1.the missing real numbers in place of the asterisks):Find the real numbers missing from the following equation (that is, rewrite the equation with2 -3-11 5 6* 61 *1-1** 2Each of the nine entries of the 3 x 3 matrix product in Problem 1 is a sum of products of entries of the othertwo matrices, and there are nine missing numbers (asterisks), so one systematic way to answer this questionis to solve a system of nine equations in nine variables. There is a much faster, less systematic way: try tofind individual missing numbers one at a time!

Answered: 1. Find the real numbers missing from…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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1.
the missing real numbers in place of the asterisks):
Find the real numbers missing from the following equation (that is, rewrite the equation with
2 -3
-11 5 6
* 6
1 *
1
-1
*
* 2
Each of the nine entries of the 3 x 3 matrix product in Problem 1 is a sum of products of entries of the other
two matrices, and there are nine missing numbers (asterisks), so one systematic way to answer this question
is to solve a system of nine equations in nine variables. There is a much faster, less systematic way: try to
find individual missing numbers one at a time!

Expert Solution

Step 1

Naming the missing numbers and rewrite the given equation as follows:

$[\begin{matrix} 2 & - 3 \\ 1 & a \\ - 1 & b \end{matrix}] [\begin{matrix} - 1 & c & 6 \\ d & 1 & e \end{matrix}] = [\begin{matrix} - 11 & 5 & 6 \\ - 1 & f & g \\ h & i & 2 \end{matrix}] L e t A = [\begin{matrix} 2 & - 3 \\ 1 & a \\ - 1 & b \end{matrix}], B = [\begin{matrix} - 1 & c & 6 \\ d & 1 & e \end{matrix}] a n d C = [\begin{matrix} - 11 & 5 & 6 \\ - 1 & f & g \\ h & i & 2 \end{matrix}]$

The sum of products of first row of A and first column of B will be the first element in C. That is,

$(2) (- 1) + (- 3) (d) = - 11 - 2 - 3 d = - 11 3 d = 11 - 2 3 d = 9 d = \frac{9}{3} d = 3$

Thus, the given equation becomes,

$[\begin{matrix} 2 & - 3 \\ 1 & a \\ - 1 & b \end{matrix}] [\begin{matrix} - 1 & c & 6 \\ 3 & 1 & e \end{matrix}] = [\begin{matrix} - 11 & 5 & 6 \\ - 1 & f & g \\ h & i & 2 \end{matrix}]$

Similarly, sum of products of first row of A and second column of B will be the element first row and second column in C. That is,

$(2) (c) + (- 3) (1) = 5 2 c - 3 = 5 2 c = 5 + 3 2 c = 8 c = \frac{8}{2} c = 4$

Thus, the given equation becomes,

$[\begin{matrix} 2 & - 3 \\ 1 & a \\ - 1 & b \end{matrix}] [\begin{matrix} - 1 & 4 & 6 \\ 3 & 1 & e \end{matrix}] = [\begin{matrix} - 11 & 5 & 6 \\ - 1 & f & g \\ h & i & 2 \end{matrix}]$

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