Problem 26 - Calculus for Biology and Medicine

Question

Let $$A=\left[\begin{array}{rr} -1 & 0 \\ 1 & 2 \end{array}\right], \quad B=\left[\begin{array}{rr} 2 & 0 \\ -1 & -1 \end{array}\right], \quad C=\left[\begin{array}{rr} 1 & 2 \\ 0 & -1 \end{array}\right]$$ Show that \(A(B+C)=A B+A C\).

Step-by-Step Solution

Answer

The identity \(A(B+C) = AB + AC\) is verified as both operations give the same matrix result.

Step 1: Compute B + C

First, add matrices \(B\) and \(C\). To add two matrices, add their corresponding elements.\[B + C = \begin{bmatrix} 2 & 0 \ -1 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 2 \ 0 & -1 \end{bmatrix} = \begin{bmatrix} 2+1 & 0+2 \ -1+0 & -1+(-1) \end{bmatrix} = \begin{bmatrix} 3 & 2 \ -1 & -2 \end{bmatrix}\]

Step 2: Compute A(B + C)

Next, perform the multiplication of matrix \(A\) with the result from Step 1, \(B+C\). Use the rule for matrix multiplication: the element in row \(i\), column \(j\) of the product corresponds to the dot product of row \(i\) of \(A\) and column \(j\) of \(B+C\).\[A(B+C) = \begin{bmatrix} -1 & 0 \ 1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 2 \ -1 & -2 \end{bmatrix}\]Calculate each element:\[\text{First row, first column: } -1 \cdot 3 + 0 \cdot (-1) = -3\]\[\text{First row, second column: } -1 \cdot 2 + 0 \cdot (-2) = -2\]\[\text{Second row, first column: } 1 \cdot 3 + 2 \cdot (-1) = 3 - 2 = 1\]\[\text{Second row, second column: } 1 \cdot 2 + 2 \cdot (-2) = 2 - 4 = -2\]Thus,\[A(B+C) = \begin{bmatrix} -3 & -2 \ 1 & -2 \end{bmatrix}\]

Step 3: Compute AB

Compute the product of matrices \(A\) and \(B\).\[AB = \begin{bmatrix} -1 & 0 \ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 0 \ -1 & -1 \end{bmatrix}\]Calculate each element:\[\text{First row, first column: } -1 \cdot 2 + 0 \cdot (-1) = -2\]\[\text{First row, second column: } -1 \cdot 0 + 0 \cdot (-1) = 0\]\[\text{Second row, first column: } 1 \cdot 2 + 2 \cdot (-1) = 2 - 2 = 0\]\[\text{Second row, second column: } 1 \cdot 0 + 2 \cdot (-1) = -2\]Thus,\[AB = \begin{bmatrix} -2 & 0 \ 0 & -2 \end{bmatrix}\]

Step 4: Compute AC

Compute the product of matrices \(A\) and \(C\).\[AC = \begin{bmatrix} -1 & 0 \ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 0 & -1 \end{bmatrix}\]Calculate each element:\[\text{First row, first column: } -1 \cdot 1 + 0 \cdot 0 = -1\]\[\text{First row, second column: } -1 \cdot 2 + 0 \cdot (-1) = -2\]\[\text{Second row, first column: } 1 \cdot 1 + 2 \cdot 0 = 1\]\[\text{Second row, second column: } 1 \cdot 2 + 2 \cdot (-1) = 2 - 2 = 0\]Thus,\[AC = \begin{bmatrix} -1 & -2 \ 1 & 0 \end{bmatrix}\]

Step 5: Compute AB + AC

Add the results from Steps 3 and 4.\[AB + AC = \begin{bmatrix} -2 & 0 \ 0 & -2 \end{bmatrix} + \begin{bmatrix} -1 & -2 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} -2 + (-1) & 0 + (-2) \ 0 + 1 & -2 + 0 \end{bmatrix} = \begin{bmatrix} -3 & -2 \ 1 & -2 \end{bmatrix}\]

Step 6: Compare A(B+C) with AB + AC

Finally, compare the matrix obtained from Step 2 with that from Step 5.Both matrices are:\[\begin{bmatrix} -3 & -2 \ 1 & -2 \end{bmatrix}\]Since they are the same, we have shown that:\[ A(B+C) = AB + AC \]

Key Concepts

Matrix Addition

Matrix addition is one of the simplest operations you can perform with matrices. It's essentially the process of adding two matrices by adding their corresponding elements together. For two matrices to be added, they must be of the same dimension, meaning they have the same number of rows and columns.
To make things clearer, let's consider matrices \(B\) and \(C\) from the example. Both \(B\) and \(C\) are 2x2 matrices:

• Matrix \(B\) is \(\begin{bmatrix} 2 & 0 \ -1 & -1 \end{bmatrix}\)
• Matrix \(C\) is \(\begin{bmatrix} 1 & 2 \ 0 & -1 \end{bmatrix}\)

To add these matrices, simply add each corresponding element: the element in the first row and first column of \(B\) gets added to the same position in \(C\), and so on. The result is a new matrix:\[B + C = \begin{bmatrix} 2+1 & 0+2 \ -1+0 & -1+(-1) \end{bmatrix} = \begin{bmatrix} 3 & 2 \ -1 & -2 \end{bmatrix}\]

Matrix Multiplication

Unlike addition, matrix multiplication is a bit more complex and requires a specific rule to follow. You need to multiply the rows of the first matrix by the columns of the second matrix. The number of columns in the first matrix must equal the number of rows in the second matrix for the multiplication to be possible.

• For matrix \(A = \begin{bmatrix} -1 & 0 \ 1 & 2 \end{bmatrix}\) which is 2x2, and matrix \(B+C = \begin{bmatrix} 3 & 2 \ -1 & -2 \end{bmatrix}\) also 2x2, multiplication can proceed.

To compute the element in row \(i\), column \(j\) of the resulting matrix, you'll perform the dot product of row \(i\) of the first matrix and column \(j\) of the second matrix.

• First row, first column: \(-1 \cdot 3 + 0 \cdot (-1) = -3\)
• Second row, first column: \(1 \cdot 3 + 2 \cdot (-1) = 3 - 2 = 1\)

Following these steps, you can calculate that:\[A(B+C) = \begin{bmatrix} -3 & -2 \ 1 & -2 \end{bmatrix}\]This shows how important understanding the order and operation is in matrix multiplication.

Distributive Property of Matrices

The distributive property of matrices is similar to the distributive property you learn for numbers. It states that for matrices \(A\), \(B\), and \(C\), the following holds true: \[A(B+C) = AB + AC\]This means that you can distribute a matrix over a matrix addition, just like with numbers in algebra.
By calculating both sides separately, as we did in the step-by-step solution, we confirm that:

• On the left-hand side, calculate \(A(B+C)\).
• On the right-hand side, calculate \(AB + AC\).

Both results show the matrix \(\begin{bmatrix} -3 & -2 \ 1 & -2 \end{bmatrix}\), proving the property. Understanding and using this property makes complex matrix operations manageable and is a foundational concept in matrix algebra. It's essential for simplifying expressions and solving matrix equations efficiently.