Question

If \(A = \begin{bmatrix} 2 & -3 & 5
3 & 2 & -4
1 & 1 & -2 \end{bmatrix}\), find \(A^{-1}\). Using \(A^{-1}\) solve the following system of equations:
2x - 3y + 5z = 11
3x + 2y - 4z = -5
x + y - 2z = -3.

Hint:

When a question asks you to find an inverse and then solve a system of equations, always check if the coefficient matrix of the system is the same as the matrix whose inverse you just found. This saves a lot of time as you can directly use the result.

Answer 1

Step 1: Understanding the Concept:
This problem has two parts. First, we need to find the inverse of a 3x3 matrix. Second, we use this inverse to solve a system of linear equations. The system of equations is given in the form AX = B, where A is the same matrix for which we need to find the inverse. The solution is given by X = A\(^{-1}\)B.
Step 2: Key Formula or Approach:
Part 1: Finding \(A^{-1}\)
1. Calculate the determinant of A, \(|A|\).
2. Find the matrix of cofactors, C.
3. Find the adjugate of A, adj(A), which is the transpose of C.
4. The inverse is \(A^{-1} = \frac{1}{|A|}\text{adj}(A)\).
Part 2: Solving the system AX = B
1. Identify the matrices A, X, and B from the system of equations.
2. Calculate the solution using X = A\(^{-1}\)B.
Step 3: Detailed Explanation:
Part 1: Finding \(A^{-1}\)
Given \(A = \begin{bmatrix} 2 & -3 & 5
3 & 2 & -4
1 & 1 & -2 \end{bmatrix}\).
1. Determinant of A: \[ |A| = 2(2(-2) - (-4)(1)) - (-3)(3(-2) - (-4)(1)) + 5(3(1) - 2(1)) \] \[ |A| = 2(-4+4) + 3(-6+4) + 5(3-2) = 2(0) + 3(-2) + 5(1) = 0 - 6 + 5 = -1 \] Since \(|A| \neq 0\), the inverse exists.
2. Adjugate of A: The cofactor matrix C is: \[ C_{11} = 0, \quad C_{12} = -(-6+4)=2, \quad C_{13} = 3-2=1 \] \[ C_{21} = -(6-5)=-1, \quad C_{22} = -4-5=-9, \quad C_{23} = -(2-(-3))=-5 \] \[ C_{31} = 12-10=2, \quad C_{32} = -(-8-15)=23, \quad C_{33} = 4-(-9)=13 \] So, \(C = \begin{bmatrix} 0 & 2 & 1
-1 & -9 & -5
2 & 23 & 13 \end{bmatrix}\).
adj(A) = \(C^T = \begin{bmatrix} 0 & -1 & 2
2 & -9 & 23
1 & -5 & 13 \end{bmatrix}\).
3. Inverse of A: \[ A^{-1} = \frac{1}{|A|}\text{adj}(A) = \frac{1}{-1} \begin{bmatrix} 0 & -1 & 2
2 & -9 & 23
1 & -5 & 13 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2
-2 & 9 & -23
-1 & 5 & -13 \end{bmatrix} \] Part 2: Solving the System of Equations
The system is: \[ 2x - 3y + 5z = 11 \] \[ 3x + 2y - 4z = -5 \] \[ x + y - 2z = -3 \] This can be written as AX = B, where: \[ A = \begin{bmatrix} 2 & -3 & 5
3 & 2 & -4
1 & 1 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} x
y
z \end{bmatrix}, \quad B = \begin{bmatrix} 11
-5
-3 \end{bmatrix} \] The solution is X = A\(^{-1}\)B. \[ \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2
-2 & 9 & -23
-1 & 5 & -13 \end{bmatrix} \begin{bmatrix} 11
-5
-3 \end{bmatrix} \] \[ \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 0(11)+1(-5)+(-2)(-3)
-2(11)+9(-5)+(-23)(-3)
-1(11)+5(-5)+(-13)(-3) \end{bmatrix} = \begin{bmatrix} 0-5+6
-22-45+69
-11-25+39 \end{bmatrix} = \begin{bmatrix} 1
2
3 \end{bmatrix} \] Step 4: Final Answer:
The inverse is \(A^{-1} = \begin{bmatrix} 0 & 1 & -2
-2 & 9 & -23
-1 & 5 & -13 \end{bmatrix}\).
The solution to the system of equations is x = 1, y = 2, z = 3.