Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
To determine the value of [tex]\( x \)[/tex] that makes the two matrices inverses of each other for the given matrices:
[tex]\[ A = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 1 & -2 \\ x & 5 \end{bmatrix}, \][/tex]
we need to find [tex]\( x \)[/tex] such that the product of the matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is the identity matrix [tex]\( I \)[/tex]:
[tex]\[ A \cdot B = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \][/tex]
Let's multiply the matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
[tex]\[ A \cdot B = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ x & 5 \end{bmatrix} \][/tex]
Perform the matrix multiplication:
[tex]\[ A \cdot B = \begin{bmatrix} (5 \cdot 1 + 2 \cdot x) & (5 \cdot -2 + 2 \cdot 5) \\ (2 \cdot 1 + 1 \cdot x) & (2 \cdot -2 + 1 \cdot 5) \end{bmatrix} = \begin{bmatrix} 5 + 2x & -10 + 10 \\ 2 + x & -4 + 5 \end{bmatrix} = \begin{bmatrix} 5 + 2x & 0 \\ 2 + x & 1 \end{bmatrix} \][/tex]
Now, we set the product equal to the identity matrix and solve for [tex]\( x \)[/tex]:
[tex]\[ \begin{bmatrix} 5 + 2x & 0 \\ 2 + x & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \][/tex]
From the matrix equality, we get the following system of equations:
[tex]\[ 5 + 2x = 1 \quad \text{(1)} \][/tex]
[tex]\[ 2 + x = 0 \quad \text{(2)} \][/tex]
First, solve equation (2) for [tex]\( x \)[/tex]:
[tex]\[ 2 + x = 0 \][/tex]
[tex]\[ x = -2 \][/tex]
Therefore, the value of [tex]\( x \)[/tex] that makes the two matrices inverses of each other is [tex]\( x = -2 \)[/tex].
To verify, substitute [tex]\( x = -2 \)[/tex] back into [tex]\( B \)[/tex] and check that [tex]\( A \cdot B = I \)[/tex]:
[tex]\[ B = \begin{bmatrix} 1 & -2 \\ -2 & 5 \end{bmatrix} \][/tex]
[tex]\[ A \cdot B = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 5 \cdot 1 + 2 \cdot (-2) & 5 \cdot -2 + 2 \cdot 5 \\ 2 \cdot 1 + 1 \cdot (-2) & 2 \cdot -2 + 1 \cdot 5 \end{bmatrix} = \begin{bmatrix} 5 - 4 & -10 + 10 \\ 2 - 2 & -4 + 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \][/tex]
Thus, when [tex]\( x = -2 \)[/tex], the matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are indeed inverses of each other.
[tex]\[ A = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 1 & -2 \\ x & 5 \end{bmatrix}, \][/tex]
we need to find [tex]\( x \)[/tex] such that the product of the matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is the identity matrix [tex]\( I \)[/tex]:
[tex]\[ A \cdot B = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \][/tex]
Let's multiply the matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
[tex]\[ A \cdot B = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ x & 5 \end{bmatrix} \][/tex]
Perform the matrix multiplication:
[tex]\[ A \cdot B = \begin{bmatrix} (5 \cdot 1 + 2 \cdot x) & (5 \cdot -2 + 2 \cdot 5) \\ (2 \cdot 1 + 1 \cdot x) & (2 \cdot -2 + 1 \cdot 5) \end{bmatrix} = \begin{bmatrix} 5 + 2x & -10 + 10 \\ 2 + x & -4 + 5 \end{bmatrix} = \begin{bmatrix} 5 + 2x & 0 \\ 2 + x & 1 \end{bmatrix} \][/tex]
Now, we set the product equal to the identity matrix and solve for [tex]\( x \)[/tex]:
[tex]\[ \begin{bmatrix} 5 + 2x & 0 \\ 2 + x & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \][/tex]
From the matrix equality, we get the following system of equations:
[tex]\[ 5 + 2x = 1 \quad \text{(1)} \][/tex]
[tex]\[ 2 + x = 0 \quad \text{(2)} \][/tex]
First, solve equation (2) for [tex]\( x \)[/tex]:
[tex]\[ 2 + x = 0 \][/tex]
[tex]\[ x = -2 \][/tex]
Therefore, the value of [tex]\( x \)[/tex] that makes the two matrices inverses of each other is [tex]\( x = -2 \)[/tex].
To verify, substitute [tex]\( x = -2 \)[/tex] back into [tex]\( B \)[/tex] and check that [tex]\( A \cdot B = I \)[/tex]:
[tex]\[ B = \begin{bmatrix} 1 & -2 \\ -2 & 5 \end{bmatrix} \][/tex]
[tex]\[ A \cdot B = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 5 \cdot 1 + 2 \cdot (-2) & 5 \cdot -2 + 2 \cdot 5 \\ 2 \cdot 1 + 1 \cdot (-2) & 2 \cdot -2 + 1 \cdot 5 \end{bmatrix} = \begin{bmatrix} 5 - 4 & -10 + 10 \\ 2 - 2 & -4 + 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \][/tex]
Thus, when [tex]\( x = -2 \)[/tex], the matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are indeed inverses of each other.
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.