Question

Show that a) \(\mathrm{A}+\mathrm{B}=\mathrm{B}+\mathrm{A}\) where \(\begin{array}{rrlrl}\mathrm{A}= & \mid 3 & 1 & 1 \mid ; \mathrm{B}= & 14 & 2 & -1 \\ \mid 2 & -1 & 1 \mid & \mid 0 & 0\end{array}\) c) If \(\mathrm{A}\) and the zero matrix \(\left(0_{i j}\right)\) have the same size, then \(\mathrm{A}+0=\mathrm{A}\) where \(\begin{array}{rl}\mathrm{A}= & \mid 2 & 1 \\ \mid 1 & 2 \mid \\ \text { d) } \mathrm{A}+(-\mathrm{A}) & =0 \text { where }\end{array}\) \(\mathrm{A}=\mid \begin{array}{ll}2 & 1 \\ 1 & 2 \mid\end{array}\) e) (ab) \(\mathrm{A}=\mathrm{a}(\mathrm{bA})\) where \(\mathrm{a}=-5, \mathrm{~b}=3\) and f) Find \(\mathrm{B}\) if \(2 \mathrm{~A}-3 \mathrm{~B}+\mathrm{C}=0\) where \(\mathrm{A}=\mid \begin{array}{rl}-1 & 3 \mid \text { and } & \mathrm{C}=\mid-2 & -1 \mid \\ \mid 0 & 0 & \mid-1 & 1 \mid\end{array}\)

Short answer

In summary, we've shown the following properties for the given matrices: a) Commutative property of addition: A + B = B + A \[\begin{pmatrix} 3 & 1 & 1 \\ 2 & -1 & 1 \end{pmatrix}\] +\[\begin{pmatrix} 14 & 2 & -1 \\ 0 & 0 & 0 \end{pmatrix}\] = \[\begin{pmatrix} 14 & 2 & -1 \\ 0 & 0 & 0 \end{pmatrix}\] +\[\begin{pmatrix} 3 & 1 & 1 \\ 2 & -1 & 1 \end{pmatrix}\] c) Additive identity property: A + 0 = A \[\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\] +\[\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\] = \[\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\] d) Additive inverse property: A + (-A) = 0 \[\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\] +\[\begin{pmatrix} -2 & -1 \\ -1 & -2 \end{pmatrix}\] = \[\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\] e) Associative property of scalar multiplication: (ab)A = a(bA) (-15)\[\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\] = (-5)\[\begin{pmatrix} 6 & 3 \\ 3 & 6 \end{pmatrix}\] f) Solved for B in the equation 2A - 3B + C = 0, and found: B = \[\begin{pmatrix} 0 & \frac{7}{3} \\ \frac{1}{3} & -\frac{1}{3} \end{pmatrix}\]

Step-by-step solution

Add A and B.

To show that A + B = B + A, first add matrices A and B: A + B= \[\begin{pmatrix} 3 & 1 & 1 \\ 2 & -1 & 1 \end{pmatrix}\] +\[\begin{pmatrix} 14 & 2 & -1 \\ 0 & 0 & 0 \end{pmatrix}\]

Add B and A.

Next, add matrices B and A: B + A= \[\begin{pmatrix} 14 & 2 & -1 \\ 0 & 0 & 0 \end{pmatrix}\] +\[\begin{pmatrix} 3 & 1 & 1 \\ 2 & -1 & 1 \end{pmatrix}\]

Compare results and confirm commutative property.

Compare the results from steps 1 and 2 to confirm that A + B = B + A. Result of A + B: \[\begin{pmatrix} 17 & 3 & 0 \\ 2 & -1 & 1 \end{pmatrix}\] Result of B + A: \[\begin{pmatrix} 17 & 3 & 0 \\ 2 & -1 & 1 \end{pmatrix}\] As the two results are equal, we can conclude that A + B = B + A. #c) Additive identity property#

Add A and the zero matrix.

The additive identity property states that any matrix A added to the zero matrix of the same size is equal to A. To show this, add matrix A and the zero matrix: A + 0= \[\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\] +\[\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\]

Confirm that result equals A.

Check if the result of the addition equals A: Result of A + 0= \[\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\] As the result equals A, we can conclude that A + 0 = A. #d) Additive inverse property#

Add A and -A.

The additive inverse property states that any matrix A added to its negation (-A) is equal to the zero matrix. To show this, add matrix A and -A: A + (-A)= \[\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\] +\[\begin{pmatrix} -2 & -1 \\ -1 & -2 \end{pmatrix}\]

Confirm that result equals the zero matrix.

Check if the result of the addition equals the zero matrix: Result of A + (-A)= \[\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\] As the result equals the zero matrix, we can conclude that A + (-A) = 0. #e) Associative property of scalar multiplication#

Calculate (ab)A.

