Problem 8 - College Algebra with Modeling and Visualization

Question

Find the value of the constant \(k\) in \(A^{-1}\). A=\left[\begin{array}{rr} -2 & 2 \\ 1 & -2 \end{array}\right], \quad A^{-1}=\left[\begin{array}{rr} -1 & k \\ -0.5 & -1 \end{array}\right]

Step-by-Step Solution

Answer

The value of \( k \) is -1.

Step 1: Understand the relationship between a matrix and its inverse

Given matrices \( A \) and \( A^{-1} \), the product of these two matrices should be the identity matrix \( I \). So, \( A \times A^{-1} = I \), where \( I \) is a matrix with diagonal elements as 1 and the rest as 0.

Step 2: Write the equations from the matrix multiplication

Let \( A \) and \( A^{-1} \) be:\[A = \begin{pmatrix} -2 & 2 \ 1 & -2 \end{pmatrix}, \quad A^{-1} = \begin{pmatrix} -1 & k \ -0.5 & -1 \end{pmatrix}\]Perform the matrix multiplication: \[ A \times A^{-1} = \begin{pmatrix} (-2)(-1) + (2)(-0.5) & (-2)(k) + (2)(-1) \ (1)(-1) + (-2)(-0.5) & (1)(k) + (-2)(-1) \end{pmatrix} \]This should result in the identity matrix \( I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \).

Step 3: Simplify and solve the equations

Substitute and simplify the equations from Step 2:- First element: \(-2(-1) + 2(-0.5) = 2 - 1 = 1\), which matches \( I_{11} \).- Second element: \(-2k + 2(-1) = -2k - 2 = 0\), solve for \( k \): \[ -2k - 2 = 0 \Rightarrow -2k = 2 \Rightarrow k = -1 \].- Third check with \( 1(-1) + (-2)(-0.5) = -1 + 1 = 0 \), matches \( I_{21} \).- Fourth element: \( 1(k) + (-2)(-1) = k + 2 \) matches \( I_{22} \). Since \( k = -1 \), \( k + 2 = 1 \).

Step 4: Verify the solution

Verify if \( k = -1 \) satisfies all conditions for the inverse rules. Re-calculate the multiplication:\[ (-2)(-1) + 2(-0.5) = 1\], \[(-2)(-1) - 2 = 0\], \[1(-1) + 1 = 0\], and \[k + 2 = 1\]. All conditions are satisfied.

Key Concepts

Matrix Multiplication

Matrix multiplication is different from the regular multiplication we perform with numbers. It is a systematic process that allows us to manipulate matrices, which are essential in linear algebra. The key is understanding how elements interact:

• The element in the first row and first column of the resulting matrix is the sum of the products of the corresponding elements from the first row of the first matrix and the first column of the second matrix.
• This process is repeated for each position in the final matrix by sliding across rows of the first matrix and columns of the second matrix.

In our exercise, we're forming products between the matrix elements of \( A \) and its inverse \( A^{-1} \) to ensure it results in an identity matrix. When multiplying, each pair of terms from \( A \) and \( A^{-1} \) contributes to achieving the identity matrix \( I \), which plays a critical role in verifying that two matrices are indeed inverses of each other.
Matrix multiplication is not commutative, meaning \( A \times B eq B \times A \) in most cases. This unique property requires careful calculation and attention to the order of multiplication.

Identity Matrix

The identity matrix is fundamental in linear algebra and acts like the number 1 in arithmetic multiplication. Its primary role is to serve as a neutral element:

• For any matrix \( A \), multiplying \( A \) by the identity matrix \( I \) leaves \( A \) unchanged. Hence, \( A \times I = A \) and \( I \times A = A \).
• It contains 1s on its diagonal (from top left to bottom right), and all other elements are 0.
• It's denoted as \( I \) with the size depending on the context, such as \( I_{2} \) or \( I_{3} \) to indicate a 2x2 or 3x3 identity matrix respectively.

In the exercise, the task was to ensure that upon multiplying matrices \( A \) and \( A^{-1} \), the result is the identity matrix. This verifies that \( A^{-1} \) is truly the inverse of \( A \). Achieving the identity matrix confirms that all matrix calculations align with the rules of inverses.
The identity matrix plays a crucial role in more advanced topics such as solving linear equations and defining inverses, providing a stepping stone to deeper mathematical insights.

Solving Linear Equations

Solving linear equations using matrices is an efficient method, especially useful when dealing with multiple variables. The process often involves matrix inversion:

• First, express the system of equations in matrix form; typically \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the variable matrix, and \( B \) is the constants matrix.
• To find \( X \), calculate the inverse of matrix \( A \), and perform the operation \( A^{-1}B = X \).
• Compute \( X \) by matrix multiplication, ensuring each step adheres to the hierarchical operations of matrices.

In our exercise context, to ensure an accurate solution, validating that \( A \times A^{-1} = I \) serves as a critical check. This ensures that the matrix formed indeed represents the solution pathway to the linear equation problem.
Understanding how matrix inversion helps solve linear equations provides a powerful tool, making complex systems more manageable and unraveling solutions that may seem intricate at first glance.