Problem 30 - Precalculus with Limits

Question

In Exercises 29-32, solve for \(X\) in the equation, given \(A=\left[\begin{array}{r} -2 & -1 \\ 1 & 0 \\ 3 & -4 \end{array}\right]\) and \(B=\left[\begin{array}{r} 0 & 3 \\ 2 & 0 \\ -4 & -1 \end{array}\right]\) \(2X = 2A - B\)

Step-by-Step Solution

Answer

The matrix \( X \) that solves the equation is \( X = \left[\begin{array}{r} -2 & -2.5 \ 0 & 0 \ 5 & -3.5 \end{array}\right] \)

Step 1: Insert the given matrices into the equation

First, you replace \( A \) and \( B \) in the equation \( 2X = 2A - B \) with their respective matrices. So the equation becomes: \( 2X = 2 \left[\begin{array}{r} -2 & -1 \ 1 & 0 \ 3 & -4 \end{array}\right] - \left[\begin{array}{r} 0 & 3 \ 2 & 0 \ -4 & -1 \end{array}\right] \)

Step 2: Perform the matrix multiplication and subtraction

Now perform the multiplication and subtraction of the matrices. Remember that multiplication of a matrix by a scalar is done entry-wise. Subtraction of matrices is also done entry-wise. After performing the operations, your new equation becomes: \( 2X = \left[\begin{array}{r} -4 - 0 & -2 - 3 \ 2 - 2 & 0 - 0 \ 6 -- 4 & -8 -- 1\end{array}\right] = \left[\begin{array}{r} -4 & -5 \ 0 & 0 \ 10 & -7 \end{array}\right] \)

Step 3: Solve the equation for \( X \)

Finally, solve the equation \( 2X = \left[\begin{array}{r} -4 & -5 \ 0 & 0 \ 10 & -7 \end{array}\right] \) for \( X \) by dividing both sides of the equation by 2. Remember that division of a matrix by a scalar is done entry-wise. So, \( X = \left[\begin{array}{r} -4/2 & -5/2 \ 0/2 & 0/2 \ 10/2 & -7/2 \end{array}\right] = \left[\begin{array}{r} -2 & -2.5 \ 0 & 0 \ 5 & -3.5 \end{array}\right] \)

Key Concepts

Matrix Multiplication

Matrix multiplication involves combining two matrices by multiplying corresponding elements and summing the results. However, in this exercise, we focus on multiplying a matrix by a scalar. This operation is simpler as it only requires multiplying each element of the matrix by the scalar. It does not involve another matrix. For matrices, instead of multiplying elements like regular numbers, we concentrate on each cell:

• Multiply each element of the matrix by the scalar value.
• Ensure the operation is applied consistently to each matrix entry.

For example, if you have a scalar of 2 and matrix \( A \), each element in \( A \) is doubled. This operation prepares the matrices properly for further operations, like subtraction.

Matrix Subtraction

Matrix subtraction is straightforward when matrices are of the same dimensions. It involves subtracting corresponding elements from each matrix. For example, if you have two matrices \( A \) and \( B \), both with dimensions that match, you subtract each element \( a_{ij} \) of \( A \) with \( b_{ij} \) of \( B \):

• Ensure both matrices have the same number of rows and columns.
• Subtract each corresponding element.
• Result in a new matrix of the same dimensions.

In our exercise, after multiplying matrix \( A \) by 2, we then subtract matrix \( B \) to find the new matrix before solving for \( X \). Make sure each subtraction is performed carefully to avoid errors.

Scalar Multiplication

Scalar multiplication is an operation where each entry of the matrix is multiplied by a number, known as the scalar. In this exercise, before subtracting the matrices, we need to perform scalar multiplication on one of them. Here's a breakdown of how it's done:

• Identify the scalar, which is the constant number by which you multiply the matrix.
• Multiply each element of the matrix by this scalar.
• Ensure precision in calculation for each element.

For our problem, the scalar is 2. By multiplying each element of matrix \( A \) by 2, we prepare it for the next operation of subtraction against matrix \( B \). This step is crucial to ensure your matrix represents the correct values for further operations.

System of Equations

When dealing with systems of equations in matrices, you aim to find solutions for variables represented in matrix form. Here, the variable is the matrix \( X \), which we solve by isolating it on one side of the equation. To solve a matrix equation like \( 2X = C \), follow these steps:

• Isolate \( X \) by dividing each element of matrix \( C \) by the scalar from the equation, which is 2 in this case.
• Apply this division uniformly to all elements in matrix \( C \).
• The result is matrix \( X \), fully simplified.

By dividing each element in the resultant matrix by the scalar from \( 2X = C \), we determine \( X \)'s exact elements. This operation showcases how matrices can succinctly express systems of equations and how scalar operations facilitate their solutions.