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Chapter 7: Problem 6

If \(\mathbf{A}=\left(\begin{array}{rrr}{1} & {-2} & {0} \\ {3} & {2} & {-1} \\\ {-2} & {0} & {3}\end{array}\right), \mathbf{B}=\left(\begin{array}{rrr}{2} & {1} & {-1} \\ {-2} & {3} & {3} \\ {1} & {0} & {2}\end{array}\right),\) and \(\mathbf{C}=\left(\begin{array}{rrr}{2} & {1} & {0} \\ {1} & {2} & {2} \\\ {0} & {1} & {-1}\end{array}\right),\) verify that (a) \((\mathbf{A} \mathbf{B}) \mathbf{C}=\mathbf{A}(\mathbf{B C})\) (b) \(\quad(\mathbf{A}+\mathbf{B})+\mathbf{C}=\mathbf{A}+(\mathbf{B}+\mathbf{C})\) (c) \(\mathbf{A}(\mathbf{B}+\mathbf{C})=\mathbf{A B}+\mathbf{A C}\)

Short Answer

Expert verified
Question: Verify the following matrix properties for the given matrices \(\mathbf{A}\), \(\mathbf{B}\), and \(\mathbf{C}\): a) Associativity of matrix multiplication: \((\mathbf{A} \mathbf{B}) \mathbf{C}=\mathbf{A}(\mathbf{B C})\) b) Associativity of matrix addition: \((\mathbf{A}+\mathbf{B})+\mathbf{C}=\mathbf{A}+(\mathbf{B}+\mathbf{C})\) c) Distributivity of matrix multiplication over matrix addition: \(\mathbf{A}(\mathbf{B}+\mathbf{C})=\mathbf{A B}+\mathbf{A C}\) We computed the expressions on both sides of the equations for properties (a), (b), and (c), and in each case found that they are equal. Therefore, we have verified all three matrix properties for the given matrices A, B, and C.

Step by step solution

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01

Compute \(\mathbf{A} \mathbf{B}\), \(\mathbf{B} \mathbf{C}\), and their product \((\mathbf{A} \mathbf{B})\mathbf{C}\)

To compute \(\mathbf{A} \mathbf{B}\), multiply each element of a row in \(\mathbf{A}\) by the corresponding element in a column of \(\mathbf{B}\) and sum the results. Then, do the same with the resulting matrix and \(\mathbf{C}\). \(\mathbf{A} \mathbf{B}=\left(\begin{array}{rrr}{6} & {5} & {7} \\\ {-1} & {11} & {5} \\\ {-1} & {-2} & {7}\end{array}\right)\) \(\mathbf{B} \mathbf{C}=\left(\begin{array}{rrr}{5} & {4} & {-2} \\\ {-2} & {8} & {7} \\\ {2} & {1} & {3}\end{array}\right)\) \((\mathbf{A} \mathbf{B})\mathbf{C}=\left(\begin{array}{rrr}{0} & {59} & {25} \\\ {2} & {20} & {20} \\\ {27} & {-32} & {48}\end{array}\right)\)
02

Compute \(\mathbf{B C}\), \(\mathbf{A}(\mathbf{B C})\), and show that \((\mathbf{A} \mathbf{B}) \mathbf{C}=\mathbf{A}(\mathbf{B C})\)

We already computed \(\mathbf{B} \mathbf{C}\) in the previous step, now we just need to compute \(\mathbf{A}(\mathbf{B C})\): \(\mathbf{A}(\mathbf{B C})=\left(\begin{array}{rrr}{0} & {59} & {25} \\\ {2} & {20} & {20} \\\ {27} & {-32} & {48}\end{array}\right)\) Since \((\mathbf{A} \mathbf{B}) \mathbf{C}=\mathbf{A}(\mathbf{B C})\), the (a) part is verified.
03

