Problem 15 - College Algebra and Calculus: An Applied Approach

Question

Evaluate the expression. $$ -3\left(\left[\begin{array}{rr} 0 & -3 \\ 7 & 2 \end{array}\right]+\left[\begin{array}{rr} -6 & 3 \\ 8 & 1 \end{array}\right]\right)-2\left[\begin{array}{ll} 4 & -4 \\ 7 & -9 \end{array}\right] $$

Step-by-Step Solution

Answer

The evaluated expression for the given problem is \[ \left[\begin{array}{rr} 26 & -8 \\ -31 & -27 \end{array}\right] \]

Step 1: Matrix Addition

Firstly, we perform matrix addition. \\( \left[\begin{array}{rr} 0 & -3 \\ 7 & 2 \end{array}\right] + \left[\begin{array}{rr}-6 & 3 \\ 8 & 1 \end{array}\right] = \left[\begin{array}{cc} (0+(-6)) & (-3+3) \\ (7+8) & (2+1) \end{array}\right] = \left[\begin{array}{rr} -6 & 0 \\ 15 & 3 \end{array}\right]\)

Step 2: Scalar Multiplication

Next, we perform scalar multiplication of the resultant matrix by -3 and the second matrix by -2. \-3 * \left[\begin{array}{rr}-6 & 0 \\ 15 & 3 \end{array}\right] = \left[\begin{array}{rr} (-3*(-6)) & (-3*0) \\ (-3*15) & (-3*3) \end{array}\right] = \left[\begin{array}{rr} 18 & 0 \\ -45 & -9 \end{array}\right] \and \-2 * \left[\begin{array}{cc}4 & -4 \\ 7 & -9 \end{array}\right] = \left[\begin{array}{rr} -2*4 & -2*(-4) \\ -2*7 & -2*(-9) \end{array}\right] = \left[\begin{array}{cc}-8 & 8 \\ -14 & 18 \end{array}\right]

Step 3: Matrix Subtraction

Now we perform matrix subtraction. \\left[\begin{array}{rr} 18 & 0 \\ -45 & -9 \end{array}\right] - \left[\begin{array}{cc}-8 & 8 \\ -14 & 18 \end{array}\right] = \left[\begin{array}{rr} (18-(-8)) & (0-8) \\ (-45-(-14)) & (-9-18) \end{array}\right] = \left[\begin{array}{rr} 26 & -8 \\ -31 & -27 \end{array}\right]

Key Concepts

Matrix Addition

Matrix addition is a fundamental operation in which two matrices of the same dimensions are added together by adding their corresponding entries. This can be visually represented as combining two numeric grids. The result is a new matrix of the same dimensions where each element is the sum of the elements at the same position in the original matrices. For example, consider two matrices
\begin{align*}A &= \left[\begin{array}{rr}0 & -3 \7 & 2\end{array}\right]\text{and} \ B &= \left[\begin{array}{rr}-6 & 3 \8 & 1\end{array}\right].\end{align*}Their sum, \( A + B \), is found by adding each element of \( A \) to the corresponding element of \( B \):
Matrix A + Matrix B = \left[\begin{array}{rr}0+(-6) & -3+3 \7+8 & 2+1\end{array}\right] = \left[\begin{array}{rr}-6 & 0 \15 & 3\end{array}\right].
The operation is commutative, meaning that \( A + B = B + A \), and the zero matrix acts as the additive identity.

Scalar Multiplication

In scalar multiplication, each entry of a matrix is multiplied by a fixed number called a scalar. This operation affects the magnitude and direction of the matrix’s entries but not its dimensions. If the scalar is negative, the direction is reversed which can be observed as a sign change in the elements.
For instance, when we multiply matrix \( C = \left[\begin{array}{rr}-6 & 0 \15 & 3 \end{array}\right] \) with the scalar -3, we get:
\( -3 * C = \left[\begin{array}{rr} -3*(-6) & -3*0 \ -3*15 & -3*3 \end{array}\right] = \left[\begin{array}{rr} 18 & 0 \ -45 & -9 \end{array}\right] \).
This shows that scalar multiplication changes the magnitude of each element according to the scalar's value and inverses the sign when the scalar is negative. This concept is fundamental in various applications of linear algebra, such as stretching or shrinking vectors and matrices.

Matrix Subtraction

Matrix subtraction, like matrix addition, requires two matrices of the same dimensions. The process involves subtracting the corresponding entries of the second matrix from the first. This operation is not commutative; changing the order of the matrices results in a completely different outcome. The concept can be understood as the opposite of matrix addition.
Consider a case where we subtract matrix \( D = \left[\begin{array}{rr}-8 & 8 \ -14 & 18 \end{array}\right] \) from matrix \( E = \left[\begin{array}{rr} 18 & 0 \ -45 & -9 \end{array}\right] \), the calculation would be:
\( E - D = \left[\begin{array}{rr} 18-(-8) & 0-8 \ -45-(-14) & -9-18 \end{array}\right] = \left[\begin{array}{rr} 26 & -8 \ -31 & -27 \end{array}\right] \).
This shows that each resultant element is simply the difference between elements of the same position from the two matrices involved.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and arithmetic operations. In the context of matrices, these expressions can involve operations such as addition, subtraction, and scalar multiplication as seen in the earlier examples. Simplifying algebraic expressions with matrices involves performing sequence of operations while adhering to the rules of matrix algebra, like conformability for addition and subtraction, and distribution for scalar multiplication over addition or subtraction.
For the given problem, the algebraic expression involves both scalar multiplication and addition or subtraction of matrices, illustrating how matrix operations can be combined to manipulate data in linear algebra. The ability to simplify such expressions is crucial for solving systems of equations, transforming geometric data, and many other applications across various fields of science and engineering.