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Sagot :
To find the inverse of a matrix [tex]\( Y \)[/tex], we need to check whether the determinant of [tex]\( Y \)[/tex] is non-zero. If the determinant is zero, the matrix does not have an inverse. If the determinant is non-zero, we can find the inverse using the formula for the inverse of a 2x2 matrix:
[tex]\[ Y^{-1} = \frac{1}{\text{det}(Y)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \][/tex]
where [tex]\( Y = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \)[/tex] and [tex]\(\text{det}(Y) = ad - bc \)[/tex].
Let's find the inverses for each matrix given.
### 1. [tex]\( Y = \begin{bmatrix} 2 & -2 \\ -1 & 2 \end{bmatrix} \)[/tex]
Determinant [tex]\( \text{det}(Y) = 2 \cdot 2 - (-2) \cdot (-1) = 4 - 2 = 2 \)[/tex].
[tex]\[ Y^{-1} = \frac{1}{2} \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0.5 & 1 \end{bmatrix} \][/tex]
### 2. [tex]\( Y = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \)[/tex]
Determinant [tex]\( \text{det}(Y) = 1 \cdot 1 - (-1) \cdot 1 = 1 + 1 = 2 \)[/tex].
[tex]\[ Y^{-1} = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 0.5 & 0.5 \\ -0.5 & 0.5 \end{bmatrix} \][/tex]
### 3. [tex]\( Y = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \)[/tex]
Determinant [tex]\( \text{det}(Y) = 2 \cdot 0 - (-1) \cdot 1 = 0 + 1 = 1 \)[/tex].
[tex]\[ Y^{-1} = \begin{bmatrix} 0 & 1 \\ -1 & 2 \end{bmatrix} \][/tex]
### 4. [tex]\( Y = \begin{bmatrix} 2 & 3 \\ 1 & 1 \end{bmatrix} \)[/tex]
Determinant [tex]\( \text{det}(Y) = 2 \cdot 1 - 3 \cdot 1 = 2 - 3 = -1 \)[/tex].
[tex]\[ Y^{-1} = \frac{1}{-1} \begin{bmatrix} 1 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} -1 & 3 \\ 1 & -2 \end{bmatrix} \][/tex]
### 5. [tex]\( Y = \begin{bmatrix} 1 & 4 \\ 1 & 3 \end{bmatrix} \)[/tex]
Determinant [tex]\( \text{det}(Y) = 1 \cdot 3 - 4 \cdot 1 = 3 - 4 = -1 \)[/tex].
[tex]\[ Y^{-1} = \frac{1}{-1} \begin{bmatrix} 3 & -4 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} -3 & 4 \\ 1 & -1 \end{bmatrix} \][/tex]
### 6. [tex]\( Y = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)[/tex]
Determinant [tex]\( \text{det}(Y) = 1 \cdot 4 - 2 \cdot 3 = 4 - 6 = -2 \)[/tex].
[tex]\[ Y^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} \][/tex]
### 7. [tex]\( Y = \begin{bmatrix} 3 & 4 \\ 3 & 4 \end{bmatrix} \)[/tex]
Determinant [tex]\( \text{det}(Y) = 3 \cdot 4 - 4 \cdot 3 = 12 - 12 = 0 \)[/tex].
This matrix does not have an inverse because the determinant is zero, indicating that the matrix is singular.
Therefore, the inverse matrices are as follows:
1. [tex]\( Y_1^{-1} = \begin{bmatrix} 1 & 1 \\ 0.5 & 1 \end{bmatrix} \)[/tex]
2. [tex]\( Y_2^{-1} = \begin{bmatrix} 0.5 & 0.5 \\ -0.5 & 0.5 \end{bmatrix} \)[/tex]
3. [tex]\( Y_3^{-1} = \begin{bmatrix} 0 & 1 \\ -1 & 2 \end{bmatrix} \)[/tex]
4. [tex]\( Y_4^{-1} = \begin{bmatrix} -1 & 3 \\ 1 & -2 \end{bmatrix} \)[/tex]
5. [tex]\( Y_5^{-1} = \begin{bmatrix} -3 & 4 \\ 1 & -1 \end{bmatrix} \)[/tex]
6. [tex]\( Y_6^{-1} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} \)[/tex]
7. [tex]\( Y_7^{-1} = \text{no inverse (determinant is zero)} \)[/tex]
[tex]\[ Y^{-1} = \frac{1}{\text{det}(Y)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \][/tex]
where [tex]\( Y = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \)[/tex] and [tex]\(\text{det}(Y) = ad - bc \)[/tex].
