Problem 37 - Precalculus: Mathematical for Calculus

Question

Solve the matrix equation by multiplying each side by the appropriate inverse matrix. $$\left[\begin{array}{rr} 3 & -2 \\ -4 & 3 \end{array}\right]\left[\begin{array}{lll} x & y & z \\ u & v & w \end{array}\right]=\left[\begin{array}{llr} 1 & 0 & -1 \\ 2 & 1 & 3 \end{array}\right]$$

Step-by-Step Solution

Answer

The solution matrix is \( \left[\begin{array}{ccc} 7 & 2 & 3 \\ 10 & 3 & 5 \end{array}\right] \).

Step 1: Identify the Matrix Equation

The given matrix equation is \( A \cdot X = B \), where \( A = \left[\begin{array}{rr}3 & -2 \-4 & 3\end{array}\right] \), \( X = \left[\begin{array}{lll}x & y & z \u & v & w\end{array}\right] \), and \( B = \left[\begin{array}{llr}1 & 0 & -1 \2 & 1 & 3\end{array}\right] \). Our goal is to solve for \( X \).

Step 2: Find the Inverse of Matrix A

To find the inverse of matrix \( A \), we first calculate the determinant: \( \text{det}(A) = (3)(3) - (-2)(-4) = 9 - 8 = 1 \). Since the determinant is 1, we can find the inverse of \( A \), denoted as \( A^{-1} \). The inverse is given by:\[ A^{-1} = \frac{1}{\text{det}(A)} \left[\begin{array}{rr}3 & 2 \4 & 3\end{array}\right] \]Thus, \( A^{-1} = \left[\begin{array}{rr}3 & 2 \4 & 3\end{array}\right] \).

Step 3: Multiply Both Sides by the Inverse of A

To solve for \( X \), multiply both sides of the equation by \( A^{-1} \) to get:\[ A^{-1}A \cdot X = A^{-1}B \]Since \( A^{-1}A = I \) (the identity matrix), we have \( X = A^{-1}B \).

Step 4: Calculate A^{-1}B

Multiply the inverse \( A^{-1} = \left[\begin{array}{rr}3 & 2 \4 & 3\end{array}\right] \) by matrix \( B = \left[\begin{array}{llr}1 & 0 & -1 \2 & 1 & 3\end{array}\right] \):\[ \left[\begin{array}{rr}3 & 2 \4 & 3\end{array}\right] \left[\begin{array}{llr}1 & 0 & -1 \2 & 1 & 3\end{array}\right] = \left[\begin{array}{ccc}3 \cdot 1 + 2 \cdot 2 & 3 \cdot 0 + 2 \cdot 1 & 3 \cdot (-1) + 2 \cdot 3 \4 \cdot 1 + 3 \cdot 2 & 4 \cdot 0 + 3 \cdot 1 & 4 \cdot (-1) + 3 \cdot 3\end{array}\right] \]This results in the matrix \( X = \left[\begin{array}{ccc}7 & 2 & 3 \10 & 3 & 5\end{array}\right] \).

Step 5: Present the Final Solution

The matrix \( X \) is found to be \( \left[\begin{array}{ccc}7 & 2 & 3 \10 & 3 & 5\end{array}\right] \). This is the solution to the matrix equation \( A \cdot X = B \).

Key Concepts

Inverse Matrix

An inverse matrix is an essential concept when solving systems of linear equations using matrices. An inverse of a square matrix \( A \) is another matrix, \( A^{-1} \), such that when multiplied with \( A \) results in the identity matrix \( I \), which is a matrix with ones along the diagonal and zeros elsewhere. This inverse property is expressed as:

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

Finding the inverse of a matrix involves determining whether the matrix is invertible by calculating the determinant. If the determinant is not zero, the matrix has an inverse, and if it is zero, it does not.

For a 2x2 matrix \( A = \left[\begin{array}{cc} a & b \ c & d \end{array}\right] \), the inverse \( A^{-1} \) can be calculated as:
\[ A^{-1} = \frac{1}{ad - bc} \left[\begin{array}{cc} d & -b \ -c & a \end{array}\right] \]Here, \( ad - bc \) is the determinant. Notice that the positions of \( a \) and \( d \) are swapped, and \( b \) and \( c \) are negated in the inverse matrix. Finding the inverse is crucial in the methodology used to solve the original exercise by transforming the matrix equation into one that directly isolates \( X \).

Determinant of a Matrix

The determinant is a critical value computed from a square matrix. It provides important analytic information about the matrix, and it's central in determining if a matrix can have an inverse. For a 2x2 square matrix \( A = \left[\begin{array}{cc} a & b \ c & d \end{array}\right] \), the determinant is calculated as:
\[ \text{det}(A) = ad - bc \]A determinant helps us understand certain properties of a matrix:

• If the determinant is zero, the matrix is singular, meaning it does not have an inverse.
• If the determinant is non-zero, the matrix is invertible, and you can find its inverse.

In the original exercise, knowing the determinant of \( A \) was crucial for confirming it had an inverse. With a non-zero determinant of 1, the matrix \( A \) was confirmed to be invertible, allowing us to proceed with finding \( A^{-1} \). Calculating determinants is not only essential for solving matrix equations but also for applications in geometry, computing volumes, and even in differential calculus.

Matrix Multiplication

Matrix multiplication is a fundamental operation where two matrices are combined to form another matrix. It is different from simple scalar multiplication. For two matrices \( A \) and \( B \), the multiplication \( A \times B \) is possible only when the number of columns in the first matrix \( A \) matches the number of rows in the second matrix \( B \).

To find the element in the \( i^{th} \) row and \( j^{th} \) column of the resulting matrix \( C \), you multiply corresponding elements and then sum up these products:
\[ C_{ij} = \sum_{k=1}^{n} A_{ik} \times B_{kj} \]Matrix multiplication has the following properties that are often useful:

• Associative: \( (AB)C = A(BC) \)
• Distributive: \( A(B + C) = AB + AC \)
• Not Commutative: Generally, \( AB eq BA \)

In the context of the original exercise, matrix multiplication was utilized to both affirm the identity matrix after multiplying \( A^{-1} \) and \( A \), and to solve for \( X \) by calculating \( A^{-1}B \). Mastery of matrix multiplication enables the solving of complex matrix equations effectively, providing solutions in a structured, mathematical framework.