Let's show how this formula works in a specific case. Suppose A is a 3 x 3 matrix and B is a 3 x 2 matrix as in our previous example, then the result of the product AB is a 3 x 2 matrix that we can call C. Now suppose we want to find the entry on the third row in the second column of C, then we would compute: 3 €3,2 = Σa3.bk,2 k=1 =a3,101,2 + a3,2b2,2 +03,303,2- Sure enough, when we look at the long version we wrote earlier for the product AB our result matches the entry on the second row, third column. The above formula makes it possible to calculate individual matrix ele- ments, without having to compute the entire matrix. Exercise 11.2.2. (a) Let the entries of A be given by ai.j = Vi+j for 1 ≤ i, j ≤ 100. Let C = A.A (we can also write C = A²). Compute €10,10- = (b) Let the entries of A and B be given by aij = (i+j)² and bij 1≤i, j≤ 27. Let C = A B Compute €8,8. for (c) For the matrices A and B in part (b), give a general formula for Ck,k, 1 ≤ < 27 wh

Advanced Engineering Mathematics
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Chapter2: Second-order Linear Odes
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Please do Part B and C and please show step by step and explain

Let's show how this formula works in a specific case. Suppose A is a
3 x 3 matrix and B is a 3 x 2 matrix as in our previous example, then the
result of the product AB is a 3 x 2 matrix that we can call C. Now suppose
we want to find the entry on the third row in the second column of C, then
we would compute:
3
€3,2 = 93,kbk,2
k=1
=a3,101,2 +03,2b2,2 + a3,3b3,2-
Sure enough, when we look at the long version we wrote earlier for the
product AB our result matches the entry on the second row, third column.
The above formula makes it possible to calculate individual matrix ele-
ments, without having to compute the entire matrix.
Exercise 11.2.2.
(a) Let the entries of A be given by aj = Vi+j for 1 ≤i, j≤ 100. Let
C = A.A (we can also write C = A²). Compute €10,10-
(b) Let the entries of A and B be given by ai.j = (i+j)² and bij = for
1 ≤i, j≤ 27. Let C = A - B Compute €8,8.
(c) For the matrices A and B in part (b), give a general formula for ck,k,1 <
k≤ 27 where C= AB.
Transcribed Image Text:Let's show how this formula works in a specific case. Suppose A is a 3 x 3 matrix and B is a 3 x 2 matrix as in our previous example, then the result of the product AB is a 3 x 2 matrix that we can call C. Now suppose we want to find the entry on the third row in the second column of C, then we would compute: 3 €3,2 = 93,kbk,2 k=1 =a3,101,2 +03,2b2,2 + a3,3b3,2- Sure enough, when we look at the long version we wrote earlier for the product AB our result matches the entry on the second row, third column. The above formula makes it possible to calculate individual matrix ele- ments, without having to compute the entire matrix. Exercise 11.2.2. (a) Let the entries of A be given by aj = Vi+j for 1 ≤i, j≤ 100. Let C = A.A (we can also write C = A²). Compute €10,10- (b) Let the entries of A and B be given by ai.j = (i+j)² and bij = for 1 ≤i, j≤ 27. Let C = A - B Compute €8,8. (c) For the matrices A and B in part (b), give a general formula for ck,k,1 < k≤ 27 where C= AB.
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