Problem 16 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The solution is \( X = \begin{bmatrix} -2 & -\frac{7}{3} \\ \frac{1}{3} & \frac{1}{3} \end{bmatrix} \).

Step 1: Set Up the Equation

We start with the matrix equation \( 2A = B - 3X \). Our goal is to solve this equation for the unknown matrix \( X \).

Step 2: Find 2A

Calculate \( 2A \):\[ 2A = 2 \begin{bmatrix} 4 & 6 \ 1 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 12 \ 2 & 6 \end{bmatrix} \]

Step 3: Solve for 3X

Substitute \( 2A \) and \( B \) into the original equation to find \( 3X \):\[ \begin{bmatrix} 8 & 12 \ 2 & 6 \end{bmatrix} = \begin{bmatrix} 2 & 5 \ 3 & 7 \end{bmatrix} - 3X \]. Rearrange to solve for \( 3X \):\[ 3X = \begin{bmatrix} 2 & 5 \ 3 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 12 \ 2 & 6 \end{bmatrix} \]

Step 4: Calculate B - 2A

Perform the matrix subtraction \( B - 2A \):\[ B - 2A = \begin{bmatrix} 2 & 5 \ 3 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 12 \ 2 & 6 \end{bmatrix} = \begin{bmatrix} 2 - 8 & 5 - 12 \ 3 - 2 & 7 - 6 \end{bmatrix} = \begin{bmatrix} -6 & -7 \ 1 & 1 \end{bmatrix} \]

Step 5: Solve for X

Now solve for \( X \) by dividing the resulting matrix by 3:\[ X = \frac{1}{3} \begin{bmatrix} -6 & -7 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} -2 & -\frac{7}{3} \ \frac{1}{3} & \frac{1}{3} \end{bmatrix} \]

Step 6: Conclusion

The solution matrix \( X \) that satisfies the equation is \( \begin{bmatrix} -2 & -\frac{7}{3} \ \frac{1}{3} & \frac{1}{3} \end{bmatrix} \).

Key Concepts

Matrix Subtraction

Matrix subtraction is a fundamental operation where you subtract the elements of one matrix from the corresponding elements of another matrix. For matrices to be eligible for subtraction, they must be of the same dimensions, meaning they have the same number of rows and columns. This is because each element from one matrix must have a corresponding element in the other matrix for the subtraction to occur.
For example, consider two matrices:

• Matrix A: \( \begin{bmatrix} 4 & 6 \ 1 & 3 \end{bmatrix} \)
• Matrix B: \( \begin{bmatrix} 2 & 5 \ 3 & 7 \end{bmatrix} \)

To perform matrix subtraction \( A - B \), you subtract each element of matrix B from the corresponding element of matrix A:
\[A - B = \begin{bmatrix} 4 - 2 & 6 - 5 \ 1 - 3 & 3 - 7 \end{bmatrix} = \begin{bmatrix} 2 & 1 \ -2 & -4 \end{bmatrix}\] This simple operation is often used in solving equations with matrices, just as it was used in solving for the unknown matrix \( X \).

Scalar Multiplication

Scalar multiplication is another basic matrix operation where every entry in a matrix is multiplied by a constant, known as a scalar. The scalar is typically a real number and it scales the matrix. Scaling a matrix changes the size of the vector it represents but keeps the ratio between elements the same.
Let's illustrate this with a matrix A:

• Matrix A: \( \begin{bmatrix} 4 & 6 \ 1 & 3 \end{bmatrix} \)

If we multiply matrix A by a scalar 2, every element in matrix A is multiplied by 2:
\[2A = 2 \times \begin{bmatrix} 4 & 6 \ 1 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 12 \ 2 & 6 \end{bmatrix}\] This operation was used in the given exercise to calculate \(2A\), as part of the process to solve the matrix equation.
This technique is valuable when we need to scale matrices to match the operations needed for solving matrix equations or aligning them in systems aligned with vectors.

Matrix Division

Unlike numbers, division is not directly defined for matrices. When we talk about dividing a matrix by a scalar — like in the task solution — we actually mean multiplying the matrix by the reciprocal of that scalar. Therefore, matrix division by a scalar is a special case of scalar multiplication.
For example, when solving the equation for \( X \), we end with a matrix that needs to be divided by 3:\[3X = \begin{bmatrix} -6 & -7 \ 1 & 1 \end{bmatrix}\]This can be rewritten as multiplying by the reciprocal of 3, i.e., \( \frac{1}{3} \):
\[X = \frac{1}{3} \times \begin{bmatrix} -6 & -7 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} -2 & -\frac{7}{3} \ \frac{1}{3} & \frac{1}{3} \end{bmatrix}\]This is how we 'divide' a matrix by a scalar.In practice, this operation is crucial, particularly in matrix equations where solving for an unknown matrix may require 'dividing' the entire matrix by a scalar constant encountered in the problem.