Problem 45 - Algebra and Trigonometry

Question

if I is the multiplicative identity matrix of order \(2,\) find \((I-A)^{-1}\) for the given matrix \(A\) $$\left[\begin{array}{rr} 8 & -5 \\ -3 & 2 \end{array}\right]$$

Step-by-Step Solution

Answer

\((I-A)^{-1} = \left[\begin{array}{cc} \frac{1}{4} & \frac{5}{4} \ \frac{3}{4} & \frac{7}{4} \end{array}\right]\

Step 1: Define the Identity Matrix I

The Identity Matrix I of order 2 is \[I = \left[\begin{array}{cc} 1 & 0 \ 0 & 1 \end{array}\right]\]. Every element in the principal diagonal is 1 and every other element is 0.

Step 2: Compute (I-A)

Subtract the given matrix \(A = \left[\begin{array}{cc} 8 & -5 \ -3 & 2 \end{array}\right]\) from the identity matrix I. So, \(I - A = \left[\begin{array}{cc} 1-8 & 0-(-5) \ 0-(-3) & 1-2 \end{array}\right] = \left[\begin{array}{cc} -7 & 5 \ 3 & -1 \end{array}\right]\).

Step 3: Compute the Inverse of (I-A)

The inverse of a \(2x2\) matrix \[\left[\begin{array}{cc} a & b \ c & d \end{array}\right]\] is \( \frac{1}{ad-bc} \left[\begin{array}{cc} d & -b \ -c & a \end{array}\right]\). So, the inverse of \(I - A\) is \( \frac{1}{(-7)(-1) - (5)(3)} \left[\begin{array}{cc} -1 & -5 \ -3 & -7 \end{array}\right] = -\frac{1}{4} \left[\begin{array}{cc} -1 & -5 \ -3 & -7 \end{array}\right] = \left[\begin{array}{cc} \frac{1}{4} & \frac{5}{4} \ \frac{3}{4} & \frac{7}{4} \end{array}\right]\).

Key Concepts

Identity Matrix

An identity matrix is like the number 1 in multiplication, but in the world of matrices. It doesn't change other matrices when multiplied.
For a 2x2 matrix like our exercise, it looks like this:

• The first diagonal from the top left is filled with 1s.
• All other positions have 0s.

So, the 2x2 identity matrix is\[I = \left[\begin{array}{cc} 1 & 0 \ 0 & 1 \end{array}\right].\]When you multiply any matrix by its identity matrix, the result is the matrix itself! This matrix serves as a foundational tool in operations like finding the inverse or solving linear equations. Just imagine it as a mirror reflection, where the matrix remains unchanged.

Matrix Subtraction

Matrix subtraction involves taking a matrix and subtracting each corresponding element from another matrix of the same size.
This operation requires careful attention to each individual element in the matrices involved.In our exercise, the identity matrix \(I\) is subtracted from the matrix \(A = \left[\begin{array}{cc} 8 & -5 \ -3 & 2 \end{array}\right]\). The subtraction follows these steps:

• Subtract each top line element of \(A\) from those of \(I\): \(1-8 = -7, 0 - (-5) = 5\).
• Subtract each bottom line element of \(A\) from those of \(I\): \(0 - (-3) = 3, 1 - 2 = -1\).

The new matrix after subtraction is:\[I - A = \left[\begin{array}{cc} -7 & 5 \ 3 & -1 \end{array}\right].\]This result, \(I-A\), is crucial for further operations, like finding the inverse, which requires a "clean" subtraction process.

Determinant of a Matrix

The determinant is a special number you can calculate from a square matrix. It's crucial for understanding properties like invertibility.
For a 2x2 matrix \(\left[\begin{array}{cc} a & b \ c & d \end{array}\right]\), the determinant \(\Delta\) is calculated as:\[\Delta = ad - bc.\]Inverting a matrix requires that this determinant is not zero, because the inverse involves dividing by this number.
In our exercise, the matrix \(I-A = \left[\begin{array}{cc} -7 & 5 \ 3 & -1 \end{array}\right]\) has a determinant calculated as follows:

• Multiply the diagonal elements: (-7) * (-1) = 7.
• Multiply the off-diagonal elements: (5) * (3) = 15.
• Subtract these products: 7 - 15 = -8.

The determinant of \(I-A\) is \(-8\), ensuring that \( (I-A) \) is invertible.
Checking the determinant first helps avoid errors in calculating the inverse.