Problem 20 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

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

Step 1: Determine Matrix Compatibility

To solve for the matrix \(X\), first ensure that the matrices in the given equation are compatible. Notice that \(A\) is a 2x2 matrix, \(B\) is a 2x2 matrix, and we need to determine the size of \(X\). Since \(B - 3X\) on the right must equal \(2A\), \(X\) must also be a 2x2 matrix.

Step 2: Rewrite the Equation

Express the given matrix equation in a form that's easier to manipulate:\[ 2A = B - 3X \]Reordering terms, we want to isolate \(X\), so rearrange it as:\[ 3X = B - 2A \]

Step 3: Calculate \(2A\)

Calculate \(2A\), the result of multiplying every element in matrix \(A\) by 2:\[ 2A = 2 \times \begin{bmatrix} 4 & 6 \ 1 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 12 \ 2 & 6 \end{bmatrix} \]

Step 4: Calculate \(B - 2A\)

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

Step 5: Solve for \(X\)

From the equation \(3X = B - 2A\), divide each component of the resulting matrix \(B - 2A\) by 3 to solve for \(X\):\[ X = \frac{1}{3} \begin{bmatrix} -6 & -7 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} -2 & -\frac{7}{3} \ \frac{1}{3} & \frac{1}{3} \end{bmatrix} \]

Step 6: Conclusion: Verify Solution

Verify that multiplying matrix \(X\) by 3 and adding it to \(2A\) results in matrix \(B\). The calculations align, confirming that the solution is correct.

Key Concepts

Matrix Compatibility

To solve a matrix equation, ensuring matrix compatibility is crucial. It's like checking to see if puzzle pieces fit together. In our example, we have matrices \( A \) and \( B \) that are both 2x2. For the equation \( 2A = B - 3X \) to hold, matrix \( X \) must also be 2x2. This requirement allows matrix operations like addition, subtraction, and multiplication to proceed seamlessly. If these dimensions aren't compatible, the equation can't be solved; instead, it's akin to trying to stack mismatched blocks.

Matrix Multiplication

Matrix multiplication is a key operation, where you multiply rows by columns. In our exercise, we calculate \( 2A \) by scaling all elements of matrix \( A \) with 2. This operation is straightforward because we multiply each element in \( A \) by 2 to yield a new matrix. Here's a quick rundown:

• Multiply each element of the first row, \([4, 6]\), by 2 to get \([8, 12]\).
• Multiply each element of the second row, \([1, 3]\), by 2 to get \([2, 6]\).

This step doesn't fully showcase the complexity of matrix multiplication of differing matrices, but it gives insight into scaling operations which is a small yet essential aspect.

Matrix Subtraction

Matrix subtraction follows a simple rule: matching each element by their positions in the matrices and subtracting. This operation requires equal dimensions in the participating matrices.
In our solution, we subtract \( 2A \) from \( B \):

• Subtract corresponding elements of the first row \([2, 5]\) of matrix \( B \) from \([8, 12]\) to obtain \([-6, -7]\).
• Do the same for the second row \([3, 7]\) from \([2, 6]\) to get \([1, 1]\).

This step helps isolate \( X \) in the equation, efficiently allowing a pathway towards eventually solving for \( X \).

Solution Verification

Verifying our solution is akin to double-checking your work in any math problem. Once we solve for \( X \) as \( \begin{bmatrix} -2 & -\frac{7}{3} \ \frac{1}{3} & \frac{1}{3} \end{bmatrix} \), it becomes essential to ensure this matrix satisfies the original equation.
Multiplying this \( X \) by 3, and adding it to \( 2A \), should yield matrix \( B \). If:

• \( 3X \) calculation from \( \begin{bmatrix} -2 & -\frac{7}{3} \ \frac{1}{3} & \frac{1}{3} \end{bmatrix} \) results in \( \begin{bmatrix} -6 & -7 \ 1 & 1 \end{bmatrix} \)
• Adding this adjusted \( X \) with \( 2A \) equals \( B \)

If these check out, then the solution is indeed validated.