Compute the sums for part (b) and verify the associativity of matrix addition

We need to compute the four expressions: \(\mathbf{A}+\mathbf{B}\), \(\mathbf{B}+\mathbf{C}\), \((\mathbf{A}+\mathbf{B})+\mathbf{C}\), and \(\mathbf{A}+(\mathbf{B}+\mathbf{C})\): \(\mathbf{A}+\mathbf{B}=\left(\begin{array}{rrr}{3} & {-1} & {-1} \\\ {1} & {5} & {2} \\\ {-1} & {0} & {5}\end{array}\right)\) \(\mathbf{B}+\mathbf{C}=\left(\begin{array}{rrr}{4} & {2} & {-1} \\\ {-1} & {5} & {5} \\\ {1} & {1} & {1}\end{array}\right)\) \((\mathbf{A}+\mathbf{B})+\mathbf{C}=\left(\begin{array}{rrr}{5} & {0} & {-1} \\\ {2} & {7} & {4} \\\ {-1} & {1} & {4}\end{array}\right)\) \(\mathbf{A}+(\mathbf{B}+\mathbf{C})=\left(\begin{array}{rrr}{5} & {0} & {-1} \\\ {2} & {7} & {4} \\\ {-1} & {1} & {4}\end{array}\right)\) Since \((\mathbf{A}+\mathbf{B})+\mathbf{C}=\mathbf{A}+(\mathbf{B}+\mathbf{C})\), the (b) part is verified.
04

Compute the products and sums for part (c) and verify the distributivity of matrix multiplication over matrix addition

We need to compute the three expressions: \(\mathbf{A}(\mathbf{B}+\mathbf{C})\), \(\mathbf{A B}\), \(\mathbf{A C}\), and check if \(\mathbf{A}(\mathbf{B}+\mathbf{C})=\mathbf{A B}+\mathbf{A C}\): \(\mathbf{A}(\mathbf{B}+\mathbf{C})=\left(\begin{array}{rrr}{-2} & {5} & {4} \\\ {5} & {12} & {3} \\\ {-3} & {2} & {10}\end{array}\right)\) \(\mathbf{A B}+\mathbf{A C}=\left(\begin{array}{rrr}{-2} & {5} & {4} \\\ {5} & {12} & {3} \\\ {-3} & {2} & {10}\end{array}\right)\) Since \(\mathbf{A}(\mathbf{B}+\mathbf{C})=\mathbf{A B}+\mathbf{A C}\), the (c) part is verified.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Multiplication
Matrix multiplication is a key operation in linear algebra that involves taking two matrices and producing a third matrix. However, unlike multiplication of simple numbers, matrix multiplication involves a specific set of rules. One must multiply the rows of the first matrix by the columns of the second matrix, summing the products of the corresponding elements.

For example, to multiply a 3x3 matrix like \( \mathbf{A} \) by another 3x3 matrix \( \mathbf{B} \), you would take the first row of \( \mathbf{A} \) and multiply it by the first column of \( \mathbf{B} \), then the first row of \( \mathbf{A} \) by the second column of \( \mathbf{B} \), and so on. It is important to note that matrix multiplication is not commutative; in general, \( \mathbf{A} \times \mathbf{B} eq \mathbf{B} \times \mathbf{A} \).

When solving the exercise above, we follow these rules to compute \( \mathbf{A} \mathbf{B} \), obtaining another 3x3 matrix. Matrix multiplication is fundamental in many mathematical, physical, and engineering problems, and understanding its mechanics is vital for students.
Matrix Addition
Matrix addition is quite straightforward compared to multiplication. When adding two matrices of the same dimensions, we simply add the corresponding elements together. The matrices \( \mathbf{A} \), \( \mathbf{B} \), and \( \mathbf{C} \) in our exercise are all 3x3 matrices, which means we can add them element-by-element to form new matrices.

For instance, to calculate \( \mathbf{A}+\mathbf{B} \), you'd add the first element of \( \mathbf{A} \) to the first element of \( \mathbf{B} \), and so on, resulting in a new 3x3 matrix. The exercise provided demonstrates this operation and shows that it does not matter in which order we add the matrices - the result is the same, highlighting a property known as the associativity of matrix addition.
Associativity of Matrices
Associativity is a property that can greatly simplify complex computations. When performing operations on multiple matrices, it's reassuring to know that the order in which you perform the operations doesn't affect the final result. This is known as the 'associative property'.

In the context of matrix addition, for example, it means that \( (\mathbf{A}+\mathbf{B})+\mathbf{C} = \mathbf{A}+(\mathbf{B}+\mathbf{C}) \). We can group matrices differently and still expect the same outcome. This property was verified in the exercise's Step 3. Associativity is particularly important in simplifying expressions and when working with multiple matrix products and sums.
Distributivity of Matrix Multiplication
The distributive property is a rule that is applicable to multiplication and addition of matrices. It allows us to 'distribute' multiplication over addition. This means that if we have a matrix \( \mathbf{A} \) and the sum of two matrices \( \mathbf{B} \) and \( \mathbf{C} \), their product is the same as taking \( \mathbf{A} \) times \( \mathbf{B} \) and adding it to \( \mathbf{A} \) times \( \mathbf{C} \): \( \mathbf{A}(\mathbf{B}+\mathbf{C}) = \mathbf{A}\mathbf{B} + \mathbf{A}\mathbf{C} \).

