Problem 53 - Algebra for College Students

Question

Use the following matrices: $$ A=\left[\begin{array}{rr} 7 & -4 \\ 6 & 9 \end{array}\right] \quad B=\left[\begin{array}{ll} -3 & 8 \\ -5 & 7 \end{array}\right] $$ $$ C=\left[\begin{array}{ll} 8 & -2 \\ 4 & -7 \end{array}\right] $$ (a) Show that \((A B) C=A(B C)\). (b) Show that \(A(B+C)=A B+A C\). (c) Show that \((B+C) A=B A+C A\).

Step-by-Step Solution

Answer

(a) \((AB)C = A(BC)\). (b) \(A(B+C) = AB + AC\). (c) \((B+C)A = BA + CA\).

Step 1: Calculate AB

To find the product \( AB \), multiply the matrices: \[ A = \begin{bmatrix} 7 & -4 \ 6 & 9 \end{bmatrix}, B = \begin{bmatrix} -3 & 8 \ -5 & 7 \end{bmatrix} \]The product is calculated as: \[AB = \begin{bmatrix} (7)(-3) + (-4)(-5) & (7)(8) + (-4)(7) \ (6)(-3) + (9)(-5) & (6)(8) + (9)(7) \end{bmatrix} = \begin{bmatrix} -21 + 20 & 56 - 28 \ -18 - 45 & 48 + 63 \end{bmatrix} = \begin{bmatrix} -1 & 28 \ -63 & 111 \end{bmatrix} \]

Step 2: Calculate (AB)C

Multiply the product \( AB \) with matrix \( C \):\[C = \begin{bmatrix} 8 & -2 \ 4 & -7 \end{bmatrix}\]Calculate:\[(AB)C = \begin{bmatrix} (-1)(8) + (28)(4) & (-1)(-2) + (28)(-7) \ (-63)(8) + (111)(4) & (-63)(-2) + (111)(-7) \end{bmatrix} = \begin{bmatrix} -8 + 112 & 2 - 196 \ -504 + 444 & 126 - 777 \end{bmatrix} = \begin{bmatrix} 104 & -194 \ -60 & -651 \end{bmatrix} \]

Step 3: Calculate BC

To find \( BC \), multiply matrices \( B \) and \( C \):\[B = \begin{bmatrix} -3 & 8 \ -5 & 7 \end{bmatrix}, C = \begin{bmatrix} 8 & -2 \ 4 & -7 \end{bmatrix}\]Using matrix multiplication:\[BC = \begin{bmatrix} (-3)(8) + (8)(4) & (-3)(-2) + (8)(-7) \ (-5)(8) + (7)(4) & (-5)(-2) + (7)(-7) \end{bmatrix} = \begin{bmatrix} -24 + 32 & 6 - 56 \ -40 + 28 & 10 - 49 \end{bmatrix} = \begin{bmatrix} 8 & -50 \ -12 & -39 \end{bmatrix} \]

Step 4: Calculate A(BC)

Multiply matrix \( A \) with the product \( BC \):\[A = \begin{bmatrix} 7 & -4 \ 6 & 9 \end{bmatrix}, BC = \begin{bmatrix} 8 & -50 \ -12 & -39 \end{bmatrix}\]Perform the multiplication:\[A(BC) = \begin{bmatrix} (7)(8) + (-4)(-12) & (7)(-50) + (-4)(-39) \ (6)(8) + (9)(-12) & (6)(-50) + (9)(-39) \end{bmatrix} = \begin{bmatrix} 56 + 48 & -350 + 156 \ 48 - 108 & -300 - 351 \end{bmatrix} = \begin{bmatrix} 104 & -194 \ -60 & -651 \end{bmatrix} \]

Step 5: Validate Associative Property

Since \((AB)C = \begin{bmatrix} 104 & -194 \ -60 & -651 \end{bmatrix}\) and \(A(BC) = \begin{bmatrix} 104 & -194 \ -60 & -651 \end{bmatrix}\), we have verified that \((AB)C = A(BC) \). Therefore, the associative property holds for matrices \( A \), \( B \), and \( C \).

