Problem 20 - Precalculus: Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

The solution for matrix \(X\) is \(\begin{bmatrix} 4 & 7 \\ 7 & 4 \\ 2 & 2 \end{bmatrix}\).

Step 1: Simplify the Equation

Start by simplifying the given matrix equation. Divide both sides of the equation by 5 to isolate the matrix expression containing the unknown. This gives us:\[X - C = \frac{1}{5}D\]

Step 2: Compute \(\frac{1}{5}D\)

Calculate \(\frac{1}{5}D\) by multiplying each element of matrix \(D\) by \(\frac{1}{5}\):\[\frac{1}{5}D = \begin{bmatrix} 10/5 & 20/5 \ 30/5 & 20/5 \ 10/5 & 0/5 \end{bmatrix} = \begin{bmatrix} 2 & 4 \ 6 & 4 \ 2 & 0 \end{bmatrix}\]

Step 3: Set Up the Equation for \(X\)

Now we have:\[X - \begin{bmatrix} 2 & 3 \ 1 & 0 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \ 6 & 4 \ 2 & 0 \end{bmatrix}\]To solve for \(X\), add matrix \(C\) to both sides of the equation.

Step 4: Add Matrices \(C\) and \(\frac{1}{5}D\)

Add matrix \(C\) to the matrix \(\frac{1}{5}D\):\[X = \begin{bmatrix} 2 & 4 \ 6 & 4 \ 2 & 0 \end{bmatrix} + \begin{bmatrix} 2 & 3 \ 1 & 0 \ 0 & 2 \end{bmatrix}\]Now, perform element-wise addition:\[X = \begin{bmatrix} 2+2 & 4+3 \ 6+1 & 4+0 \ 2+0 & 0+2 \end{bmatrix} = \begin{bmatrix} 4 & 7 \ 7 & 4 \ 2 & 2 \end{bmatrix}\]

Step 5: Check the Solution

Confirm that our solution for \(X\) satisfies the original matrix equation by substituting back. Since all operations align correctly with addition and multiplication of matrices, our solution is correct for the provided equation.

Key Concepts

Matrix Addition

Matrix addition is a fundamental operation where two matrices of the same dimensions are combined to form a new matrix. Each element of the resulting matrix is the sum of the corresponding elements of the original matrices. For instance, if you have matrices:

• Matrix A: \( \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix} \)
• Matrix B: \( \begin{bmatrix} b_{11} & b_{12} \ b_{21} & b_{22} \end{bmatrix} \)

The addition of A and B is found by adding each element individually, so \(A + B = \begin{bmatrix} a_{11} + b_{11} & a_{12} + b_{12} \ a_{21} + b_{21} & a_{22} + b_{22} \end{bmatrix}\). This element-wise addition is straightforward as long as the matrices are of the same size. If they aren't, the addition is not defined and cannot be performed.

Matrix Multiplication

Matrix multiplication involves multiplying rows by columns, and it is a bit more involved than addition. When multiplying matrices, the element in the i-th row and j-th column of the resulting matrix is computed by taking the dot product of the i-th row of the first matrix and the j-th column of the second matrix. For multiplication to be possible, the number of columns in the first matrix must equal the number of rows in the second matrix. Here’s a simple case:

• If matrix A is \(2 \times 3\) (2 rows, 3 columns)
• And matrix B is \(3 \times 2\)

Then matrix C = AB will be a \(2 \times 2\) matrix. It’s important to follow the rules of matrix multiplication, as mismatched dimensions often lead to errors or undefined operations in matrix equations.

Matrix Simplification

Simplifying a matrix equation is akin to simplifying algebraic expressions. You often need to perform operations like scaling a matrix or performing element-wise operations to isolate the unknown matrix. In an equation like \(5(X - C) = D\), you first simplify by eliminating the scalar via division. This involves multiplying each element of the matrix by the reciprocal of the scalar (1/5 in this case). For example, if \(D = \begin{bmatrix} 10 & 20 \ 30 & 20 \ 10 & 0 \end{bmatrix}\), then \(\frac{1}{5}D = \begin{bmatrix} 2 & 4 \ 6 & 4 \ 2 & 0 \end{bmatrix}\). Simplifying matrix equations helps in preparing them for further operations like matrix addition to solve for the unknown.

Matrix Solution

Finding the solution to a matrix equation often involves isolating the unknown matrix, similar to solving for a variable in algebra. After simplifying the matrix equation, as seen in previous sections, the next step is to use matrix addition or subtraction to find the unknown matrix. Consider the matrix equation \(X - \begin{bmatrix} 2 & 3 \ 1 & 0 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \ 6 & 4 \ 2 & 0 \end{bmatrix}\), where you solve for \(X\) by adding the matrix \(\begin{bmatrix} 2 & 3 \ 1 & 0 \ 0 & 2 \end{bmatrix}\) to both sides. The resulting operation gives you \(X = \begin{bmatrix} 4 & 7 \ 7 & 4 \ 2 & 2 \end{bmatrix}\). Each step involves careful element-wise addition, ensuring each part of the equation remains balanced. This guarantees the solution is valid within the context of the given matrix equation.