matrix:
[1−120−1−1102]\begin{bmatrix} 1 & -1 & 2 \\ 0 & -1 & -1 \\ 1 & 0 & 2 \end{bmatrix}
Calculate:
det(A)=1((−1)(2)−(−1)(0))−(−1)(0(2)−(−1)(1))+2(0(0)−(−1)(1))\text{det}(A) = 1((-1)(2) - (-1)(0)) - (-1)(0(2) - (-1)(1)) + 2(0(0) - (-1)(1))
=1(−2−0)+1(0+1)+2(0+1)= 1(-2 - 0) + 1(0 + 1) + 2(0 + 1)
=−2+1+2= -2 + 1 + 2
=1= 1

STEP 4

Find the matrix of minors. The minor of an element is the determinant of the 2×22 \times 2 matrix that remains after removing the row and column of that element.
For element a11a_{11}, the minor is:
∣−1−102∣=(−1)(2)−(−1)(0)=−2\begin{vmatrix} -1 & -1 \\ 0 & 2 \end{vmatrix} = (-1)(2) - (-1)(0) = -2
For element a12a_{12}, the minor is:
∣0−112∣=(0)(2)−(−1)(1)=1\begin{vmatrix} 0 & -1 \\ 1 & 2 \end{vmatrix} = (0)(2) - (-1)(1) = 1
For element a13a_{13}, the minor is:
∣0−110∣=(0)(0)−(−1)(1)=1\begin{vmatrix} 0 & -1 \\ 1 & 0 \end{vmatrix} = (0)(0) - (-1)(1) = 1
Continue this process for all elements to get the matrix of minors:
[−211−20110−1]\begin{bmatrix} -2 & 1 & 1 \\ -2 & 0 & 1 \\ 1 & 0 & -1 \end{bmatrix}

STEP 5

Find the matrix of cofactors by applying the checkerboard pattern of signs:
[+−+−+−+−+]\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}
Apply this pattern to the matrix of minors:
[−2−1120−110−1]\begin{bmatrix} -2 & -1 & 1 \\ 2 & 0 & -1 \\ 1 & 0 & -1 \end{bmatrix}

STEP 6

Find the adjugate by transposing the cofactor matrix:
[−221−1001−1−1]\begin{bmatrix} -2 & 2 & 1 \\ -1 & 0 & 0 \\ 1 & -1 & -1 \end{bmatrix}

STEP 7

Calculate the inverse using the formula:
A−1=1det(A)⋅adjugate(A)A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adjugate}(A)
Since det(A)=1\text{det}(A) = 1, the inverse is simply the adjugate:
A−1=[−221−1001−1−1]A^{-1} = \begin{bmatrix} -2 & 2 & 1 \\ -1 & 0 & 0 \\ 1 & -1 & -1 \end{bmatrix}
The inverse of the given matrix is:
[−221−1001−1−1]\boxed{\begin{bmatrix} -2 & 2 & 1 \\ -1 & 0 & 0 \\ 1 & -1 & -1 \end{bmatrix}}

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