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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

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\(A + 2B\), \(3C - E\), \(CB\), \(EB\).

Suppose the transfer function W(s) in Exercise 19 is invertible for some s. It can be showed that the inverse transfer function \(W{\left( s \right)^{ - {\bf{1}}}}\), which transforms outputs into inputs, is the Schur complement of \(A - BC - s{I_n}\) for the matrix below. Find the Sachur complement. See Exercise 15.

\(\left[ {\begin{array}{*{20}{c}}{A - BC - s{I_n}}&B\\{ - C}&{{I_m}}\end{array}} \right]\)

In exercise 5 and 6, compute the product \(AB\) in two ways: (a) by the definition, where \(A{b_{\bf{1}}}\) and \(A{b_{\bf{2}}}\) are computed separately, and (b) by the row-column rule for computing \(AB\).

\(A = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}&{\bf{2}}\\{\bf{5}}&{\bf{4}}\\{\bf{2}}&{ - {\bf{3}}}\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}&{ - {\bf{2}}}\\{ - {\bf{2}}}&{\bf{1}}\end{aligned}} \right)\)

Suppose block matrix \(A\) on the left side of (7) is invertible and \({A_{{\bf{11}}}}\) is invertible. Show that the Schur component \(S\) of \({A_{{\bf{11}}}}\) is invertible. [Hint: The outside factors on the right side of (7) are always invertible. Verify this.] When \(A\) and \({A_{{\bf{11}}}}\) are invertible, (7) leads to a formula for \({A^{ - {\bf{1}}}}\), using \({S^{ - {\bf{1}}}}\) \(A_{{\bf{11}}}^{ - {\bf{1}}}\), and the other entries in \(A\).

3. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{5}}\\{ - {\bf{7}}}&{ - {\bf{5}}}\end{aligned}} \right)\).

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