Let's find the inverses for each matrix given.
### 1. [tex]\( Y = \begin{bmatrix} 2 & -2 \\ -1 & 2 \end{bmatrix} \)[/tex]
Determinant [tex]\( \text{det}(Y) = 2 \cdot 2 - (-2) \cdot (-1) = 4 - 2 = 2 \)[/tex].
[tex]\[ Y^{-1} = \frac{1}{2} \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0.5 & 1 \end{bmatrix} \][/tex]
### 2. [tex]\( Y = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \)[/tex]
Determinant [tex]\( \text{det}(Y) = 1 \cdot 1 - (-1) \cdot 1 = 1 + 1 = 2 \)[/tex].
[tex]\[ Y^{-1} = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 0.5 & 0.5 \\ -0.5 & 0.5 \end{bmatrix} \][/tex]
### 3. [tex]\( Y = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \)[/tex]
Determinant [tex]\( \text{det}(Y) = 2 \cdot 0 - (-1) \cdot 1 = 0 + 1 = 1 \)[/tex].
[tex]\[ Y^{-1} = \begin{bmatrix} 0 & 1 \\ -1 & 2 \end{bmatrix} \][/tex]
### 4. [tex]\( Y = \begin{bmatrix} 2 & 3 \\ 1 & 1 \end{bmatrix} \)[/tex]
Determinant [tex]\( \text{det}(Y) = 2 \cdot 1 - 3 \cdot 1 = 2 - 3 = -1 \)[/tex].
[tex]\[ Y^{-1} = \frac{1}{-1} \begin{bmatrix} 1 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} -1 & 3 \\ 1 & -2 \end{bmatrix} \][/tex]
### 5. [tex]\( Y = \begin{bmatrix} 1 & 4 \\ 1 & 3 \end{bmatrix} \)[/tex]
Determinant [tex]\( \text{det}(Y) = 1 \cdot 3 - 4 \cdot 1 = 3 - 4 = -1 \)[/tex].
[tex]\[ Y^{-1} = \frac{1}{-1} \begin{bmatrix} 3 & -4 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} -3 & 4 \\ 1 & -1 \end{bmatrix} \][/tex]
### 6. [tex]\( Y = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)[/tex]
Determinant [tex]\( \text{det}(Y) = 1 \cdot 4 - 2 \cdot 3 = 4 - 6 = -2 \)[/tex].
[tex]\[ Y^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} \][/tex]
### 7. [tex]\( Y = \begin{bmatrix} 3 & 4 \\ 3 & 4 \end{bmatrix} \)[/tex]
Determinant [tex]\( \text{det}(Y) = 3 \cdot 4 - 4 \cdot 3 = 12 - 12 = 0 \)[/tex].
This matrix does not have an inverse because the determinant is zero, indicating that the matrix is singular.
Therefore, the inverse matrices are as follows:
1. [tex]\( Y_1^{-1} = \begin{bmatrix} 1 & 1 \\ 0.5 & 1 \end{bmatrix} \)[/tex]
2. [tex]\( Y_2^{-1} = \begin{bmatrix} 0.5 & 0.5 \\ -0.5 & 0.5 \end{bmatrix} \)[/tex]
3. [tex]\( Y_3^{-1} = \begin{bmatrix} 0 & 1 \\ -1 & 2 \end{bmatrix} \)[/tex]
4. [tex]\( Y_4^{-1} = \begin{bmatrix} -1 & 3 \\ 1 & -2 \end{bmatrix} \)[/tex]
5. [tex]\( Y_5^{-1} = \begin{bmatrix} -3 & 4 \\ 1 & -1 \end{bmatrix} \)[/tex]
6. [tex]\( Y_6^{-1} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} \)[/tex]
7. [tex]\( Y_7^{-1} = \text{no inverse (determinant is zero)} \)[/tex]
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