Problem 25 - 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 \((A+B) C=A C+B C\).

Step-by-Step Solution

Answer

The equality \((A+B)C = AC + BC\) holds true, confirmed by calculations.

Step 1: Add Matrices A and B

To find \((A + B)\), add corresponding elements of matrices \(A\) and \(B\). Matrix \(A\) is given as \(\begin{bmatrix} -1 & 0 \ 1 & 2 \end{bmatrix}\) and matrix \(B\) is \(\begin{bmatrix} 2 & 0 \ -1 & -1 \end{bmatrix}\). Therefore, \(A + B = \begin{bmatrix} -1+2 & 0+0 \ 1+(-1) & 2+(-1) \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\).

Step 2: Calculate Product (A+B)C

Now calculate \((A + B)C\) using matrix multiplication on \(\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\) and \(\begin{bmatrix} 1 & 2 \ 0 & -1 \end{bmatrix}\). The result is: \((A+B)C = \begin{bmatrix} 1 \cdot 1 + 0 \cdot 0 & 1 \cdot 2 + 0 \cdot (-1) \ 0 \cdot 1 + 1 \cdot 0 & 0 \cdot 2 + 1 \cdot (-1) \end{bmatrix} = \begin{bmatrix} 1 & 2 \ 0 & -1 \end{bmatrix}\).

Step 3: Calculate Product AC

Multiply matrix \(A\) by matrix \(C\) to find \(AC\). \(A = \begin{bmatrix} -1 & 0 \ 1 & 2 \end{bmatrix}\) and \(C = \begin{bmatrix} 1 & 2 \ 0 & -1 \end{bmatrix}\). Perform the multiplication: \(AC = \begin{bmatrix} -1 \cdot 1 + 0 \cdot 0 & -1 \cdot 2 + 0 \cdot (-1) \ 1 \cdot 1 + 2 \cdot 0 & 1 \cdot 2 + 2 \cdot (-1) \end{bmatrix} = \begin{bmatrix} -1 & -2 \ 1 & 0 \end{bmatrix}\).

Step 4: Calculate Product BC

Multiply matrix \(B\) by matrix \(C\) to find \(BC\). \(B = \begin{bmatrix} 2 & 0 \ -1 & -1 \end{bmatrix}\) and \(C = \begin{bmatrix} 1 & 2 \ 0 & -1 \end{bmatrix}\). The multiplication gives: \(BC = \begin{bmatrix} 2 \cdot 1 + 0 \cdot 0 & 2 \cdot 2 + 0 \cdot (-1) \ -1 \cdot 1 + (-1) \cdot 0 & -1 \cdot 2 + (-1) \cdot (-1) \end{bmatrix} = \begin{bmatrix} 2 & 4 \ -1 & -1 \end{bmatrix}\).

Step 5: Calculate AC + BC

Now add the results from Step 3 and Step 4 to find \(AC + BC = \begin{bmatrix} -1 & -2 \ 1 & 0 \end{bmatrix} + \begin{bmatrix} 2 & 4 \ -1 & -1 \end{bmatrix}\). This addition results in: \(AC + BC = \begin{bmatrix} -1+2 & -2+4 \ 1+(-1) & 0+(-1) \end{bmatrix} = \begin{bmatrix} 1 & 2 \ 0 & -1 \end{bmatrix}\).

Step 6: Verify Equality

Compare the results from Step 2 and Step 5. Both \((A+B)C\) and \(AC + BC\) yield \(\begin{bmatrix} 1 & 2 \ 0 & -1 \end{bmatrix}\), confirming that \((A+B)C = AC + BC\). The equality holds, demonstrating the distributive property of matrix multiplication.

Key Concepts

Matrix Addition

Matrix addition is a straightforward process. It's like adding two sets of numbers, but in this case, you add two matrices. To do this, you take corresponding elements from each matrix and add them together. For example, if we have two matrices, A and B, they must be of the same size. Let's say both are 2x2 matrices. This means each matrix has two rows and two columns.

The operation is done element by element. So, you take the element in the first row and first column of matrix A, and add it to the element in the first row and first column of matrix B. Repeat this step for all corresponding elements. The result is a new matrix, which is the sum of the two matrices.

In the context of our exercise, when we added matrices A and B, we performed the operation:

• \[\begin{bmatrix} -1 & 0 \ 1 & 2 \end{bmatrix} + \begin{bmatrix} 2 & 0 \ -1 & -1 \end{bmatrix} = \begin{bmatrix} -1+2 & 0+0 \ 1+(-1) & 2+(-1) \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \] .

With matrix addition, remember: you can only add matrices of the same dimensions. It's like adding numbers, but in a square grid!

Distributive Property

The distributive property is vital in simplifying expressions, especially in matrix algebra. In terms of matrices, it states that for any three matrices, A, B, and C, the following equation always holds: \[ (A + B)C = AC + BC \]

This property is much like distributing multiplication over addition in regular arithmetic, where, for numbers x, y, and z,

• \[x(y + z) = xy + xz\]

applies. It's a neat trick that allows you to break down complex matrix operations into simpler parts.

In the given exercise, we've demonstrated this property. When we calculated

• \[(A + B)C = AC + BC \]

We first computed

• \[(A + B)C\]
• Then \[(AC) + (BC)\]

Both yielded the same result:

• \[(AC + BC) = \begin{bmatrix} 1 & 2 \ 0 & -1 \end{bmatrix}\] .

This showed that despite involving more steps when handled separately, the final combined operation through distributivity proves consistent. The distributive property is a powerful tool in matrix algebra and can simplify calculations greatly, making complex problems more manageable.

Matrix Algebra

Matrix algebra involves operations similar to those of regular algebra, but it's applied to matrices. Basic operations include addition, subtraction, and multiplication. However, unlike regular numbers, matrix multiplication is not commutative, meaning

• \[AB eq BA \]

in general. Multiplying matrices involves an intricate process. You take the rows of the first matrix and "dot" them with the columns of the second matrix. This means multiplying corresponding elements and summarizing the products.

Matrix algebra is fundamental in many fields like physics, computer graphics, and data science. It allows for solving systems of equations, transformations, and more complex calculations. Each matrix operation has specific rules and requirements, like equal dimensions for addition and specific compatibility for multiplication.

In our exercise, we're showing one of the vital properties of matrix multiplication through the distributive property. By working through the multiplication of matrices separately and proving the equality with distributive law, we're using concepts central to matrix algebra. As you practice, the subtleties and specific rules of matrix algebra will become more intuitive, harnessing its power for solving diverse problems.
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