Find a matrix B such that AB = I, where 6 1 A = 1 0 0 1 and I = -6 6 a b Hint: Let B =c d
Find a matrix B such that AB = I, where 6 1 A = 1 0 0 1 and I = -6 6 a b Hint: Let B =c d
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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![### Finding the Inverse of a Matrix
**Problem Statement:**
Find a matrix \( B \) such that \( AB = I \), where
\[
A = \begin{bmatrix}
6 & 1 \\
-6 & 6
\end{bmatrix}
\]
and
\[
I = \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}.
\]
**Hint:**
Let \( B = \begin{bmatrix}
a & b \\
c & d
\end{bmatrix} \).
### Explanation:
In this problem, you need to find the matrix \( B \) that, when multiplied by matrix \( A \), results in the identity matrix \( I \). The identity matrix \( I \) is the matrix that does not change any matrix it multiplies with, which in 2D looks like
\[
I = \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}.
\]
To find \( B \), set \( B \) as
\[
B = \begin{bmatrix}
a & b \\
c & d
\end{bmatrix}.
\]
You will need to identify the elements \(a\), \(b\), \(c\), and \(d\) such that \( AB = I \).
### Steps to Solve:
1. Multiply \( A \) by \( B \).
2. Set the resulting matrix equal to the identity matrix \( I \).
3. Solve the resulting system of equations to find the values of \( a \), \( b \), \( c \), and \( d \).
### Multiplication:
\[
AB = \begin{bmatrix}
6 & 1 \\
-6 & 6
\end{bmatrix} \begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
= \begin{bmatrix}
6a + c & 6b + d\\
-6a + 6c & -6b + 6d
\end{bmatrix}.
\]
### System of Equations:
Since \( AB = I \),
\[ \begin{bmatrix}
6a + c & 6b + d \\
-6a + 6c & -6b + 6d
\end{b](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4e80d594-14a8-4713-8ea5-1fda205a6b86%2F24aeaef6-0df7-41b6-ab03-1f6257d44ee5%2Fyofslos.png&w=3840&q=75)
Transcribed Image Text:### Finding the Inverse of a Matrix
**Problem Statement:**
Find a matrix \( B \) such that \( AB = I \), where
\[
A = \begin{bmatrix}
6 & 1 \\
-6 & 6
\end{bmatrix}
\]
and
\[
I = \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}.
\]
**Hint:**
Let \( B = \begin{bmatrix}
a & b \\
c & d
\end{bmatrix} \).
### Explanation:
In this problem, you need to find the matrix \( B \) that, when multiplied by matrix \( A \), results in the identity matrix \( I \). The identity matrix \( I \) is the matrix that does not change any matrix it multiplies with, which in 2D looks like
\[
I = \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}.
\]
To find \( B \), set \( B \) as
\[
B = \begin{bmatrix}
a & b \\
c & d
\end{bmatrix}.
\]
You will need to identify the elements \(a\), \(b\), \(c\), and \(d\) such that \( AB = I \).
### Steps to Solve:
1. Multiply \( A \) by \( B \).
2. Set the resulting matrix equal to the identity matrix \( I \).
3. Solve the resulting system of equations to find the values of \( a \), \( b \), \( c \), and \( d \).
### Multiplication:
\[
AB = \begin{bmatrix}
6 & 1 \\
-6 & 6
\end{bmatrix} \begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
= \begin{bmatrix}
6a + c & 6b + d\\
-6a + 6c & -6b + 6d
\end{bmatrix}.
\]
### System of Equations:
Since \( AB = I \),
\[ \begin{bmatrix}
6a + c & 6b + d \\
-6a + 6c & -6b + 6d
\end{b
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