Step 6: Calculate B+C

Add matrices \( B \) and \( C \):\[B = \begin{bmatrix} -3 & 8 \ -5 & 7 \end{bmatrix}, C = \begin{bmatrix} 8 & -2 \ 4 & -7 \end{bmatrix}\]Resulting in:\[B+C = \begin{bmatrix} -3 + 8 & 8 - 2 \ -5 + 4 & 7 - 7 \end{bmatrix} = \begin{bmatrix} 5 & 6 \ -1 & 0 \end{bmatrix} \]

Step 7: Calculate A(B+C)

Multiply matrix \( A \) with \( B+C \):\[A = \begin{bmatrix} 7 & -4 \ 6 & 9 \end{bmatrix}, B+C = \begin{bmatrix} 5 & 6 \ -1 & 0 \end{bmatrix}\]Result:\[A(B+C) = \begin{bmatrix} (7)(5) + (-4)(-1) & (7)(6) + (-4)(0) \ (6)(5) + (9)(-1) & (6)(6) + (9)(0) \end{bmatrix} = \begin{bmatrix} 35 + 4 & 42 + 0 \ 30 - 9 & 36 + 0 \end{bmatrix} = \begin{bmatrix} 39 & 42 \ 21 & 36 \end{bmatrix} \]

Step 8: Calculate AB + AC

Calculate \( AC \) first:\[C = \begin{bmatrix} 8 & -2 \ 4 & -7 \end{bmatrix}\]Multiply:\[AC = \begin{bmatrix} 7 & -4 \ 6 & 9 \end{bmatrix} \begin{bmatrix} 8 & -2 \ 4 & -7 \end{bmatrix} = \begin{bmatrix} (7)(8) + (-4)(4) & (7)(-2) + (-4)(-7) \ (6)(8) + (9)(4) & (6)(-2) + (9)(-7) \end{bmatrix} = \begin{bmatrix} 56 - 16 & -14 + 28 \ 48 + 36 & -12 - 63 \end{bmatrix} = \begin{bmatrix} 40 & 14 \ 84 & -75 \end{bmatrix} \]Add \( AB \) and \( AC \):\[AB = \begin{bmatrix} -1 & 28 \ -63 & 111 \end{bmatrix}, AC = \begin{bmatrix} 40 & 14 \ 84 & -75 \end{bmatrix}\]\[AB + AC = \begin{bmatrix} -1 + 40 & 28 + 14 \ -63 + 84 & 111 - 75 \end{bmatrix} = \begin{bmatrix} 39 & 42 \ 21 & 36 \end{bmatrix} \]

Step 9: Validate Distributive Property for A

Since \( A(B+C) = \begin{bmatrix} 39 & 42 \ 21 & 36 \end{bmatrix} \) and \( AB + AC = \begin{bmatrix} 39 & 42 \ 21 & 36 \end{bmatrix} \), this verifies the distributive property: \( A(B+C) = AB + AC \).

Step 10: Calculate (B+C)A

Multiply \( B+C \) with \( A \):\[B+C = \begin{bmatrix} 5 & 6 \ -1 & 0 \end{bmatrix}, A = \begin{bmatrix} 7 & -4 \ 6 & 9 \end{bmatrix}\]\[(B+C)A = \begin{bmatrix} (5)(7) + (6)(6) & (5)(-4) + (6)(9) \ (-1)(7) + (0)(6) & (-1)(-4) + (0)(9) \end{bmatrix} = \begin{bmatrix} 35 + 36 & -20 + 54 \ -7 + 0 & 4 + 0 \end{bmatrix} = \begin{bmatrix} 71 & 34 \ -7 & 4 \end{bmatrix} \]

Step 11: Calculate BA and CA

Compute the products separately:\[BA = \begin{bmatrix} -3 & 8 \ -5 & 7 \end{bmatrix} \begin{bmatrix} 7 & -4 \ 6 & 9 \end{bmatrix} = \begin{bmatrix} (-3)(7) + (8)(6) & (-3)(-4) + (8)(9) \ (-5)(7) + (7)(6) & (-5)(-4) + (7)(9) \end{bmatrix} = \begin{bmatrix} -21 + 48 & 12 + 72 \ -35 + 42 & 20 + 63 \end{bmatrix} = \begin{bmatrix} 27 & 84 \ 7 & 83 \end{bmatrix} \]\[CA = \begin{bmatrix} 8 & -2 \ 4 & -7 \end{bmatrix} \begin{bmatrix} 7 & -4 \ 6 & 9 \end{bmatrix} = \begin{bmatrix} (8)(7) + (-2)(6) & (8)(-4) + (-2)(9) \ (4)(7) + (-7)(6) & (4)(-4) + (-7)(9) \end{bmatrix} = \begin{bmatrix} 56 - 12 & -32 - 18 \ 28 - 42 & -16 - 63 \end{bmatrix} = \begin{bmatrix} 44 & -50 \ -14 & -79 \end{bmatrix} \]

