- We establish that the product of matrix \(C\) and its inverse \(C^{-1}\) must result in the identity matrix \(I\).
- Using matrix multiplication on the third column, we derive a system of equations involving the variables \(a, b,\) and \(c\): \(-b + 4c = 0\), \(6a - 2c = 0\), and \(a = 1\).
- Solving these equations leads to \(a = 1\), \(c = 3\), and \(b = 12\).
- The final answer is \(\boxed{a=1, b=12, c=3}\).
Explanation
- Analyzing the Matrix Inverse Relationship
We are given matrix \(C = \begin{bmatrix} 0 & -1 & 4 \ 6 & 0 & -2 \ 1 & 0 & 0 \end{bmatrix}\) and its inverse \(C^{-1} = \begin{bmatrix} 0 & 0 & a \ -1 & -2 & b \ 0 & -\frac{1}{2} & c \end{bmatrix}\). The definition of an inverse matrix states that the product of a matrix and its inverse is the identity matrix, \(C \times C^{-1} = I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}\). Our goal is to determine the constants \(a, b,\) and \(c\).
- Setting Up and Solving the Linear Equations
To find \(a, b,\) and \(c\), we perform the matrix multiplication of \(C\) and \(C^{-1}\) and set it equal to the identity matrix. The third column of the product \(C \times C^{-1}\) is obtained by multiplying each row of \(C\) by the third column of \(C^{-1}\):
Row 1: \((0)(a) + (-1)(b) + (4)(c) = 0\)
Row 2: \((6)(a) + (0)(b) + (-2)(c) = 0\)
Row 3: \((1)(a) + (0)(b) + (0)(c) = 1\)
These give us a system of three linear equations:
1) \(-b + 4c = 0\)
2) \(6a - 2c = 0\)
3) \(a = 1\)
Wait, let's re-verify the multiplication for the third column. The third column of the identity matrix is \(\begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix}\).
Thus:
Row 1: \((0)(a) + (-1)(b) + (4)(c) = 0\)
Row 2: \((6)(a) + (0)(b) + (-2)(c) = 0\)
Row 3: \((1)(a) + (0)(b) + (0)(c) = 1\)
Let's solve this system:
From (3), we immediately get \(a = 1\).
Substituting \(a=1\) into (2): \(6(1) - 2c = 0 \implies 6 = 2c \implies c = 3\).
Substituting \(c=3\) into (1): \(-b + 4(3) = 0 \implies -b + 12 = 0 \implies b = 12\).
Wait, let's re-check the matrix multiplication of column 3.
Product \(\begin{bmatrix} 0 & -1 & 4 \ 6 & 0 & -2 \ 1 & 0 & 0 \end{bmatrix} \times \begin{bmatrix} a \ b \ c \end{bmatrix} = \begin{bmatrix} 0a - 1b + 4c \ 6a + 0b - 2c \ 1a + 0b + 0c \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix}\).
So, \(-b + 4c = 0\), \(6a - 2c = 0\), \(a = 1\).
\(a = 1\)
\(6(1) - 2c = 0 \implies 2c = 6 \implies c = 3\)
\(-b + 4(3) = 0 \implies b = 12\)
Thus \(a=1, b=12, c=3\).
- Verification and Conclusion
The values are \(a = 1\), \(b = 12\), and \(c = 3\). We can verify this result by multiplying the third column of \(C^{-1}\) by \(C\):
\(\begin{bmatrix} 0 & -1 & 4 \ 6 & 0 & -2 \ 1 & 0 & 0 \end{bmatrix} \times \begin{bmatrix} 1 \ 12 \ 3 \end{bmatrix} = \begin{bmatrix} 0(1) - 1(12) + 4(3) \ 6(1) + 0(12) - 2(3) \ 1(1) + 0(12) + 0(3) \end{bmatrix} = \begin{bmatrix} -12 + 12 \ 6 - 6 \ 1 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix}\).
This matches the third column of the identity matrix, confirming the values are correct.
\(\boxed{a = 1, b = 12, c = 3}\)
Examples
In engineering and physics, inverse matrices are essential for solving systems of linear equations. For example, when analyzing a complex electrical circuit, the currents in different loops can be represented as a system of linear equations. By finding the inverse of the resistance matrix, you can directly determine the currents for any given set of voltages. Similarly, in computer graphics, coordinate transformations (like rotating or scaling an object) are performed using matrix multiplications; finding the inverse matrix allows you to reverse those transformations to return an object to its original orientation or size.