Question

Matrices \(A\) and \(B\) are given below. Find \(X\) that satisfies the equation. $$ A=\left[\begin{array}{cc} 3 & -1 \\ 2 & 5 \end{array}\right] \quad B=\left[\begin{array}{cc} 1 & 7 \\ 3 & -4 \end{array}\right] $$ $$ 2 A+X=B $$

Short answer

Matrix \(X = \left[\begin{array}{cc} -5 & 9 \\ -1 & -14 \end{array}\right]\).

Step-by-step solution

Understand the Equation

The problem requires us to find matrix \(X\) such that when we double matrix \(A\) and add \(X\), the result is matrix \(B\). The equation is \(2A + X = B\).

Solve for X

To isolate matrix \(X\), we rearrange the equation: \(X = B - 2A\). This allows us to calculate \(X\) directly by first finding \(2A\) and then subtracting it from \(B\).

Calculate 2A

Multiply matrix \(A\) by 2. \[2 \times \left[\begin{array}{cc} 3 & -1 \ 2 & 5 \end{array}\right] = \left[\begin{array}{cc} 6 & -2 \ 4 & 10 \end{array}\right]\]

Subtract 2A from B

Subtract the resulting matrix \(2A\) from matrix \(B\). \[X = \left[\begin{array}{cc} 1 & 7 \ 3 & -4 \end{array}\right] - \left[\begin{array}{cc} 6 & -2 \ 4 & 10 \end{array}\right] = \left[\begin{array}{cc} 1-6 & 7-(-2) \ 3-4 & -4-10 \end{array}\right] = \left[\begin{array}{cc} -5 & 9 \ -1 & -14 \end{array}\right]\]

Verify the Solution

Re-check the calculations by adding \(2A\) and \(X\) to see if the resulting matrix is \(B\). Calculating \(2A + X\): \[\left[\begin{array}{cc} 6 & -2 \ 4 & 10 \end{array}\right] + \left[\begin{array}{cc} -5 & 9 \ -1 & -14 \end{array}\right] = \left[\begin{array}{cc} 6 - 5 & -2 + 9 \ 4 - 1 & 10 - 14 \end{array}\right] = \left[\begin{array}{cc} 1 & 7 \ 3 & -4 \end{array}\right]\]This confirms our answer is correct as the result matches \(B\).

Key concepts

Matrix Subtraction

Matrix subtraction involves element-wise subtraction of corresponding elements from the matrices. It is only possible to subtract matrices of the same dimensions. This means both the number of rows and columns in the matrices must be the same for subtraction to work. For instance, in our exercise, matrices \( B \) and \( 2A \) are both 2x2 matrices, allowing us to subtract them.

• Arrange the matrices horizontally to align corresponding elements.
• Subtract each element of the second matrix from the corresponding element of the first matrix.
• The resulting matrix, \( X \), is another matrix of the same dimension.

For example, the top-left element of \( X \) is computed as follows: \[ X_{11} = B_{11} - 2A_{11} = 1 - 6 = -5 \]This process is repeated for every element of the matrices involved.

Matrix Multiplication

Matrix multiplication is generally not as straightforward as multiplication of numbers. The number of columns in the first matrix must match the number of rows in the second matrix for multiplication to be possible. However, when multiplying a matrix by a scalar (a constant value), as seen in our exercise, it's much simpler.

• Every element of the matrix is multiplied by this scalar.
• The result is a matrix of the same dimensions as the original.

For example, when multiplying matrix \( A \) by 2, each element of \( A \) is multiplied by 2:\[ 2 \times A = \begin{bmatrix} 2 \times 3 & 2 \times -1 \ 2 \times 2 & 2 \times 5 \end{bmatrix} = \begin{bmatrix} 6 & -2 \ 4 & 10 \end{bmatrix} \]This process allows us to easily scale matrices, which is a crucial step in solving equations involving matrices.

System of Linear Equations

Matrices are powerful tools for solving systems of linear equations. In scenarios like our exercise, matrices can represent the coefficients and constants of systems of equations. Solving these involves performing operations such as matrix addition, subtraction, and multiplication to simplify or manipulate the equations into a desired form.

• Equations like \( 2A + X = B \) illustrate how matrices can simplify complex systems.
• These matrix equations can be solved by rearranging and performing arithmetic operations.

In our exercise, matrix \( X \) is isolated by first expressing it in terms of other known matrices, i.e., rearranging to \( X = B - 2A \). Subsequently, matrix subtraction is used to compute the solution. This technique helps efficiently solve systems that might otherwise involve lengthy algebraic manipulations.