Problem 22 - 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. $$ 2 X+5 A=B $$

Step-by-Step Solution

Answer

The X matrix is: \(\left[\begin{array}{rr} {5} & {17} \ {-5} & {22.5} \ {-11} & {-2} \end{array}\right]\)

Step 1: Multiplication of Matrix A by Scalar

First, multiply matrix A by the scalar 5. This is done by multiplying each element of the matrix by 5. \nSo, \(5A = 5\left[\begin{array}{rr} {-3} & {-7} \ {2} & {-9} \ {5} & {0} \end{array}\right] = \left[\begin{array}{rr} {-15} & {-35} \ {10} & {-45} \ {25} & {0} \end{array}\right]\)

Step 2: Calculation of 5A + B

Now, add the resultant matrix from Step 1 to matrix B. Add corresponding elements of both matrices to get the result. \nSo, \(5A + 2X = B \Rightarrow 2X = B - 5A\)\nCalculate \(B - 5A = \left[\begin{array}{rr} {-5} & {-1} \ {0} & {0} \ {3} & {-4} \end{array}\right] - \left[\begin{array}{rr} {-15} & {-35} \ {10} & {-45} \ {25} & {0} \end{array}\right] = \left[\begin{array}{rr} {10} & {34} \ {-10} & {45} \ {-22} & {-4} \end{array}\right]\)

Step 3: Solve for X

Since \(2X = B - 5A\), we can find X by dividing each element of the matrix by 2. \nSo, \(X = \frac{1}{2} \times (B - 5A)\)\nCalculate \(X = \frac{1}{2} \left[\begin{array}{rr} {10} & {34} \ {-10} & {45} \ {-22} & {-4} \end{array}\right] = \left[\begin{array}{rr} {5} & {17} \ {-5} & {22.5} \ {-11} & {-2} \end{array}\right]\)

Key Concepts

matrix multiplication

Matrix multiplication can be tricky at first, but it’s a fundamental operation in linear algebra. It involves multiplying rows by columns. Let's start by considering two matrices, say matrix A of size \( m \times n \) and matrix B of size \( n \times p \). The resulting product, matrix C, will have a dimension of \( m \times p \). Note that column numbers in A must match row numbers in B.
Each element in the product matrix is calculated by taking the dot product of the corresponding row from matrix A and column from matrix B.

• Consistency in dimensions is crucial for matrix multiplication. If this condition is not met, multiplication cannot proceed.
• The product matrix's dimensions are determined by the outer dimensions of the matrices being multiplied.

Understanding the indexing helps in aligning the computation process and magnifies the importance of carefully arranging matrix elements during multiplication.

scalar multiplication

Scalar multiplication involves multiplying each element of a matrix by a single number (scalar). It is a straightforward yet essential operation that scales the matrix. In our example, matrix A was scaled by 5 in the equation \(2X + 5A = B\). Here's a breakdown of scalar multiplication:

• You multiply every element within the matrix by the scalar.
• The result is a new matrix of the same dimensions, but with each value adjusted by the scalar.
• This operation can amplify, reduce, or reverse the sign of the matrix elements, depending on whether the scalar is positive, less than one, or negative.

In matrix equations, scalar multiplication often prepares matrices for further manipulations, such as solving for variables or adding and subtracting matrices.

addition and subtraction of matrices

Adding and subtracting matrices involve element-wise operations between matrices of identical dimensions. For the matrix equation \(2X + 5A = B\), after scalar multiplication, subtraction is performed as follows:

• Ensure both matrices have the same number of rows and columns.
• Add or subtract each corresponding element from the matrices to get the result of the operation.
• In our example, the subtraction \(B - 5A\) is executed by subtracting elements of \(5A\) from corresponding elements in \(B\).

These operations are fundamental for solving matrix equations as they enable simplification of expressions and isolation of specific terms, such as solving for matrix X here.

solving for variables in matrices

When solving for variables in matrices, we aim to isolate the unknown matrix by algebraic manipulation of equations. In the equation \(2X + 5A = B\), we need to solve for \(X\).
Here's how to proceed:

• First, isolate terms involving the unknown matrix X by subtracting any additive terms (Here, subtract \(5A\) from \(B\)).
• With terms involving X isolated, simplify the expression further by performing any scalar multiplication or division needed to finalize \(X\) (Here, divide by 2).

This process mirrors solving for variables in traditional algebra but involves matrix operations such as scalar multiplication, matrix addition, and matrix inversion where applicable.
Through practice, solving matrix equations can become intuitive, providing a versatile tool for many areas of study in mathematics and engineering.