Question

In each case, solve the systems of equations by finding the inverse of the coefficient matrix. $$ \begin{array}{llrl} \text { a. } & 3 x-y=5 & & \text { b. } & 2 x-3 y=0 \\ & 2 x+2 y=1 & & & x-4 y=1 \\ \text { c. } x+y+2 z= & 5 & \text { d. } & x+4 y+2 z=1 \\ & x+y+z=0 & & 2 x+3 y+3 z=-1 \\ & x+2 y+4 z=-2 & & 4 x+y+4 z=0 \end{array} $$

Short answer

Solve each system using the inverse of the coefficient matrix: (a) and (b) use 2x2 inverses, (c) and (d) use larger matrix inverses.

Step-by-step solution

Define the System of Equations

Identify the coefficients and the constants for each system of equations. We'll solve (a), (b), (c), and (d) separately.For (a): 1. Equation 1: \(3x - y = 5\) 2. Equation 2: \(2x + 2y = 1\)For (b): 1. Equation 1: \(2x - 3y = 0\)2. Equation 2: \(x - 4y = 1\)For (c): 1. Equation 1: \(x + y + 2z = 5\)2. Equation 2: \(x + y + z = 0\)3. Equation 3: \(x + 2y + 4z = -2\)For (d): 1. Equation 1: \(x + 4y + 2z = 1\)2. Equation 2: \(2x + 3y + 3z = -1\)3. Equation 3: \(4x + y + 4z = 0\)

Form the Coefficient Matrix and Constant Matrix

For each system, extract the coefficients to form a matrix \(A\) and constants to form a matrix \(B\).For (a):\(A = \begin{pmatrix} 3 & -1 \ 2 & 2 \end{pmatrix}, B = \begin{pmatrix} 5 \ 1 \end{pmatrix}\)For (b):\(A = \begin{pmatrix} 2 & -3 \ 1 & -4 \end{pmatrix}, B = \begin{pmatrix} 0 \ 1 \end{pmatrix}\)For (c):\(A = \begin{pmatrix} 1 & 1 & 2 \ 1 & 1 & 1 \ 1 & 2 & 4 \end{pmatrix}, B = \begin{pmatrix} 5 \ 0 \ -2 \end{pmatrix}\)For (d):\(A = \begin{pmatrix} 1 & 4 & 2 \ 2 & 3 & 3 \ 4 & 1 & 4 \end{pmatrix}, B = \begin{pmatrix} 1 \ -1 \ 0 \end{pmatrix}\)

Calculate the Inverse of the Coefficient Matrix

Compute the inverse of matrix \(A\) for each case, denoted as \(A^{-1}\). The inverse is calculated using the formula for 2x2 matrices:\[ A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \]For larger matrices, use the row reduction method or determinants and adjugate method.For (a): Check if the determinant is non-zero.\( \text{det}(A) = 3(2) - (-1)(2) = 6 + 2 = 8 eq 0\) For (b):\( \text{det}(A) = 2(-4) - (-3)(1) = -8 + 3 = -5 eq 0\)For (c) and (d), use computational tools or manual calculation to check if the determinant is non-zero and find the inverse.

Solve for the Variable Matrix

Apply the inverse matrix to find the solution vector \(X\) using \(X = A^{-1} B\).For (a): Calculate \(X = A^{-1} B\).For (b): Calculate \(X = A^{-1} B\).For (c) and (d), calculate \(X = A^{-1} B\) using the inverses obtained in step 3.

Verify the Solutions

Substitute the solutions back into the original equations to confirm they satisfy all equations. For each set of solutions, cross-verify by plugging the values of \(x, y, z\) back into the equations for (a), (b), (c), and (d) ensuring each equality holds true.

Key concepts

Coefficient Matrix

Understanding the coefficient matrix is essential when solving systems of equations using matrices. Imagine you have a few equations, each with variables; the coefficient matrix is a way of organizing these coefficients neatly into a grid format. Each row represents one equation, while each column corresponds to one variable. For example, in system (a):

• Equation 1: \(3x - y = 5\)
• Equation 2: \(2x + 2y = 1\)

The coefficient matrix \(A\) for these equations is:\[A = \begin{pmatrix} 3 & -1 \ 2 & 2 \end{pmatrix}\]This matrix forms the backbone of our calculations, as everything hinges on it when we compute solutions using the inverse matrix method.

Inverse Matrix

The inverse matrix is a powerful tool for solving systems of linear equations. If you think of the coefficient matrix as a grid of numbers, the inverse matrix sort of 'undoes' the effects of the original matrix. When you multiply a matrix by its inverse, you end up with the identity matrix, which is similar to multiplying a number by its reciprocal to get 1. For a square matrix \(A\), its inverse \(A^{-1}\) helps solve equations like \(AX = B\) by transforming it into \(X = A^{-1}B\). This means we can directly find the variable matrix \(X\), which holds our solutions.It's important to remember that only matrices with non-zero determinants have inverses. Hence, checking the determinant first is crucial. For example, the determinant in system (a) is 8, which is not zero, confirming that an inverse exists. Calculating the inverse can be straightforward for 2x2 matrices using the formula:\[A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\]

Determinant

The determinant is a special number calculated from a square matrix. It's incredibly important because it tells us whether a matrix has an inverse. When the determinant is zero, the matrix doesn't have an inverse, making it impossible to solve the system using matrix inversion. For a 2x2 matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is computed as:\[a \cdot d - b \cdot c\]If this number isn't zero, you're in luck because it means you can find an inverse.In our example with system (a), the determinant of the coefficient matrix \(A\) is computed as:\[3 \cdot 2 - (-1) \cdot 2 = 6 + 2 = 8\]This non-zero result confirms that we can safely proceed to calculate the inverse matrix to find solutions for the set of equations.

Row Reduction Method

The row reduction method, also known as Gaussian elimination, is a way to solve systems of equations by simplifying the coefficient matrix step by step. By performing a series of row operations, you transform the matrix into a simpler form, usually the row echelon form or reduced row echelon form. This method is particularly handy when finding the inverse of larger matrices, where hand calculations become cumbersome. Here's a quick rundown of what you might do:

• Swap rows, if necessary, to position a non-zero element at the pivot position.
• Multiply rows by constants to get leading ones.
• Subtract multiples of rows to eliminate other values in a column.

In systems with three or more variables, such as in equations (c) and (d), row reduction can help streamline finding the inverse and thus solve the equations efficiently. While the method can be intensive, it's invaluable for larger systems.