Answer :
To find the adjugate and inverse of the matrix [tex]\( A = \left[\begin{array}{cc}2 & 3 \\ 4 & -5\end{array}\right] \)[/tex], we can follow the steps below:
### Step 1: Calculate the Determinant of Matrix [tex]\( A \)[/tex]
The determinant of a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( A = \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] \)[/tex] is given by:
[tex]\[ \det(A) = ad - bc \][/tex]
For our matrix [tex]\( A \)[/tex]:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = 3 \)[/tex]
- [tex]\( c = 4 \)[/tex]
- [tex]\( d = -5 \)[/tex]
So, we have:
[tex]\[ \det(A) = (2 \cdot -5) - (3 \cdot 4) = -10 - 12 = -22 \][/tex]
### Step 2: Check if Matrix [tex]\( A \)[/tex] is Invertible
A matrix is invertible if its determinant is not zero. Since [tex]\(\det(A) = -22 \neq 0\)[/tex], matrix [tex]\( A \)[/tex] is invertible.
### Step 3: Calculate the Adjugate (Adjoint) of Matrix [tex]\( A \)[/tex]
The adjugate of a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( A \)[/tex] is given by the transpose of its cofactor matrix.
The cofactor matrix [tex]\( \text{C} \)[/tex] of matrix [tex]\( A \)[/tex] is:
[tex]\[ \text{C} = \left[\begin{array}{cc} C_{11} & C_{12} \\ C_{21} & C_{22} \end{array}\right] \][/tex]
where [tex]\( C_{ij} \)[/tex] is the cofactor of the element at position [tex]\( (i, j) \)[/tex].
For [tex]\( A \)[/tex]:
1. [tex]\( C_{11} = \det \left(\left[\begin{array}{cc}-5\end{array}\right]\right) = -5 \)[/tex]
2. [tex]\( C_{12} = -\det \left(\left[\begin{array}{cc}4\end{array}\right]\right) = -4 \)[/tex]
3. [tex]\( C_{21} = -\det \left(\left[\begin{array}{cc}3\end{array}\right]\right) = -3 \)[/tex]
4. [tex]\( C_{22} = \det \left(\left[\begin{array}{cc}2\end{array}\right]\right) = 2 \)[/tex]
So, the cofactor matrix is:
[tex]\[ \text{C} = \left[\begin{array}{cc} -5 & -4 \\ -3 & 2 \end{array}\right] \][/tex]
The adjugate of [tex]\( A \)[/tex] is the transpose of the cofactor matrix:
[tex]\[ \text{adj}(A) = \left[\begin{array}{cc} -5 & -3 \\ -4 & 2 \end{array}\right] \][/tex]
### Step 4: Calculate the Inverse of Matrix [tex]\( A \)[/tex]
The inverse of matrix [tex]\( A \)[/tex] is given by:
[tex]\[ A^{-1} = \frac{1}{\det(A)} \text{adj}(A) \][/tex]
Substituting the determinant and the adjugate into this formula, we get:
[tex]\[ A^{-1} = \frac{1}{-22} \left[\begin{array}{cc} -5 & -3 \\ -4 & 2 \end{array}\right] = \left[\begin{array}{cc} \frac{5}{22} & \frac{3}{22} \\ \frac{4}{22} & -\frac{2}{22} \end{array}\right] \][/tex]
Simplify the fractions:
[tex]\[ A^{-1} = \left[\begin{array}{cc} 0.22727273 & 0.13636364 \\ 0.18181818 & -0.09090909 \end{array}\right] \][/tex]
To summarize, the results are:
1. Determinant of [tex]\( A \)[/tex]: [tex]\(-22\)[/tex]
2. Adjugate of [tex]\( A \)[/tex]: [tex]\(\left[\begin{array}{cc} -5 & -3 \\ -4 & 2 \end{array}\right]\)[/tex]
3. Inverse of [tex]\( A \)[/tex]: [tex]\(\left[\begin{array}{cc} 0.22727273 & 0.13636364 \\ 0.18181818 & -0.09090909 \end{array}\right]\)[/tex]
### Step 1: Calculate the Determinant of Matrix [tex]\( A \)[/tex]
The determinant of a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( A = \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] \)[/tex] is given by:
[tex]\[ \det(A) = ad - bc \][/tex]
For our matrix [tex]\( A \)[/tex]:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = 3 \)[/tex]
- [tex]\( c = 4 \)[/tex]
- [tex]\( d = -5 \)[/tex]
So, we have:
[tex]\[ \det(A) = (2 \cdot -5) - (3 \cdot 4) = -10 - 12 = -22 \][/tex]
### Step 2: Check if Matrix [tex]\( A \)[/tex] is Invertible
A matrix is invertible if its determinant is not zero. Since [tex]\(\det(A) = -22 \neq 0\)[/tex], matrix [tex]\( A \)[/tex] is invertible.
