Problem 21 - Algebra and Trigonometry

Question

Let $$ A=\left[\begin{array}{rr} {-3} & {-7} \\ {2} & {-9} \\ {5} & {0} \end{array}\right] \text { and } B=\left[\begin{array}{rr} {-5} & {-1} \\ {0} & {0} \\ {3} & {-4} \end{array}\right] $$ Solve each matrix equation for X. $$ 3 X+2 A=B $$

Step-by-Step Solution

Answer

The solution to the given matrix equation \(3X + 2A = B\) is \(X = \[\begin{array}{rr}{1/3} & {13/3} \\ {-4/3} & {6} \\ {-7/3} & {-4/3}\end{array}\]\).

Step 1: Simplify Equation

The first step is to simplify the matrix equation. Here, use the given equation \(3X + 2A = B\). First, you need to calculate the value of \(2A\). Multiply every element in matrix A by 2.

Step 2: Calculate 2A

After multiplying every element we get: \(2A = \[\begin{array}{rr}{-6} & {-14} \\ {4} & {-18} \\ {10} & {0}\end{array}\]\).

Step 3: Modifying the Equation

Substitute the value of \(2A\) into the given equation \(3X + 2A = B\). This gives \(3X = B - 2A\).

Step 4: Calculate B - 2A

Subtract matrix 2A from B, element by element, and get: \(B - 2A = \[\begin{array}{rr}{-5-(-6)} & {-1-(-14)} \\ {0-(4)} & {0-(-18)} \\ {3-(10)} & {-4-(0)}\end{array}\] = \[\begin{array}{rr}{1} & {13} \\ {-4} & {18} \\ {-7} & {-4}\end{array}\] \).

Step 5: Solving for X

The last step is to solve for X. According to the equation from Step 3: \(3X = B - 2A\). Divide each element in the \(B - 2A\) matrix by 3 to find the value of X.

Step 6: Calculate X

The calculation gives: \(X = \[\begin{array}{rr}{1/3} & {13/3} \\ {-4/3} & {6} \\ {-7/3} & {-4/3}\end{array}\] \).

Key Concepts

Matrix Algebra

Matrix algebra is a significant area of mathematics that deals with the study of matrices and the various operations that can be applied to them. It serves as a foundation for many areas of mathematics and applications in physics, engineering, and computer science. One of the fundamental operations in matrix algebra is solving matrix equations, which involves finding a matrix that satisfies an equation of the form, for example, 'AX + B = C'.

The process requires familiarity with matrix addition, subtraction, scalar multiplication, and other more complex operations like matrix multiplication and matrix inversion. Since matrices represent linear transformations or systems of linear equations, mastering the concept of matrix algebra allows us to solve a wide range of practical problems such as those found in electric circuits, structural engineering, and economic models.

Matrix Addition and Subtraction

Matrix addition and subtraction are basic arithmetic operations that are performed element-wise on matrices of the same dimensions. When adding or subtracting matrices, such as matrices 'A' and 'B' in our exercise, you simply combine corresponding elements from each matrix to produce a new matrix.

For instance, if you have two 3x2 matrices to add or subtract, you'll perform the operation on each corresponding pair of elements, resulting in another 3x2 matrix. It's important to note that matrix addition is commutative (\(A + B = B + A\) ) and associative (\( (A + B) + C = A + (B + C) \) ). In contrast, these properties do not necessarily apply to matrix subtraction. These operations often serve as intermediate steps in solving more complex matrix equations.

Scalar Multiplication of Matrices

Scalar multiplication of matrices is another elementary operation that entails multiplying every element within a matrix by a single number, known as a scalar. This scalar could be any real or complex number.

For example, if we multiply matrix 'A' by the scalar number 2, each element in matrix 'A' is doubled. This operation is useful in modifying equations or scaling transformations represented by matrices. Scalar multiplication is distributive, which means multiplying a scalar with the sum of two matrices is the same as adding the scalar multiples of each matrix separately (\(a(A + B) = aA + aB\) ).

In the context of solving matrix equations, scalar multiplication allows us to isolate terms and move towards finding the solution of the matrix 'X', as seen in our step-by-step solution to the given exercise.