Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

Find [tex]\( M^{-1} \)[/tex] and show that [tex]\( M^{-1} M = I \)[/tex].

Given:
[tex]\[ M = \begin{bmatrix} 1 & 2 \\ -3 & -7 \end{bmatrix} \][/tex]

Find the value in the first row and second column of the product [tex]\( M^{-1} M \)[/tex] using matrix multiplication. Select the correct expression below and fill in the answer box to complete your work.

A. [tex]\( (7 \cdot 1) + (2 \cdot -3) = \square \)[/tex]
B. [tex]\( (7 \cdot 2) + (2 \cdot -7) = \square \)[/tex]
C. [tex]\( (-3 \cdot 2) + (-1 \cdot -7) = \square \)[/tex]
D. [tex]\( (-3 \cdot 1) + (-1 \cdot -3) = \square \)[/tex]

Find the value in the second row and first column of the product [tex]\( M^{-1} M \)[/tex] using matrix multiplication. Select the correct expression below and fill in the answer box to complete your work.

A. [tex]\( (-3 \cdot 2) + (-1 \cdot -7) = \square \)[/tex]
B. [tex]\( (7 \cdot 1) + (2 \cdot -3) = \square \)[/tex]


Sagot :

To solve this problem, let's approach each part step by step.

1. Find the inverse of [tex]\( M \)[/tex]:
Given matrix [tex]\( M \)[/tex]:
[tex]\[ M = \begin{pmatrix} 1 & 2 \\ -3 & -7 \end{pmatrix} \][/tex]
The inverse of a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] can be found using the formula:
[tex]\[ M^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For our matrix, [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], [tex]\( c = -3 \)[/tex], [tex]\( d = -7 \)[/tex], so:
[tex]\[ ad - bc = (1 \cdot -7) - (2 \cdot -3) = -7 + 6 = -1 \][/tex]
Thus,
[tex]\[ M^{-1} = \frac{1}{-1} \begin{pmatrix} -7 & -2 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} 7 & 2 \\ -3 & -1 \end{pmatrix} \][/tex]

2. Show that [tex]\( M^{-1}M = I \)[/tex]:
To verify, we multiply [tex]\( M^{-1} \)[/tex] by [tex]\( M \)[/tex]:
[tex]\[ M^{-1}M = \begin{pmatrix} 7 & 2 \\ -3 & -1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ -3 & -7 \end{pmatrix} \][/tex]
Perform the matrix multiplication:
[tex]\[ \begin{pmatrix} 7 \cdot 1 + 2 \cdot -3 & 7 \cdot 2 + 2 \cdot -7 \\ -3 \cdot 1 + -1 \cdot -3 & -3 \cdot 2 + -1 \cdot -7 \end{pmatrix} = \begin{pmatrix} 7 - 6 & 14 - 14 \\ -3 + 3 & -6 + 7 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \][/tex]
This verifies that [tex]\( M^{-1} M \)[/tex] is indeed the identity matrix [tex]\( I \)[/tex].

3. Evaluate specific expressions:
- For [tex]\( (7 \cdot 1) + (2 \cdot -3) \)[/tex] corresponding to expression A:
[tex]\[ (7 \cdot 1) + (2 \cdot -3) = 7 - 6 = 1 \][/tex]
Therefore, the value in the first row, second column, corresponding to option A, is:
[tex]\[ \boxed{1} \][/tex]

- For [tex]\( (-3 \cdot 2) + (-1 \cdot -7) \)[/tex] corresponding to expression A:
[tex]\[ (-3 \cdot 2) + (-1 \cdot -7) = -6 + 7 = 1 \][/tex]
Therefore, the value in the second row, first column, corresponding to option A, is:
[tex]\[ \boxed{1} \][/tex]

In summary:
- The inverse of [tex]\( M \)[/tex] is [tex]\( \begin{pmatrix} 7 & 2 \\ -3 & -1 \end{pmatrix} \)[/tex].
- [tex]\( M^{-1} M \)[/tex] results in the identity matrix [tex]\( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)[/tex].
- The simplified value of expression A is [tex]\( 1 \)[/tex].
- The simplified value of expression A for the second row, first column, is [tex]\( 1 \)[/tex].