Problem 39 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The solution matrix is \(\begin{bmatrix} 7 & 2 & 3 \\ 10 & 3 & 5 \end{bmatrix}\).

Step 1: Identify the Matrix

We begin by identifying the provided matrices in the equation. The equation is given as \(\begin{bmatrix} 3 & -2 \ -4 & 3 \end{bmatrix} \begin{bmatrix} x & y & z \ u & v & w \end{bmatrix} = \begin{bmatrix} 1 & 0 & -1 \ 2 & 1 & 3 \end{bmatrix} \). The first matrix is a 2x2 matrix and the second matrix is a 2x3 matrix.

Step 2: Find the Inverse of the First Matrix

To solve for the unknown matrix, we need to find the inverse of the 2x2 matrix \(\begin{bmatrix} 3 & -2 \ -4 & 3 \end{bmatrix} \). The formula for the inverse of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix} \) is \(\frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \) if \(ad - bc eq 0\). Therefore, calculate: \(3 \cdot 3 - (-2) \cdot (-4) = 9 - 8 = 1 \). The inverse is: \(\begin{bmatrix} 3 & 2 \ 4 & 3 \end{bmatrix} \).

Step 3: Multiply by the Inverse Matrix on Both Sides

Multiply both sides of the equation by the inverse of the 2x2 matrix from the left. Calculate: \(\begin{bmatrix} 3 & 2 \ 4 & 3 \end{bmatrix} \begin{bmatrix} 3 & -2 \ -4 & 3 \end{bmatrix} \begin{bmatrix} x & y & z \ u & v & w \end{bmatrix} = \begin{bmatrix} 3 & 2 \ 4 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & -1 \ 2 & 1 & 3 \end{bmatrix} \). Simplify to get \(\begin{bmatrix} x & y & z \ u & v & w \end{bmatrix} = \begin{bmatrix} 3 & 2 \ 4 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & -1 \ 2 & 1 & 3 \end{bmatrix} \).

Step 4: Compute the Resulting Product

Now, perform the matrix multiplication \(\begin{bmatrix} 3 & 2 \ 4 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & -1 \ 2 & 1 & 3 \end{bmatrix} \) to obtain the solution matrix. Calculate: \(\begin{bmatrix} 3\cdot1 + 2\cdot2 & 3\cdot0 + 2\cdot1 & 3\cdot(-1) + 2\cdot3 \ 4\cdot1 + 3\cdot2 & 4\cdot0 + 3\cdot1 & 4\cdot(-1) + 3\cdot3 \end{bmatrix} = \begin{bmatrix} 7 & 2 & 3 \ 10 & 3 & 5 \end{bmatrix} \).

Step 5: Verify the Result

The final matrix \(\begin{bmatrix} 7 & 2 & 3 \ 10 & 3 & 5 \end{bmatrix} \) is the answer. Ensure that each calculation step is verified for accuracy, confirming that all computations have resulted in the correct solution.

Key Concepts

Inverse Matrix

The inverse of a matrix is a powerful tool in solving matrix equations. Imagine you have a matrix equation such as \( A \cdot X = B \), where \( A \) and \( B \) are known matrices, and \( X \) is the matrix we want to find. To isolate \( X \), we need the inverse of matrix \( A \), designated as \( A^{-1} \), to both sides of the equation: \( A^{-1} \cdot (A \cdot X) = A^{-1} \cdot B \). This simplifies to \( X = A^{-1} \cdot B \), because \( A^{-1} \cdot A \) is the identity matrix, essentially "canceling out" \( A \).

For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is calculated as \( \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \), provided that \( ad - bc eq 0 \). This value \( (ad - bc) \) is called the determinant, and if it's zero, the matrix has no inverse. Thus, calculating the inverse allows for transforming matrix equations into simpler, more solvable forms.

Matrix Multiplication

Matrix multiplication is central in matrix equations. Unlike regular multiplication, matrix multiplication is not commutative; meaning \( A \cdot B eq B \cdot A \), in most cases. To multiply two matrices, say matrix \( A \) and matrix \( B \), the number of columns in \( A \) must match the number of rows in \( B \).

The resulting matrix's element in row \( i \) and column \( j \) is the dot product of the \( i \)-th row of the first matrix and the \( j \)-th column of the second. For example, if \( A \) is a 2x2 matrix and \( B \) is a 2x3 matrix, their product will be a 2x3 matrix. Perform this process element-wise across the entire matrices to complete the multiplication.

• Use the formula \( (A \cdot B)_{ij} = \sum_k A_{ik}B_{kj} \) to compute elements.
• Check dimensions carefully to ensure compatibility.
• Each resulting element represents a summation of products between corresponding row and column elements.

2x2 Matrices

A 2x2 matrix is a basic form of a matrix comprising two rows and two columns, structured as \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \). They are straightforward to handle compared to larger matrices but still encapsulate all the basic properties of matrix operations.

Key properties include:

• Determinant: computed as \( ad - bc \). It is pivotal in determining whether an inverse exists.
• Inverse: given by \( \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \) if \( ad - bc eq 0 \).
• Simplicity: easy to visualize and compute, making them ideal for learning foundational principles of linear algebra.

In practice, using 2x2 matrices simplifies many larger real-world problems into manageable steps and acts as an entry point to more intricate matrix operations like those involving larger systems.

Solving Matrix Equations

Solving matrix equations extends the concept of solving standard algebraic equations to systems involving matrices. The essence lies in using inverses to isolate and find unknown matrices.

Given an equation like \( A \cdot X = B \), we solve for matrix \( X \) by finding \( X = A^{-1} \cdot B \). Steps to solve such equations include:

• Identify matrices involved, confirming dimensions and compatibility.
• Calculate the inverse for the provided matrix, ensuring its determinant is non-zero.
• Multiply both sides of the equation by this inverse, simplifying the equation.
• Conduct matrix multiplications accurately, following the rules of dot products for elements.

Verification is crucial post-calculation, ensuring all steps reflect correct arithmetic and logical consistency, thereby delivering reliable solutions to complex sets of equations.