Step 12: Calculate BA + CA

Add the results of \( BA \) and \( CA \):\[BA = \begin{bmatrix} 27 & 84 \ 7 & 83 \end{bmatrix}, CA = \begin{bmatrix} 44 & -50 \ -14 & -79 \end{bmatrix}\]\[BA + CA = \begin{bmatrix} 27 + 44 & 84 - 50 \ 7 - 14 & 83 - 79 \end{bmatrix} = \begin{bmatrix} 71 & 34 \ -7 & 4 \end{bmatrix} \]

Step 13: Validate Distributive Property for (B+C)

Since \((B+C)A = \begin{bmatrix} 71 & 34 \ -7 & 4 \end{bmatrix}\) and \( BA + CA = \begin{bmatrix} 71 & 34 \ -7 & 4 \end{bmatrix}\), we confirm that \((B+C)A = BA + CA \).

Key Concepts

Matrix Multiplication

Matrix multiplication is a process of multiplying two matrices to produce a third matrix. For two matrices, A and B, to be multiplied, the number of columns in matrix A must equal the number of rows in matrix B. This allows each element of the resulting matrix to be calculated as the sum of the products of row elements from the first matrix and the column elements of the second matrix.

Consider matrices A and B:

• If matrix A is of size \(m \times n\) and matrix B is \(n \times p\), then the resulting matrix will have a size of \(m \times p\).
• The formula used for each element \(c_{ij}\) in the product matrix C is: \[\sum_{k=1}^{n} (a_{ik} \cdot b_{kj})\]
• For our problem, to multiply matrix \(A = \begin{bmatrix} 7 & -4 \ 6 & 9 \end{bmatrix}\) with matrix \(B = \begin{bmatrix} -3 & 8 \ -5 & 7 \end{bmatrix}\), calculate each element:
  • First row, first column: \((-21 + 20 = -1)\)
  • First row, second column: \((56 - 28 = 28)\)
  • Second row, first column: \((-18 - 45 = -63)\)
  • Second row, second column: \((48 + 63 = 111)\)
  Thus, producing the matrix \(AB = \begin{bmatrix} -1 & 28 \ -63 & 111 \end{bmatrix}\).

Matrix multiplication is not commutative, meaning \(AB eq BA\), but it is associative, as discussed next.

Associative Property of Matrices

The associative property in matrix multiplication states that when multiplying three matrices together, the way in which the matrices are grouped does not change the resulting product. This is represented as:

• \((AB)C = A(BC)\)
• This means we can first multiply A and B, then multiply the result by C, or we can first multiply B and C, then multiply the result by A.

In our solution, matrices \(A = \begin{bmatrix} 7 & -4 \ 6 & 9 \end{bmatrix}\), \(B = \begin{bmatrix} -3 & 8 \ -5 & 7 \end{bmatrix}\), and \(C = \begin{bmatrix} 8 & -2 \ 4 & -7 \end{bmatrix}\) were used to verify this property:

• First, we calculated \((AB)C\), which resulted in: \(\begin{bmatrix} 104 & -194 \ -60 & -651 \end{bmatrix}\)
• Then, we computed \(A(BC)\), which also amounted to \(\begin{bmatrix} 104 & -194 \ -60 & -651 \end{bmatrix}\)

Since \((AB)C = A(BC)\) in both calculations, the associative property holds in this case. This concept is crucial for understanding that while the order of matrix multiplication is important (as seen in non-commutative property), the grouping of matrices in multiplication isn't.

Distributive Property of Matrices

The distributive property of matrices allows us to distribute scalar multiplication over matrix addition. It can be noted as:

• For matrices A, B, and C: \( A(B+C) = AB + AC \)
• Similarly: \((B+C)A = BA + CA\)

For the matrices in the problem, we demonstrated the distributive property as follows:

• We first computed \( B+C = \begin{bmatrix} 5 & 6 \ -1 & 0 \end{bmatrix} \)
• Then we found \( A(B+C) \) resulting in \( \begin{bmatrix} 39 & 42 \ 21 & 36 \end{bmatrix} \)
• Independently, we calculated both \( AB + AC \) which also resulted in \( \begin{bmatrix} 39 & 42 \ 21 & 36 \end{bmatrix} \)
• Similarly, for \((B+C)A\) which resulted in \( \begin{bmatrix} 71 & 34 \ -7 & 4 \end{bmatrix} \) and \( BA + CA \) resulting in the same \( \begin{bmatrix} 71 & 34 \ -7 & 4 \end{bmatrix} \)

These examples show that the property holds true in both multiplication orders for matrix addition, demonstrating that matrix multiplication indeed follows the distributive law.