### Step 3: Calculate the Adjugate (Adjoint) of Matrix [tex]\( A \)[/tex]
The adjugate of a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( A \)[/tex] is given by the transpose of its cofactor matrix.
The cofactor matrix [tex]\( \text{C} \)[/tex] of matrix [tex]\( A \)[/tex] is:
[tex]\[ \text{C} = \left[\begin{array}{cc} C_{11} & C_{12} \\ C_{21} & C_{22} \end{array}\right] \][/tex]
where [tex]\( C_{ij} \)[/tex] is the cofactor of the element at position [tex]\( (i, j) \)[/tex].
For [tex]\( A \)[/tex]:
1. [tex]\( C_{11} = \det \left(\left[\begin{array}{cc}-5\end{array}\right]\right) = -5 \)[/tex]
2. [tex]\( C_{12} = -\det \left(\left[\begin{array}{cc}4\end{array}\right]\right) = -4 \)[/tex]
3. [tex]\( C_{21} = -\det \left(\left[\begin{array}{cc}3\end{array}\right]\right) = -3 \)[/tex]
4. [tex]\( C_{22} = \det \left(\left[\begin{array}{cc}2\end{array}\right]\right) = 2 \)[/tex]
So, the cofactor matrix is:
[tex]\[ \text{C} = \left[\begin{array}{cc} -5 & -4 \\ -3 & 2 \end{array}\right] \][/tex]
The adjugate of [tex]\( A \)[/tex] is the transpose of the cofactor matrix:
[tex]\[ \text{adj}(A) = \left[\begin{array}{cc} -5 & -3 \\ -4 & 2 \end{array}\right] \][/tex]
### Step 4: Calculate the Inverse of Matrix [tex]\( A \)[/tex]
The inverse of matrix [tex]\( A \)[/tex] is given by:
[tex]\[ A^{-1} = \frac{1}{\det(A)} \text{adj}(A) \][/tex]
Substituting the determinant and the adjugate into this formula, we get:
[tex]\[ A^{-1} = \frac{1}{-22} \left[\begin{array}{cc} -5 & -3 \\ -4 & 2 \end{array}\right] = \left[\begin{array}{cc} \frac{5}{22} & \frac{3}{22} \\ \frac{4}{22} & -\frac{2}{22} \end{array}\right] \][/tex]
Simplify the fractions:
[tex]\[ A^{-1} = \left[\begin{array}{cc} 0.22727273 & 0.13636364 \\ 0.18181818 & -0.09090909 \end{array}\right] \][/tex]
To summarize, the results are:
1. Determinant of [tex]\( A \)[/tex]: [tex]\(-22\)[/tex]
2. Adjugate of [tex]\( A \)[/tex]: [tex]\(\left[\begin{array}{cc} -5 & -3 \\ -4 & 2 \end{array}\right]\)[/tex]
3. Inverse of [tex]\( A \)[/tex]: [tex]\(\left[\begin{array}{cc} 0.22727273 & 0.13636364 \\ 0.18181818 & -0.09090909 \end{array}\right]\)[/tex]
