Problem 21
Question
In Exercises \(17-26,\) 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
Verified Answer
To solve for \(X\), we first isolated \(X\) by moving all other terms to the other side of the equation. We then performed matrix subtraction and scalar division to get our solution matrix \(X\).
1Step 1: Understand the Matrix Equation
The given matrix equation is \(3X + 2A = B\). This equation follows the same general rules as normal algebra equations, but for matrices. Our goal is to isolate \(X\) on one side of the equation.
2Step 2: Express X in terms of A and B
Let's start by moving the term with \(A\) to the other side of the equation, to isolate terms with \(X\) on one side. We achieve this by subtracting \(2A\) from both sides of the equation: \(3X = B - 2A\).
3Step 3: Compute B - 2A
We calculate \(2A\) by multiplying each element within matrix \(A\) by the scalar \(2\). Then, subtract this new matrix from \(B\) by subtracting corresponding elements in the two matrices (matrix subtraction)
4Step 4: Solve for X
Now, we have an equation where \(3X\) equals a certain matrix. So, to isolate \(X\), we need to divide this whole matrix by \(3\). we will perform scalar matrix division, dividing each element of the matrix by 3.
5Step 5: Present X
Once we have done all the calculations, we are able to present the solution matrix \(X\).
Key Concepts
Matrix SubtractionScalar MultiplicationSolving Matrix EquationsMatrix Algebra
Matrix Subtraction
Matrix subtraction is an operation where we subtract corresponding elements of two matrices. To perform matrix subtraction, both matrices involved must have the same dimensions.
For instance, consider the matrices from the exercise:
For instance, consider the matrices from the exercise:
- Matrix A: \[\begin{bmatrix}-3 & -7 \2 & -9 \5 & 0\end{bmatrix}\]
- Matrix B: \[\begin{bmatrix}-5 & -1 \0 & 0 \3 & -4\end{bmatrix}\]
Scalar Multiplication
Scalar multiplication involves multiplying every element within a matrix by a scalar (a constant number). This operation is straightforward yet significant in matrix algebra. In our exercise, we needed to perform scalar multiplication of matrix \(A\) by \(2\) to form \(2A\).Consider matrix \(A\):\[\begin{bmatrix}-3 & -7 \2 & -9 \5 & 0\end{bmatrix}\]By multiplying each element by \(2\), the matrix \(2A\) becomes:\[\begin{bmatrix}-6 & -14 \4 & -18 \10 & 0\end{bmatrix}\]Each element from the original matrix is transformed by the scalar, allowing us to manipulate the matrix as a single entity. This operation is crucial in expressing matrix equations where terms need to be isolated, such as the step \(3X = B - 2A\) in the given problem.
Solving Matrix Equations
Solving matrix equations involves isolating the matrix variable, similar to solving for a variable in algebraic equations. In the given exercise, we had the equation \(3X + 2A = B\).
Our goal was to solve for \(X\). Here's how:- First, express \(X\) by rearranging the equation: subtract \(2A\) from both sides to isolate terms with \(X\): \[3X = B - 2A\]- Calculate \(B - 2A\), using matrix subtraction after scalar multiplication.- Divide every element in the resulting matrix by \(3\) to complete the isolation of matrix \(X\).This step-by-step process demonstrates the principles of matrix algebra, ensuring that operations maintain matrix balance, just as we do with numbers in regular equation solving.
Our goal was to solve for \(X\). Here's how:- First, express \(X\) by rearranging the equation: subtract \(2A\) from both sides to isolate terms with \(X\): \[3X = B - 2A\]- Calculate \(B - 2A\), using matrix subtraction after scalar multiplication.- Divide every element in the resulting matrix by \(3\) to complete the isolation of matrix \(X\).This step-by-step process demonstrates the principles of matrix algebra, ensuring that operations maintain matrix balance, just as we do with numbers in regular equation solving.
Matrix Algebra
Matrix algebra is the study and application of algebraic operations within the framework of matrix structures. It enables the solving of complex systems and operations using matrices instead of individual numbers. In this exercise:
- We utilized the fundamental operations of scalar multiplication and matrix subtraction.
- Each operation obeys specific rules, ensuring that the structural integrity of matrices remains unchanged while performing arithmetic on their components.
Understanding matrix algebra allows us to solve equations involving matrices, similar to the way traditional algebra helps solve equations with numbers. This is particularly useful in fields like engineering, physics, and computer science, where matrices are used to represent data, transform spaces, and model various systems and interactions.
Other exercises in this chapter
Problem 20
Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$\begin{aligned}w+x-y+z &=-2 \\\2 w-x+2 y-z &=7 \
View solution Problem 20
In Exercises \(19-24\), perform each matrix row operation and write the new matrix. \(\left[\begin{array}{rrr|r}3 & -12 & 6 & 9 \\ 1 & -4 & 4 & 0 \\ 2 & 0 & 7 &
View solution Problem 21
Use Cramer's rule to solve each system or to determine that the system is inconsistent or contains dependent equations. $$ \begin{aligned}&2 x=3 y+2\\\&5 x=51-4
View solution Problem 21
find \(A^{-1}\) by forming \([A | I]\) and then using row operations to obtain [ \(I | B],\) where \(A^{-1}=[B]\) Check that \(A A^{-1}=I\) and \(A^{-1} A=I\) $
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