Problem 19 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The solution is \( X = \begin{bmatrix} 0 & -5 \\ -25 & -20 \\ -10 & 10 \end{bmatrix} \).

Step 1: Understand the Equation

The equation given is \( \frac{1}{5}(X+D)=C \). We need to solve for the unknown matrix \( X \).

Step 2: Remove the Fraction

To eliminate the fraction, multiply both sides of the equation by 5 to obtain: \( X + D = 5C \).

Step 3: Compute the Matrix 5C

Compute the product of the scalar 5 with the matrix \( C \). This gives: \[ 5C = 5 \times \begin{bmatrix} 2 & 3 \ 1 & 0 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} 10 & 15 \ 5 & 0 \ 0 & 10 \end{bmatrix} \]

Step 4: Solve for X

We know \( X + D = 5C \). Thus, solving for \( X \) gives us \( X = 5C - D \).

Step 5: Compute X

Subtract matrix \( D \) from matrix \( 5C \):\[ X = \begin{bmatrix} 10 & 15 \ 5 & 0 \ 0 & 10 \end{bmatrix} - \begin{bmatrix} 10 & 20 \ 30 & 20 \ 10 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -5 \ -25 & -20 \ -10 & 10 \end{bmatrix} \]

Step 6: Verify Matrix Compatibility

Verify that \( X = \begin{bmatrix} 0 & -5 \ -25 & -20 \ -10 & 10 \end{bmatrix} \) is a 3x2 matrix, which is compatible with matrices \( D \) and \( 5C \), as both are also 3x2 matrices.

Key Concepts

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra, crucial for solving matrix equations. It involves multiplying two matrices to produce a third matrix. However, not every pair of matrices can be multiplied. The number of columns in the first matrix must equal the number of rows in the second matrix for multiplication to be possible.

When multiplying matrices, each element in the resulting matrix is the sum of products of the corresponding elements. For example, if we have two matrices, say matrix A with dimensions (m x n) and matrix B with dimensions (n x p), their resulting matrix C will have dimensions (m x p). Each element in matrix C, say located at row i and column j, is calculated as:

• \( C_{ij} = A_{i1} \times B_{1j} + A_{i2} \times B_{2j} + \ldots + A_{in} \times B_{nj} \)

Matrix multiplication is not commutative, which means that \( AB \) does not generally equal \( BA \). It's essential to adhere to the order of multiplication as prescribed in problems.

In our context, while solving matrix equation \( X = 5C - D \), we did not directly multiply matrices. Yet, understanding matrix multiplication assures that our operations fit within the rules, especially when involving combinations or transformations through matrix multiplications.

Scalar Multiplication

Scalar multiplication is one of the simpler operations involving matrices, where each entry of a matrix is multiplied by a constant, known as a scalar. This process changes each element of the matrix but retains its size and shape.

To perform scalar multiplication on a matrix \( C \) and a scalar \( 5 \), we multiply each element of the matrix by 5. If matrix \( C \) is:

• \( C = \begin{bmatrix} 2 & 3 \ 1 & 0 \ 0 & 2 \end{bmatrix} \)

Then multiplying by scalars results in:

• \( 5 \times C = \begin{bmatrix} 10 & 15 \ 5 & 0 \ 0 & 10 \end{bmatrix} \)

Each element within matrix \( C \) is simply scaled by the factor 5. Such scalar multiplications are helpful when equations require simplifications or scaling of matrix components.

Matrix Subtraction

Matrix subtraction involves element-wise subtraction between two matrices of the same dimensions. It allows us to find the difference between corresponding elements of each matrix.

To subtract matrix \( D \) from matrix \( 5C \), having the same dimensions is a prerequisite. Both should be 3x2 matrices for this operation to be valid. The subtraction formula is simple:

• \( (5C - D)_{ij} = (5C)_{ij} - D_{ij} \)

Given matrices:

• \( 5C = \begin{bmatrix} 10 & 15 \ 5 & 0 \ 0 & 10 \end{bmatrix} \)
• \( D = \begin{bmatrix} 10 & 20 \ 30 & 20 \ 10 & 0 \end{bmatrix} \)

Subtracting \( D \) from \( 5C \) results in:

• \( X = \begin{bmatrix} 0 & -5 \ -25 & -20 \ -10 & 10 \end{bmatrix} \)

Each element of matrix \( 5C \) is subtracted by the corresponding element in matrix \( D \), yielding the new matrix \( X \). This operation helps you to solve the matrix equation by isolating the unknown matrix \( X \).