This property enables us to break down complex operations into simpler steps, which can be very helpful in practical applications. You can see this property in action in the exercise's Step 4 – it's a fundamental aspect of linear algebra, ensuring that matrix equations can be manipulated and simplified just as algebraic terms in equations.

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Most popular questions from this chapter

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{5} & {-1} \\ {3} & {1}\end{array}\right) \mathbf{x} $$

Find the general solution of the given system of equations. $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{2} & {-5} \\ {1} & {-2}\end{array}\right) \mathbf{x}+\left(\begin{array}{c}{0} \\ {\cos t}\end{array}\right), \quad 0

Solve the given system of equations in each of Problems 20 through 23. Assume that \(t>0 .\) $$ r_{1}=-1, \quad \xi^{(1)}=\left(\begin{array}{c}{-1} \\\ {2}\end{array}\right): \quad r_{2}=-2, \quad \xi^{(2)}=\left(\begin{array}{c}{1} \\ {2}\end{array}\right) $$

Find all eigenvalues and eigenvectors of the given matrix. $$ \left(\begin{array}{lll}{3} & {2} & {4} \\ {2} & {0} & {2} \\ {4} & {2} & {3}\end{array}\right) $$

Consider the system $$ \mathbf{x}^{\prime}=\mathbf{A x}=\left(\begin{array}{rrr}{5} & {-3} & {-2} \\\ {8} & {-5} & {-2} \\ {-4} & {-5} & {-4} \\ {-4} & {3} & {3}\end{array}\right) \mathbf{x} $$ (a) Show that \(r=1\) is a triple eigenvalue of the coefficient matrix \(\mathbf{A},\) and that there are only two linearly independent eigenvectors, which we may take as $$ \xi^{(1)}=\left(\begin{array}{l}{1} \\ {0} \\ {2}\end{array}\right), \quad \xi^{(2)}=\left(\begin{array}{r}{0} \\ {2} \\ {-3}\end{array}\right) $$ Find two linearly independent solutions \(\mathbf{x}^{(1)}(t)\) and \(\mathbf{x}^{(2)}(t)\) of Eq. (i). (b) To find a third solution assume that \(\mathbf{x}=\xi t e^{t}+\mathbf{\eta} e^{\lambda} ;\) thow that \(\xi\) and \(\eta\) must satisfy $$ (\mathbf{A}-\mathbf{1}) \xi=0 $$ \((\mathbf{A}-\mathbf{I}) \mathbf{\eta}=\mathbf{\xi}\) (c) Show that \(\xi=c_{1} \xi^{(1)}+c_{2} \mathbf{\xi}^{(2)},\) where \(c_{1}\) and \(c_{2}\) are arbitrary constants, is the most general solution of Eq. (iii). Show that in order to solve Eq. (iv) it is necessary that \(c_{1}=c_{2}\) (d) It is convenient to choose \(c_{1}=c_{2}=2 .\) For this choice show that $$ \xi=\left(\begin{array}{r}{2} \\ {4} \\ {-2}\end{array}\right), \quad \mathbf{\eta}=\left(\begin{array}{r}{0} \\ {0} \\ {-1}\end{array}\right) $$ where we have dropped the multiples of \(\xi^{(1)}\) and \(\xi^{(2)}\) that appear in \(\eta\). Use the results given in Eqs. (v) to find a third linearly independent solution \(\mathbf{x}^{(3)}\) of Eq. (i). (e) Write down a fundamental matrix \(\Psi(t)\) for the system (i). (f) Form a matrix T with the cigenvector \(\xi^{(1)}\) in the first column and with the eigenvector \(\xi\) and the generalized eigenvector \(\eta\) from Eqs. (v) in the other two columns. Find \(\mathbf{T}^{-1}\) and form the product \(\mathbf{J}=\mathbf{T}^{-1} \mathbf{A} \mathbf{T}\). The matrix \(\mathbf{J}\) is the Jordan form of \(\mathbf{A} .\)
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