The associative property states (ab)A = a(bA). To show this, first calculate (ab)A with a = -5, b = 3, and A as provided: (ab)A = (-5)(3)A= \[-15\] \[\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\]

Calculate a(bA).

Next, calculate a(bA): a(bA) = -5(3A)= \[-5\] \[\begin{pmatrix} 6 & 3 \\ 3 & 6 \end{pmatrix}\]

Compare results and confirm associative property.

Compare the results from steps 1 and 2 to confirm that (ab)A = a(bA). Result of (ab)A: \[\begin{pmatrix} -30 & -15 \\ -15 & -30 \end{pmatrix}\] Result of a(bA): \[\begin{pmatrix} -30 & -15 \\ -15 & -30 \end{pmatrix}\] As the two results are equal, we can conclude that (ab)A = a(bA). #f) Solve for B in the equation 2A - 3B + C = 0#

Rearrange the equation to solve for B.

First, rearrange the equation to isolate B: 3B = 2A - C Let's substitute A and C with the given matrices: 3B = 2 \[\begin{pmatrix} -1 & 3 \\ 0 & 0 \end{pmatrix}\] - \[\begin{pmatrix} -2 & -1 \\ -1 & 1 \end{pmatrix}\]

Multiply A by 2 and subtract C from the result.

Multiply A by 2 and subtract C from the result: 3B = \[\begin{pmatrix} -2 & 6 \\ 0 & 0 \end{pmatrix}\] -\[\begin{pmatrix} -2 & -1 \\ -1 & 1 \end{pmatrix}\] =\[\begin{pmatrix} 0 & 7 \\ 1 & -1 \end{pmatrix}\]

Divide by 3 to find B.

Finally, divide the result by 3 to find matrix B: B = \[\begin{pmatrix} 0 & 7 \\ 1 & -1 \end{pmatrix}\]/3 B = \[\begin{pmatrix} 0 & \frac{7}{3} \\ \frac{1}{3} & -\frac{1}{3} \end{pmatrix}\] We found matrix B, as required.

Key concepts

Commutative Property of Addition

In mathematics, the commutative property of addition is a fundamental principle that applies to matrices just as it does to regular numbers. It states that you can change the order of the matrices being added without changing the result. Specifically, for two matrices \( A \) and \( B \), if both are the same size, then \( A + B = B + A \). This concept is crucial because it assures us that the sum remains consistent regardless of the order in which we perform the operation.
For example, when you have matrices \( A \) and \( B \):

• Adding \( A + B \) gives a result matrix.
• Adding \( B + A \) yields the same result as \( A + B \).

Using this property saves time and effort when solving algebraic expressions that involve multiple matrix additions.

Additive Identity

The additive identity property is a touchstone in both arithmetic and matrix operations. This property states that adding the zero matrix to any given matrix \( A \) yields the original matrix \( A \), which translates to \( A + 0 = A \). The zero matrix is like the number zero in regular arithmetic operations; when added, it does not change the original value.
In the context of matrices:

• \( A \) is the given matrix, such as \( \begin{pmatrix} 2 & 1 \ 1 & 2 \end{pmatrix} \).
• The zero matrix is \( \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} \) of the same size.

When you add them, the original structure of \( A \) remains intact, emphasizing that the presence of a neutral element (zero) in addition doesn't alter the original matrix.

Additive Inverse

The additive inverse property is another critical concept in matrix operations. It involves finding a matrix that "undoes" another through addition, which means that when you add a matrix \( A \) to its additive inverse \( -A \), the result is the zero matrix. This is expressed as \( A + (-A) = 0 \), where \(-A\) is the matrix with all elements being the negatives of \( A \).
This property is crucial when solving equations, as it allows cancelling terms effectively. For example:

• If \( A \) is \( \begin{pmatrix} 2 & 1 \ 1 & 2 \end{pmatrix} \), then \(-A\) will be \( \begin{pmatrix} -2 & -1 \ -1 & -2 \end{pmatrix} \).

Adding \( A \) and \( -A \) results in the zero matrix \( \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} \), demonstrating how each element neutralizes its counterpart.

Associative Property of Scalar Multiplication

The associative property of scalar multiplication is an important concept when applied to matrices involving scalar factors. It indicates that, for any scalar values \( a \) and \( b \), and a matrix \( A \), the expression \((ab)A = a(bA)\) holds true.
This essentially means that the manner of grouping the scalars when they multiply the matrix \( A \) does not matter, allowing for flexible computational order:

• Calculate \((ab)A\) by finding the product \( ab \) first, then multiplying it by \( A \).
• Alternatively, compute \( bA \) first, then multiply the result by \( a \).

Both methods produce the same matrix result, illustrating that the operation's sequence might be altered without affecting the outcome. For instance, with \( a = -5 \), \( b = 3 \), and a given matrix \( A \), confirming that each approach resolves equivalently validates the property.