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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

Let \(S = \left( {\begin{aligned}{*{20}{c}}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\end{aligned}} \right)\). Compute \({S^k}\) for \(k = {\bf{2}},...,{\bf{6}}\).

Let \(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\), and \(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\). Compute \(AD\) and \(DA\). Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Find a \(3 \times 3\) matrix B, not the identity matrix or the zero matrix, such that \(AB = BA\).

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation. Explain why T is both one-to-one and onto \({\mathbb{R}^n}\). Use equations (1) and (2). Then give a second explanation using one or more theorems.

[M] For block operations, it may be necessary to access or enter submatrices of a large matrix. Describe the functions or commands of your matrix program that accomplish the following tasks. Suppose A is a \(20 \times 30\) matrix.

  1. Display the submatrix of Afrom rows 15 to 20 and columns 5 to 10.
  2. Insert a \(5 \times 10\) matrix B into A, beginning at row 10 and column 20.
  3. Create a \(50 \times 50\) matrix of the form \(B = \left[ {\begin{array}{*{20}{c}}A&0\\0&{{A^T}}\end{array}} \right]\).

[Note: It may not be necessary to specify the zero blocks in B.]

In exercise 11 and 12, mark each statement True or False.Justify each answer.

a. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}&{{A_{\bf{2}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), with \({A_{\bf{1}}}\) and \({A_{\bf{2}}}\) the same sizes as \({B_{\bf{1}}}\) and \({B_{\bf{2}}}\), respectively then \(A + B = \left[ {\begin{array}{*{20}{c}}{{A_1} + {B_1}}&{{A_{\bf{2}}} + {B_{\bf{2}}}}\end{array}} \right]\).

b. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_1}}\\{{B_{\bf{2}}}}\end{array}} \right]\), then the partitions of \(A\) and \(B\) are comfortable for block multiplication.

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