Answer :
To verify that [tex]\( A \cdot (\operatorname{adj} A) = (\operatorname{adj} A) \cdot A = |A| I \)[/tex] for [tex]\( A = \begin{pmatrix} 5 & 3 \\ 4 & 2 \end{pmatrix} \)[/tex], we will go through the calculation of the determinant, the adjugate matrix, and the necessary matrix products.
1. Calculate the Determinant of [tex]\( A \)[/tex]:
The determinant of a 2x2 matrix [tex]\( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] is given by:
[tex]\[ |A| = ad - bc \][/tex]
For [tex]\( A = \begin{pmatrix} 5 & 3 \\ 4 & 2 \end{pmatrix} \)[/tex]:
[tex]\[ |A| = (5 \times 2) - (3 \times 4) = 10 - 12 = -2 \][/tex]
2. Calculate the Adjugate of [tex]\( A \)[/tex]:
The adjugate matrix (also known as the adjoint) of a 2x2 matrix [tex]\( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] is given by:
[tex]\[ \operatorname{adj} A = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For [tex]\( A = \begin{pmatrix} 5 & 3 \\ 4 & 2 \end{pmatrix} \)[/tex]:
[tex]\[ \operatorname{adj} A = \begin{pmatrix} 2 & -3 \\ -4 & 5 \end{pmatrix} \][/tex]
3. Calculate the Product [tex]\( A \cdot (\operatorname{adj} A) \)[/tex]:
We need to multiply matrix [tex]\( A \)[/tex] with its adjugate matrix:
[tex]\[ A = \begin{pmatrix} 5 & 3 \\ 4 & 2 \end{pmatrix}, \quad \operatorname{adj} A = \begin{pmatrix} 2 & -3 \\ -4 & 5 \end{pmatrix} \][/tex]
[tex]\[ A \cdot (\operatorname{adj} A) = \begin{pmatrix} 5 & 3 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} 2 & -3 \\ -4 & 5 \end{pmatrix} \][/tex]
Calculate the entries:
[tex]\[ \begin{pmatrix} (5 \times 2 + 3 \times -4) & (5 \times -3 + 3 \times 5) \\ (4 \times 2 + 2 \times -4) & (4 \times -3 + 2 \times 5) \end{pmatrix} = \begin{pmatrix} 10 - 12 & -15 + 15 \\ 8 - 8 & -12 + 10 \end{pmatrix} = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} \][/tex]
4. Calculate the Product [tex]\( (\operatorname{adj} A) \cdot A \)[/tex]:
Similarly, compute the adjugate matrix multiplied by [tex]\( A \)[/tex]:
[tex]\[ (\operatorname{adj} A) \cdot A = \begin{pmatrix} 2 & -3 \\ -4 & 5 \end{pmatrix} \begin{pmatrix} 5 & 3 \\ 4 & 2 \end{pmatrix} \][/tex]
Calculate the entries:
[tex]\[ \begin{pmatrix} (2 \times 5 + -3 \times 4) & (2 \times 3 + -3 \times 2) \\ (-4 \times 5 + 5 \times 4) & (-4 \times 3 + 5 \times 2) \end{pmatrix} = \begin{pmatrix} 10 - 12 & 6 - 6 \\ -20 + 20 & -12 + 10 \end{pmatrix} = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} \][/tex]
5. Identity Matrix Multiplied by Determinant of [tex]\( A \)[/tex]:
[tex]\[ |A| I = -2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} \][/tex]
Now we have verified that:
[tex]\[ A \cdot (\operatorname{adj} A) = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} \][/tex]
[tex]\[ (\operatorname{adj} A) \cdot A = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} \][/tex]
[tex]\[ |A| I = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} \][/tex]
Thus, \( A \cdot (\operatorname{adj} A) = (\operatorname{adj} A) \cdot A = |A| I \wrew verified.
1. Calculate the Determinant of [tex]\( A \)[/tex]:
The determinant of a 2x2 matrix [tex]\( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] is given by:
[tex]\[ |A| = ad - bc \][/tex]
For [tex]\( A = \begin{pmatrix} 5 & 3 \\ 4 & 2 \end{pmatrix} \)[/tex]:
[tex]\[ |A| = (5 \times 2) - (3 \times 4) = 10 - 12 = -2 \][/tex]
2. Calculate the Adjugate of [tex]\( A \)[/tex]:
The adjugate matrix (also known as the adjoint) of a 2x2 matrix [tex]\( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] is given by:
[tex]\[ \operatorname{adj} A = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For [tex]\( A = \begin{pmatrix} 5 & 3 \\ 4 & 2 \end{pmatrix} \)[/tex]:
[tex]\[ \operatorname{adj} A = \begin{pmatrix} 2 & -3 \\ -4 & 5 \end{pmatrix} \][/tex]
3. Calculate the Product [tex]\( A \cdot (\operatorname{adj} A) \)[/tex]:
We need to multiply matrix [tex]\( A \)[/tex] with its adjugate matrix:
[tex]\[ A = \begin{pmatrix} 5 & 3 \\ 4 & 2 \end{pmatrix}, \quad \operatorname{adj} A = \begin{pmatrix} 2 & -3 \\ -4 & 5 \end{pmatrix} \][/tex]
[tex]\[ A \cdot (\operatorname{adj} A) = \begin{pmatrix} 5 & 3 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} 2 & -3 \\ -4 & 5 \end{pmatrix} \][/tex]
Calculate the entries:
[tex]\[ \begin{pmatrix} (5 \times 2 + 3 \times -4) & (5 \times -3 + 3 \times 5) \\ (4 \times 2 + 2 \times -4) & (4 \times -3 + 2 \times 5) \end{pmatrix} = \begin{pmatrix} 10 - 12 & -15 + 15 \\ 8 - 8 & -12 + 10 \end{pmatrix} = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} \][/tex]
4. Calculate the Product [tex]\( (\operatorname{adj} A) \cdot A \)[/tex]:
Similarly, compute the adjugate matrix multiplied by [tex]\( A \)[/tex]:
[tex]\[ (\operatorname{adj} A) \cdot A = \begin{pmatrix} 2 & -3 \\ -4 & 5 \end{pmatrix} \begin{pmatrix} 5 & 3 \\ 4 & 2 \end{pmatrix} \][/tex]
Calculate the entries:
[tex]\[ \begin{pmatrix} (2 \times 5 + -3 \times 4) & (2 \times 3 + -3 \times 2) \\ (-4 \times 5 + 5 \times 4) & (-4 \times 3 + 5 \times 2) \end{pmatrix} = \begin{pmatrix} 10 - 12 & 6 - 6 \\ -20 + 20 & -12 + 10 \end{pmatrix} = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} \][/tex]
5. Identity Matrix Multiplied by Determinant of [tex]\( A \)[/tex]:
[tex]\[ |A| I = -2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} \][/tex]
Now we have verified that:
[tex]\[ A \cdot (\operatorname{adj} A) = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} \][/tex]
[tex]\[ (\operatorname{adj} A) \cdot A = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} \][/tex]
[tex]\[ |A| I = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} \][/tex]
Thus, \( A \cdot (\operatorname{adj} A) = (\operatorname{adj} A) \cdot A = |A| I \wrew verified.
