Question

Solve for \(X\) when $$A=\left[\begin{array}{rr} -2 & -1 \\ 1 & 0 \\ 3 & -4 \end{array}\right] \text { and } B=\left[\begin{array}{rr} 0 & 3 \\ 2 & 0 \\ -4 & -1 \end{array}\right].$$ $$ X=3 \dot{A}-2 \bar{B} $$

Short answer

The matrix \(X\) is \[\[ -6, -9 \] , \[ -1, 0 \] , \[ 17, -10 \]\]

Step-by-step solution

Apply the scalar multiplication on matrix A and B

Firstly, we multiply each element of matrix \(A\) and \(B\) by 3 and 2 respectively. The result is: \(3A = \[\[ -6, -3 \] , \[ 3, 0 \] , \[ 9, -12 \]\] \) and \(2B = \[\[ 0, 6 \] , \[ 4, 0 \] , \[ -8, -2 \]\] \)

Subtract Matrix B from Matrix A

Since the equation is \(X=3A - 2B\), we subtract each corresponding element in 2B from 3A. Take the first element from 3A and subtract the first element from 2B and so on. The answer will be: \(X = \[\[ -6, -9 \] , \[ -1, 0 \] , \[ 17, -10 \]\] \)

Key concepts

Scalar Multiplication

Scalar multiplication in matrices involves multiplying each element of a matrix by a scalar, which is simply a single number. When you perform scalar multiplication, every element in the matrix is affected by the same scalar.
For example, consider matrix \( A \):

• \( A = \begin{bmatrix} -2 & -1 \ 1 & 0 \ 3 & -4 \end{bmatrix} \)

If we multiply \( A \) by 3 (the scalar), every element in \( A \) gets multiplied by 3:

• \( 3A = \begin{bmatrix} 3(-2) & 3(-1) \ 3(1) & 3(0) \ 3(3) & 3(-4) \end{bmatrix} = \begin{bmatrix} -6 & -3 \ 3 & 0 \ 9 & -12 \end{bmatrix} \).

Scalar multiplication is a straightforward yet crucial operation in matrix algebra. It serves as a building block for more complex operations, like matrix addition and subtraction.

Matrix Subtraction

Matrix subtraction is very similar to regular subtraction but is performed element-wise for matrices of the same dimensions. When we talk about subtracting matrices, we refer to subtracting corresponding elements of those matrices.
As a rule of thumb:- You can only subtract matrices if they are of the same size.Looking at the problem, we have the matrices \( 3A \) and \( 2B \):

• \( 3A = \begin{bmatrix} -6 & -3 \ 3 & 0 \ 9 & -12 \end{bmatrix} \)
• \( 2B = \begin{bmatrix} 0 & 6 \ 4 & 0 \ -8 & -2 \end{bmatrix} \)

Each element in \( 2B \) is subtracted from its corresponding element in \( 3A \):

• Top left: \(-6 - 0 = -6\)
• Top right: \(-3 - 6 = -9\)
• Middle left: \(3 - 4 = -1\)
• Middle right: \(0 - 0 = 0\)
• Bottom left: \(9 - (-8) = 17\)
• Bottom right: \(-12 - (-2) = -10\)

Thus, the resulting matrix from subtracting \( 2B \) from \( 3A \) is:\[ X = \begin{bmatrix} -6 & -9 \ -1 & 0 \ 17 & -10 \end{bmatrix} \].

Solving for X

Solving for \( X \) in matrix equations involves using the basic operations of scalar multiplication and matrix subtraction. Given the equation:\[ X = 3A - 2B \]our goal is to find matrix \( X \).
First, apply scalar multiplication to the matrices \( A \) and \( B \). Multiply \( A \) by 3 and \( B \) by 2:

• \( 3A = \begin{bmatrix} -6 & -3 \ 3 & 0 \ 9 & -12 \end{bmatrix} \)
• \( 2B = \begin{bmatrix} 0 & 6 \ 4 & 0 \ -8 & -2 \end{bmatrix} \)

Next, perform matrix subtraction of \( 2B \) from \( 3A \). Align and subtract each corresponding element to find \( X \):

• \( X = \begin{bmatrix} -6 - 0 & -3 - 6 \ 3 - 4 & 0 - 0 \ 9 - (-8) & -12 - (-2) \end{bmatrix} \)
• This simplifies to \( X = \begin{bmatrix} -6 & -9 \ -1 & 0 \ 17 & -10 \end{bmatrix} \)

Breaking it down ensures that we're following the correct sequence and operations to solve for \( X \). Understanding this step allows us to solve similar problems with confidence.