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Prove Theorem 2(b) and 2(c). Use the row-column rule. The \(\left( {i,j} \right)\)- entry in \(A\left( {B + C} \right)\) can be written as \({a_{i1}}\left( {{b_{1j}} + {c_{1j}}} \right) + ... + {a_{in}}\left( {{b_{nj}} + {c_{nj}}} \right)\) or \(\sum\limits_{k = 1}^n {{a_{ik}}\left( {{b_{kj}} + {c_{kj}}} \right)} \).

Short Answer

Expert verified

Theorem 2(b) and theorem 2(c) are proved.

Step by step solution

01

The row-column rule

If the product AB is defined, the entry in row \(i\) and column \(j\) of ABis the sum of the products of corresponding entries from the row \(i\)of Aand column \(j\) of B. If \({\left( {AB} \right)_{ij}}\) denotes the \(\left( {i,j} \right)\)- entry in AB, and if Ais a \(m \times n\) matrix, then

\({\left( {AB} \right)_{ij}} = {a_{i1}}{b_{1j}} + {a_{i2}}{b_{2j}} + ... + {a_{in}}{b_{nj}}\).

02

Prove theorem 2(b)

Theorem 2states that Abe a \(m \times n\) matrix let Band Chave sizes for which the indicated sums and products are defined.

  1. \(A\left( {BC} \right) = \left( {AB} \right)C\) (associative law of multiplication)
  2. \(A\left( {B + C} \right) = AB + AC\) (left distributive law)
  3. \(\left( {B + C} \right)A = BA + CA\) (right distributive law)

The \(\left( {i,j} \right)\)- entry in \(A\left( {B + C} \right)\) can be written as \({a_{i1}}\left( {{b_{1j}} + {c_{1j}}} \right) + ... + {a_{in}}\left( {{b_{nj}} + {c_{nj}}} \right)\).

The \(\left( {i,j} \right)\)- entry of \(A\left( {B + C} \right)\) equals to the \(\left( {i,j} \right)\)- entry of \(AB + AC\) since \(\sum\limits_{k = 1}^n {{a_{ik}}\left( {{b_{kj}} + {c_{kj}}} \right)} = \sum\limits_{k = 1}^n {{a_{ik}}{b_{kj}}} + \sum\limits_{k = 1}^n {{a_{ik}}{c_{kj}}} \).

03

Prove theorem 2(c)

The \(\left( {i,j} \right)\)- entry of \(\left( {B + C} \right)A\) equals to the \(\left( {i,j} \right)\)- entry of \(BA + CA\) since \(\sum\limits_{k = 1}^n {\left( {{b_{ik}} + {c_{ik}}} \right){a_{kj}}} = \sum\limits_{k = 1}^n {{b_{ik}}{a_{kj}}} + \sum\limits_{k = 1}^n {{c_{ik}}{a_{kj}}} \).

Hence, the theorems 2(b) and 2(c) are proved.

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Most popular questions from this chapter

(M) Read the documentation for your matrix program, and write the commands that will produce the following matrices (without keying in each entry of the matrix).

  1. A \({\bf{5}} \times {\bf{6}}\) matrix of zeros
  2. A \({\bf{3}} \times {\bf{5}}\) matrix of ones
  3. The \({\bf{6}} \times {\bf{6}}\) identity matrix
  4. A \({\bf{5}} \times {\bf{5}}\) diagonal matrix, with diagonal entries 3, 5, 7, 2, 4

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{ - {\bf{3}}}\\{ - {\bf{4}}}&{\bf{6}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{4}}\\{\bf{5}}&{\bf{5}}\end{aligned}} \right)\) and \(C = \left( {\begin{aligned}{*{20}{c}}{\bf{5}}&{ - {\bf{2}}}\\{\bf{3}}&{\bf{1}}\end{aligned}} \right)\). Verfiy that \(AB = AC\) and yet \(B \ne C\).

Generalize the idea of Exercise 21(a) [not 21(b)] by constructing a \(5 \times 5\) matrix \(M = \left[ {\begin{array}{*{20}{c}}A&0\\C&D\end{array}} \right]\) such that \({M^2} = I\). Make C a nonzero \(2 \times 3\) matrix. Show that your construction works.

Exercises 15 and 16 concern arbitrary matrices A, B, and Cfor which the indicated sums and products are defined. Mark each statement True or False. Justify each answer.

15. a. If A and B are \({\bf{2}} \times {\bf{2}}\) with columns \({{\bf{a}}_1},{{\bf{a}}_2}\) and \({{\bf{b}}_1},{{\bf{b}}_2}\) respectively, then \(AB = \left( {\begin{aligned}{*{20}{c}}{{{\bf{a}}_1}{{\bf{b}}_1}}&{{{\bf{a}}_2}{{\bf{b}}_2}}\end{aligned}} \right)\).

b. Each column of ABis a linear combination of the columns of Busing weights from the corresponding column of A.

c. \(AB + AC = A\left( {B + C} \right)\)

d. \({A^T} + {B^T} = {\left( {A + B} \right)^T}\)

e. The transpose of a product of matrices equals the product of their transposes in the same order.

Suppose Aand Bare \(n \times n\), Bis invertible, and ABis invertible. Show that Ais invertible. (Hint: Let C=AB, and solve this equation for A.)

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