Problem 38 - Precalculus: Mathematical for Calculus

Question

Solve the matrix equation by multiplying each side by the appropriate inverse matrix. $$\left[\begin{array}{rrr} 0 & -2 & 2 \\ 3 & 1 & 3 \\ 1 & -2 & 3 \end{array}\right]\left[\begin{array}{ll} x & u \\ y & v \\ z & w \end{array}\right]=\left[\begin{array}{lr} 3 & 6 \\ 6 & 12 \\ 0 & 0 \end{array}\right]$$

Step-by-Step Solution

Answer

Find the inverse of matrix A and multiply it by matrix C to solve for B.

Step 1: Identify the Matrices

The given matrix equation is \( A \cdot B = C \), where \( A = \begin{bmatrix} 0 & -2 & 2 \ 3 & 1 & 3 \ 1 & -2 & 3 \end{bmatrix} \), \( B = \begin{bmatrix} x & u \ y & v \ z & w \end{bmatrix} \), and \( C = \begin{bmatrix} 3 & 6 \ 6 & 12 \ 0 & 0 \end{bmatrix} \). We need to solve for matrix \( B \).

Step 2: Find the Inverse of Matrix A

To isolate \( B \), we need the inverse of matrix \( A \). Calculate the determinant of \( A \), which is \( \ |A| = 0(1 \cdot 3 - 3(-2)) - (-2)(3 \cdot 3 - 1(-2)) + 2(3(-2) - 1(1)) = -20. \). Then, calculate the adjugate matrix and divide it by the determinant to find \( A^{-1} \).

Step 3: Verify Matrix Inverse

The inverse matrix \( A^{-1} \) should satisfy \( A \cdot A^{-1} = I \, \) where \( I \) is the identity matrix. Verify the calculations to ensure \( A^{-1} \) is correct. If the calculation matches, move to the next step.

Step 4: Multiply Each Side by Inverse Matrix

Multiply both sides of the equation \( A \cdot B = C \) by \( A^{-1} \) on the left: \ \( A^{-1} \cdot A \cdot B = A^{-1} \cdot C \. \) This simplifies to \( I \cdot B = A^{-1} \cdot C \, \) leading to \( B = A^{-1} \cdot C \).

Step 5: Calculate B

Now calculate \( B \) by multiplying \( A^{-1} \) by \( C \). Perform each matrix multiplication operation. This gives the solution for \( B \) as a 3x2 matrix, which will be the final result.

Key Concepts

Inverse Matrix

An inverse matrix is a crucial concept when solving matrix equations. It is akin to finding a reciprocal in arithmetic. For a square matrix \( A \), its inverse \( A^{-1} \) exists if and only if \( A \) is invertible, which means its determinant is not zero. The inverse, when multiplied by the original matrix, yields the identity matrix:

• \( A \cdot A^{-1} = I \)
• \( A^{-1} \cdot A = I \)

To calculate the inverse, you must first find the determinant of the matrix. If the determinant is zero, the matrix has no inverse. Otherwise, use the formula involving the adjugate (or adjoint) of the matrix divided by the determinant. Thus, knowing how to find the inverse is essential for solving matrix equations, like in the given exercise, where isolating \( B \) involved using \( A^{-1} \).

Determinant

The determinant is a special number that can be calculated from a square matrix. It provides essential information about the matrix, such as whether the matrix is invertible. For a 3x3 matrix \( A \), the determinant is calculated through a specific formula involving its elements:\[ |A| = a(ei-fh) - b(di-fg) + c(dh-eg) \]Where \( a, b, c, d, e, f, g, h, i \) are the elements of the matrix in row-major order. In simpler terms, it's a bit like a weighted sum of the matrix's minors with alternating signs. In the exercise, calculating the determinant of matrix \( A \) was necessary to find its inverse, as it helped verify \( A \) had an inverse by ensuring it's not zero. If the determinant is non-zero, the matrix is invertible. Otherwise, the matrix equation cannot be solved using inverses.

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra, essential for transforming data represented in matrices. When multiplying matrix \( A \) by matrix \( B \), the result is matrix \( C \), where each element \( c_{ij} \) of \( C \) is the dot product of the \( i^{th} \) row of \( A \) and the \( j^{th} \) column of \( B \). For two matrices \( A \) of size \( m \times n \) and \( B \) of size \( n \times p \), the resulting matrix \( C \) is of size \( m \times p \):

• \( c_{ij} = \sum_{k=1}^{n} a_{ik} \cdot b_{kj} \)

This operation is not commutative, meaning \( A \cdot B eq B \cdot A \). In the exercise, after obtaining \( A^{-1} \), matrix multiplication was used to solve for \( B \) by multiplying \( A^{-1} \) with matrix \( C \). The careful application of matrix multiplication ensures accurate results, facilitating the solution of complex matrix equations.