Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
To find the inverse of the matrix [tex]\( A \)[/tex], we need to follow a few key steps. The matrix [tex]\( A \)[/tex] is given as:
[tex]\[ A = \begin{pmatrix} 6 & 1 \\ 11 & 2 \end{pmatrix} \][/tex]
First, we calculate the determinant of [tex]\( A \)[/tex]. For a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex], the determinant is given by:
[tex]\[ \text{det}(A) = ad - bc \][/tex]
Plugging in the values from matrix [tex]\( A \)[/tex]:
[tex]\[ \text{det}(A) = (6 \times 2) - (1 \times 11) = 12 - 11 = 1 \][/tex]
Since the determinant is 1, which is non-zero, the matrix [tex]\( A \)[/tex] is invertible.
Next, for a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex], the inverse [tex]\( A^{-1} \)[/tex] is given by:
[tex]\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
Applying this to matrix [tex]\( A \)[/tex]:
[tex]\[ A^{-1} = \frac{1}{1} \begin{pmatrix} 2 & -1 \\ -11 & 6 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -11 & 6 \end{pmatrix} \][/tex]
So, the inverse of matrix [tex]\( A \)[/tex] should be:
[tex]\[ A^{-1} = \begin{pmatrix} 2 & -1 \\ -11 & 6 \end{pmatrix} \][/tex]
However, the numerical values may contain minor discrepancies due to floating-point arithmetic precision inherent in calculations. Specifically, the result we found is:
[tex]\[ A^{-1} = \begin{pmatrix} 2.0000000000000018 & -1.0000000000000009 \\ -11.00000000000001 & 6.000000000000005 \end{pmatrix} \][/tex]
Thus, the inverse of the matrix [tex]\( A \)[/tex] is precisely:
[tex]\[ A^{-1} = \begin{pmatrix} 2.0000000000000018 & -1.0000000000000009 \\ -11.00000000000001 & 6.000000000000005 \end{pmatrix} \][/tex]
[tex]\[ A = \begin{pmatrix} 6 & 1 \\ 11 & 2 \end{pmatrix} \][/tex]
First, we calculate the determinant of [tex]\( A \)[/tex]. For a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex], the determinant is given by:
[tex]\[ \text{det}(A) = ad - bc \][/tex]
Plugging in the values from matrix [tex]\( A \)[/tex]:
[tex]\[ \text{det}(A) = (6 \times 2) - (1 \times 11) = 12 - 11 = 1 \][/tex]
Since the determinant is 1, which is non-zero, the matrix [tex]\( A \)[/tex] is invertible.
Next, for a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex], the inverse [tex]\( A^{-1} \)[/tex] is given by:
[tex]\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
Applying this to matrix [tex]\( A \)[/tex]:
[tex]\[ A^{-1} = \frac{1}{1} \begin{pmatrix} 2 & -1 \\ -11 & 6 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -11 & 6 \end{pmatrix} \][/tex]
So, the inverse of matrix [tex]\( A \)[/tex] should be:
[tex]\[ A^{-1} = \begin{pmatrix} 2 & -1 \\ -11 & 6 \end{pmatrix} \][/tex]
However, the numerical values may contain minor discrepancies due to floating-point arithmetic precision inherent in calculations. Specifically, the result we found is:
[tex]\[ A^{-1} = \begin{pmatrix} 2.0000000000000018 & -1.0000000000000009 \\ -11.00000000000001 & 6.000000000000005 \end{pmatrix} \][/tex]
Thus, the inverse of the matrix [tex]\( A \)[/tex] is precisely:
[tex]\[ A^{-1} = \begin{pmatrix} 2.0000000000000018 & -1.0000000000000009 \\ -11.00000000000001 & 6.000000000000005 \end{pmatrix} \][/tex]
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.