Problem 11 - Precalculus: Mathematical for Calculus

Question

Solve the matrix equation for the unknown matrix \(X\) or explain why no solution exists. $$\begin{aligned} &A=\left[\begin{array}{ll} 4 & 6 \\\1 & 3\end{array}\right] \quad B=\left[\begin{array}{ll}2 & 5 \\\3 & 7\end{array}\right]\\\ &C=\left[\begin{array}{ll} 2 & 3 \\\1 & 0 \\\0 & 2\end{array}\right] \quad D=\left[\begin{array}{lr}10 & 20 \\\30 & 20 \\\10 & 0\end{array}\right]\end{aligned}$$ $$2 X+A=B$$

Step-by-Step Solution

Answer

The solution for matrix \(X\) is \(\begin{pmatrix} -1 & -0.5 \\ 1 & 2 \end{pmatrix}\).

Step 1: Extract the Given Equation

We have the matrix equation \(2X + A = B\). Our task is to solve for the unknown matrix \(X\).

Step 2: Substitute Matrices into Equation

Substitute matrix \(A\) and \(B\) into the equation:\[2X + \begin{pmatrix} 4 & 6 \ 1 & 3 \end{pmatrix} = \begin{pmatrix} 2 & 5 \ 3 & 7 \end{pmatrix}\]

Step 3: Rearrange to Solve for 2X

Isolate \(2X\) on one side by subtracting matrix \(A\) from both sides:\[2X = \begin{pmatrix} 2 & 5 \ 3 & 7 \end{pmatrix} - \begin{pmatrix} 4 & 6 \ 1 & 3 \end{pmatrix}\]

Step 4: Perform Matrix Subtraction

Subtract matrix \(A\) from \(B\):\[2X = \begin{pmatrix} 2 - 4 & 5 - 6 \ 3 - 1 & 7 - 3 \end{pmatrix} = \begin{pmatrix} -2 & -1 \ 2 & 4 \end{pmatrix}\]

Step 5: Solve for Matrix X

To find \(X\), divide each element of the resulting matrix by 2:\[X = \frac{1}{2} \begin{pmatrix} -2 & -1 \ 2 & 4 \end{pmatrix} = \begin{pmatrix} -1 & -0.5 \ 1 & 2 \end{pmatrix}\]

Step 6: Confirm Solution

Verify the solution by substituting \(X\) back into the original equation:\[2 \begin{pmatrix} -1 & -0.5 \ 1 & 2 \end{pmatrix} + \begin{pmatrix} 4 & 6 \ 1 & 3 \end{pmatrix} = \begin{pmatrix} -2 & -1 \ 2 & 4 \end{pmatrix} + \begin{pmatrix} 4 & 6 \ 1 & 3 \end{pmatrix} = \begin{pmatrix} 2 & 5 \ 3 & 7 \end{pmatrix}\] which matches \(B\). Hence, our solution for \(X\) is correct.

Key Concepts

Matrix Subtraction

Matrix subtraction is similar to normal arithmetic subtraction, but you need to handle each element in the matrices one by one. To subtract two matrices, say matrix \( A \) and matrix \( B \), they must have the same dimensions. This means that both matrices must have the same number of rows and columns.

Here is how you subtract two matrices:

• Subtract the corresponding elements from each matrix.
• For example, if you have two matrices \( A = \begin{pmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{pmatrix} \) and \( B = \begin{pmatrix} b_{11} & b_{12} \ b_{21} & b_{22} \end{pmatrix} \), the result of \( A - B \) is \( \begin{pmatrix} a_{11} - b_{11} & a_{12} - b_{12} \ a_{21} - b_{21} & a_{22} - b_{22} \end{pmatrix} \).

With matrix subtraction, if the matrices have different dimensions, subtraction is not possible. In our example, subtracting matrix \( A \) from matrix \( B \) means calculating each element like so: first element of \( B \) minus the first element of \( A \), second element of \( B \) minus the second element of \( A \), and continue this for all elements.

Matrix Multiplication

Matrix multiplication can often be tricky to grasp at first, but it is an essential operation in linear algebra. It is not like multiplying numbers; you use the rows of the first matrix and the columns of the second matrix.

For example, if you have matrix \( A = \begin{pmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{pmatrix} \) and matrix \( B = \begin{pmatrix} b_{11} & b_{12} \ b_{21} & b_{22} \end{pmatrix} \), the product \( AB \) is calculated by:

• Taking the dot product of rows of \( A \) with columns of \( B \).
• The result will be a new matrix where each element is the sum of products of corresponding elements.

To perform the multiplication, remember:

• The number of columns in the first matrix must match the number of rows in the second matrix.
• The resulting matrix will have dimensions equal to the number of rows of the first matrix by the number of columns of the second matrix.

For instance, for a single element \( c_{11} \) in the product matrix, it's computed as \( c_{11} = a_{11}b_{11} + a_{12}b_{21} \). It requires multiplying and then adding up the elements of corresponding rows and columns.

Solving Matrix Equations

Solving matrix equations generally involves finding the unknown matrix that satisfies the equation. Much like solving for \( x \) in arithmetic equations, you aim to isolate the unknown matrix on one side of the equation.

Consider an equation like \( 2X + A = B \), where \( X \) is your unknown matrix.

• First step is usually to isolate the term with \( X \) on one side, done by performing operations such as subtraction.
• In our exercise, you subtract matrix \( A \) from \( B \) to isolate \( 2X \), which gives \( 2X = B - A \).

Next, you solve for \( X \) by making all elements of the matrix equation equal through applicable arithmetic operations.

• Divide each element by the coefficient multiplying \( X \), here it is 2, to solve for \( X \).
• This division results in the elements for matrix \( X \).

You can verify your solution by substituting \( X \) back into the original equation to ensure it satisfies \( B \). This re-checking is a critical final step to confirm the accuracy of your solution.