LINEAR ALGEBRA PRACTICE EXERCISES ELEMENTARY MATRICES |
Reminder: an Elementary Matrix is an Identity matrix
on which a single row operation has been performed.
example: the elementary matrix A = is I 2 x 2 with row 2 added to row 1.
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QUESTIONS
1) B =
a) Find 3 elementary matrices E 1, E 2, E 3, such that E 3 $ E 2 $ E 1 $ B = I.
b) Find 3 elementary matrices D 1, D 2, D 3, such that D 1 $ D 2 $ D 3 = B.
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2) C =
a) Find 4 elementary matrices E 1, E 2, E 3, E 4, such that E 4 $ E 3 $ E 2 $ E 1 $ C = I.
b) Find 4 elementary matrices D 1, D 2, D 3, D 4 such that D 1 $ D 2 $ D 3 $ D 4 = C.
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3) A = and B =
Find 2 elementary matrices E 1 and E 2 so that E 2 $ E 1 $ A = B
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4) A = and B =
Find 2 elementary matrices E 1 and E 2 so that E 1 $ A = E 2 $ B
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5) If A = use elementary matrices to find A - 1 if it exists.
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SOLUTIONS
1) B =
a) The 3 row operations we want to perform on B to obtain I are:
i) row 2 minus 2(row 1) | ii) -1/3(row 2) | iii) row 1 minus 3(row 2) |
E 1 = ![]() so E 1 B = |
E 2 = ![]() so E 2 E 1 B = |
E 3 = ![]() so E 3 E 2 E 1 B = |
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b) now we want to do the inverse of the 3 row operations in the reverse order*.
So the 3 row operations we want to perform on I to obtain B are:
iii) row 2 plus 2(row 1) | ii) - 3 (row 2) | i) row 1 plus 3(row 2) |
D 1 = ![]() so D 1 D 2 D3 = |
D 2 = ![]() so D 2 D3 = |
D 3 = ![]() |
* Note: we did these operations in the reverse order because of how the question is stated.
Had we been asked to find these matrices such that D 3 $ D 2 $ D 1 = B, we
would have done the operations in the same order we used in part (a).
Remember that matrix multiplication on the right and on the left are not
necessarily equivalent operations.
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2) C =
a) We want to turn C into I so here's the 4 operations we should do:
i) switch row 3 and row 1 E 1 = so E 1 C = |
ii) take ½ (row 2) E 2 = so E 2 E 1 C = |
iii) row 3 minus 6(row 2) E 3 = so E 3 E 2 E 1 C = |
iv) row 2 minus 2(row 3) E 4 = so E 4 E 3 E 2 E 1 C = |
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b) Just as in question 1, we want the inverse operations in the reverse order.
iv) switch row 3 and row 1 D 1 = |
iii) take 2 (row 2) E 2 = |
ii) row 3 plus 6(row 2) E 3 = |
i) row 2 plus 2(row 3) E 4 = |
** Note that the order is reversed. the 1st move is in the fourth table cell.
3) A = and B =
Since row 2 in both matrices is identical, we need to change row 1 and row 3.
To change Row 3 of A to Row 3 of B, we take row 3 minus 2(row 1).
This makes E 1 = ,
Now, to change Row 1 of A into Row 1 of B, we take row 1 minus 4(row 2)
This makes E 2 = and E 2 $ E 1 $ A = B
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4) A = and B =
It's pretty obvious that we have to switch the rows and double or halve a row.
The order we choose determines which set of answers we get.
i) If we double row 1 of A first, E 1 = , and E 1 A =
,
This will make E 2 = , and E 2 B =
so that E 1 $ A = E 2 $ B
ii) If we flip the rows of A first, E 1 = , and E 1 A =
This will make E 2 = , since we need to take ½ of row 2
E 2 B = so that E 1 $ A = E 2 $ B
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5) We set up the "super-augmented" matrix with A on the left and I on the right,
then we do row reduction. When we have I on the left, we have A - 1 on the right.
The steps which reduce A to I are:
i) switch R3 and R1 | ii) R2 - 4(R1) R3 - 3(R1) |
iii) - (R3 - R2) | iv) R1 - R3 R2 - 2(R3) |
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