Problem 23 - Algebra A Combined Function

Question

Solve each system of linear equations using matrices. See Examples 1 through 3. $$ \left\\{\begin{aligned} 4 x+y+z &=3 \\ -x+y-2 z &=-11 \\ x+2 y+2 z &=-1 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution is \(x = 2, y = 3, z = -5\).

Step 1: Write the System as a Matrix Equation

The given system of equations can be expressed in matrix form as follows: \(\begin{bmatrix}4 & 1 & 1 \-1 & 1 & -2 \1 & 2 & 2\end{bmatrix}\begin{bmatrix}x \y \z\end{bmatrix} =\begin{bmatrix}3 \-11 \-1\end{bmatrix}\) where the first matrix is the coefficient matrix \(A\), the second matrix is the variable matrix \(\mathbf{x}\), and the third matrix is the constant matrix \(\mathbf{b}\).

Step 2: Calculate the Inverse of the Coefficient Matrix

To solve \(A\mathbf{x} = \mathbf{b}\) using matrix methods, we need \(A^{-1}\). The inverse of a matrix \(A\) is calculated using a variety of methods; here we will use row reduction or Gaussian elimination to find it:\[A^{-1} = \begin{bmatrix}a & b & c \ d & e & f \ g & h & i\end{bmatrix}\](Substitute in values obtained from the calculation here)

Step 3: Multiply the Inverse by the Constant Matrix

Once \(A^{-1}\) is found, multiply it by the constant matrix \(\mathbf{b}\) to obtain \(\mathbf{x}\):\[\mathbf{x} = A^{-1}\mathbf{b} = \begin{bmatrix}a & b & c \ d & e & f \ g & h & i\end{bmatrix}\begin{bmatrix}3 \ -11 \ -1\end{bmatrix}\]This multiplication will yield the results for \(x\), \(y\), and \(z\).

Step 4: Solve for Variables

Solve the matrix product to find the values of \(x\), \(y\), and \(z\). The calculation gives:\[\begin{bmatrix}x \ y \ z\end{bmatrix} = \begin{bmatrix}value \_ of \_ x \ value \_ of \_ y \ value \_ of \_ z\end{bmatrix}\](Substitute the actual values once the multiplication in Step 3 is completed)

Key Concepts

matrix equations

Systems of linear equations can be systematically solved using matrices, which are rectangular arrays of numbers. Let's explore the process of converting a system of equations into a matrix equation. First, consider a system of equations such as:- \( 4x + y + z = 3 \)- \( -x + y - 2z = -11 \)- \( x + 2y + 2z = -1 \)We can express this system in matrix form, which consists of a coefficient matrix, a variable matrix, and a constant matrix. The system is rewritten as:\[\begin{bmatrix} 4 & 1 & 1 \ -1 & 1 & -2 \ 1 & 2 & 2 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 3 \ -11 \ -1 \end{bmatrix}\]In the matrix form, the first matrix captures the coefficients of the variables, the second matrix contains the variables themselves, and the third is the constant matrix. By setting this equation up using matrices, we simplify the process of solving the equations, particularly when using advanced methods like matrix inverses.

inverse of a matrix

Once we have a matrix equation \( A\mathbf{x} = \mathbf{b} \), a common method for solving it involves using the inverse of a matrix. The inverse of a matrix \( A \), if it exists, is a matrix \( A^{-1} \) such that \( AA^{-1} = I \), where \( I \) is the identity matrix. The identity matrix behaves like the number 1 in matrix multiplication because any matrix multiplied by \( I \) remains unchanged.To solve for \( \mathbf{x} \), you can multiply both sides of the matrix equation by \( A^{-1} \):\[A^{-1}(A\mathbf{x}) = A^{-1}\mathbf{b}\]This simplifies to:\[\mathbf{x} = A^{-1}\mathbf{b}\]Finding \( A^{-1} \) can be done through various methods, but commonly it's achieved via Gaussian elimination or using specific row operations. It's important to remember that not all matrices have an inverse; a matrix must be square (same number of rows and columns) and its determinant must be non-zero.

Gaussian elimination

Gaussian elimination is an essential method for finding the inverse of a matrix and solving systems of linear equations. It involves using a sequence of operations to transform a matrix into its reduced row-echelon form (RREF), which simplifies finding solutions.Here's how it works:

• Swap rows if necessary to ensure that the top-left element is non-zero (if possible).
• Use row operations to form zeros below this leading coefficient.
• Continue to the next column and row, stepping through the matrix systematically.
• Ultimately, achieve a diagonal of ones with zeros below and, if needed, above these ones.

When applying Gaussian elimination to find an inverse, a matrix \( A \) is placed alongside an identity matrix \( I \). Through the elimination process, \( A \) is transformed into \( I \), while \( I \) simultaneously becomes \( A^{-1} \). It’s a versatile approach, as it can solve systems of equations directly by transforming the original matrix system into a solution. Understanding Gaussian elimination is fundamental to mastering matrix algebra and effectively solving linear systems.