Question

find the inverse of the matrix \\(\

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

\\).

\\(\circ \

$$\begin{bmatrix} 3 & 2 & 6 \\\\ 1 & 1 & 2 \\\\ 2 & 2 & 5 \\end{bmatrix}$$

\\)

\\(\circ \

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

\\)

Explanation:

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<reasoning>

Calculate the determinant of the matrix

Using the Matrix Determinant knowledge point
\[
\det(A) = 1 \cdot (3 \cdot 1 - 0 \cdot (-2)) - 2 \cdot ((-1) \cdot 1 - 0 \cdot 0) + (-2) \cdot ((-1) \cdot (-2) - 3 \cdot 0)
\]
\[
\det(A) = 1 \cdot 3 - 2 \cdot (-1) - 2 \cdot 2 = 3 + 2 - 4 = 1
\]

Find the cofactor matrix elements

Using the Cofactor Matrix knowledge point
\[

$$\begin{aligned} C_{11} &= \det\begin{bmatrix} 3 & 0 \\ -2 & 1 \end{bmatrix} = 3, & C_{12} &= -\det\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} = 1, & C_{13} &= \det\begin{bmatrix} -1 & 3 \\ 0 & -2 \end{bmatrix} = 2 \\ C_{21} &= -\det\begin{bmatrix} 2 & -2 \\ -2 & 1 \end{bmatrix} = 2, & C_{22} &= \det\begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} = 1, & C_{23} &= -\det\begin{bmatrix} 1 & 2 \\ 0 & -2 \end{bmatrix} = 2 \\ C_{31} &= \det\begin{bmatrix} 2 & -2 \\ 3 & 0 \end{bmatrix} = 6, & C_{32} &= -\det\begin{bmatrix} 1 & -2 \\ -1 & 0 \end{bmatrix} = 2, & C_{33} &= \det\begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} = 5 \end{aligned}$$

\]

Construct the adjoint matrix

Using the Adjoint Matrix knowledge point
\[
\text{adj}(A) = C^T =

$$\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$$

\]

Compute the inverse matrix

Using the Matrix Inverse Formula knowledge point
\[
A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{1}

$$\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$$

=

$$\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$$

\]
</reasoning>

<answer>
<mcq-correct>(A) \(

$$\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$$

\)</mcq-correct>
<mcq-option>(B) \(

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

\)</mcq-option>
</answer>

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Answer:

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<reasoning>

Calculate the determinant of the matrix

Using the Matrix Determinant knowledge point
\[
\det(A) = 1 \cdot (3 \cdot 1 - 0 \cdot (-2)) - 2 \cdot ((-1) \cdot 1 - 0 \cdot 0) + (-2) \cdot ((-1) \cdot (-2) - 3 \cdot 0)
\]
\[
\det(A) = 1 \cdot 3 - 2 \cdot (-1) - 2 \cdot 2 = 3 + 2 - 4 = 1
\]

Find the cofactor matrix elements

Using the Cofactor Matrix knowledge point
\[

$$\begin{aligned} C_{11} &= \det\begin{bmatrix} 3 & 0 \\ -2 & 1 \end{bmatrix} = 3, & C_{12} &= -\det\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} = 1, & C_{13} &= \det\begin{bmatrix} -1 & 3 \\ 0 & -2 \end{bmatrix} = 2 \\ C_{21} &= -\det\begin{bmatrix} 2 & -2 \\ -2 & 1 \end{bmatrix} = 2, & C_{22} &= \det\begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} = 1, & C_{23} &= -\det\begin{bmatrix} 1 & 2 \\ 0 & -2 \end{bmatrix} = 2 \\ C_{31} &= \det\begin{bmatrix} 2 & -2 \\ 3 & 0 \end{bmatrix} = 6, & C_{32} &= -\det\begin{bmatrix} 1 & -2 \\ -1 & 0 \end{bmatrix} = 2, & C_{33} &= \det\begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} = 5 \end{aligned}$$

\]

Construct the adjoint matrix

Using the Adjoint Matrix knowledge point
\[
\text{adj}(A) = C^T =

$$\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$$

\]

Compute the inverse matrix

Using the Matrix Inverse Formula knowledge point
\[
A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{1}

$$\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$$

=

$$\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$$

\]
</reasoning>

<answer>
<mcq-correct>(A) \(

$$\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$$

\)</mcq-correct>
<mcq-option>(B) \(

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

\)</mcq-option>
</answer>

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