Practice for Test 5, M110.
(1) Perform the indicated row operation for the matrix:
-2r1 + r2 = r2
(2) Use a matrix to solve the system of linear equations:
(3) Use the TI-83 and matrices to solve the system of linear equations:
(4) Perform the matrix operation, if possible
(a) 2[A] + 5[B], where [A] = , [B] = .
(b) *
(5) By hand, find A-1, the inverse matrix of A =
(6) Use the TI-83 to find the inverse of A = .
(7) Find the determinant of the square matrix:
(a) (b)
(8) Calculate:
(9) Use the binomial theorem (and Pascal's triangle) to find:
(a) (x + y)4 (b) (3x + 2)3.
(10) An employee code consists of 2 letters followed by 4 digits. How many such codes are possible if:
(a) repeated digits are and letters are allowed?
(b) No repeated letters or digits are allowed.
(11) A group of 8 workers aer available, from which 3 are chosen for a crew. In how many ways can the crew be chosen if (a) one worker is a foreman, one is assistant, and the other is just spying for the bossman? (b) the crew is unranked?
Solutions:
(1) Multiply everything in the first row by -2, adding it to the number below it in the second row:
-2r1 + r2 = r2
(2) First, pack the system into it's associated augmented matrix. Then get it in echelon form by eliminating the 2 in the second row (add -2 to it.)
-2r1 + r2 = r2 . Now, unpack the matrix as equations again, last row first:
-7y = -7, (solve for y)
y = 1, (unpack the first equation to back-solve for x)
x + 3y = 7,
x + 3(1) = 7
-3 -3
x = 4. The solution to the system is the ordered pair (4,1).
(3) The augmented matrix for is . We need to enter this matrix into the TI-83. Go to the matrix menu:
: Use the right-direction button to move over to EDIT.
I'm going to enter it as [A], so hit : Type the proper dimension over the numbers at the top - it should now be '3x4': , and you will be in the matrix: Type in the entries of our augmented matrix - hit enter after each one to move to the next. Once you have entered them all, go back to the main screen: . Now we need to run the function 'rref(' on our matrix. Go back to the matrix menu: go right-direction over to the MATH sub-menu, and move down until you get to rref( : Hit :
Now we have to enter [A] into rref(,:
The matrix is in reduced-row echelon form. We can read off the answer as the right-hand column x = 1, y = -2, z = 3: (1,-2,3)
(4)(a) .
(b) You have to multiply each row of the first by each column of the second (or use the TI-83)
=
=.
(5) - put the matrix on the left-side and the identity matrix on the right-hand side of an augmented matrix, then use the elimination method to move the identity matrix to the right-hand side (eliminate the 2 first, then the second one in the first row, then make the lead numbers both 1).
-2r1 + r2= r2 -r2 + r1 = r1 .
The matrix on the right is now A-1 = .
(6) First, enter the matrix into [A] like we did in problem 3 (note that this matrix has 3 rows, and 3 columns, so you'll have to have 3x3 at the top). Then go back to the main screen . Now bring the matrix [A] to the main screen with .
Now just hit the key and . Change the answers to fractions with : This is A-1
(7) (a) the determinant is the difference of the cross-products:
(2)(4) - (1)(3) = 5.
(b) The determinant is
i - j + k=
i( 1(-1) - (2)(4) ) - j( (3)(-1)-(2)(2) ) + k( (3)(4) - (2)(1)) =
-9i +7j + 10k.
(8) . So
(9) Up to the fourth tier, Pascal's triangle looks like:
1 1
1 2 1
1 3 3 1
1 4 6 4 1
(a) (x + y)4 =
(b) (3x + 2)3 =
(10) (a) 'Let me count the ways' - here, we're going to use the multiplication principle - put in a Blank for each digit and letter:
______
l1 l2 d1 d2 d3 d4 , then fill it in with the available number of choices:
26 26 10 10 10 10 #codes = 262 * 104.
l1 l2 d1 d2 d3 d4
(b) 'No repeats cuts down on the choices after one or more had been made:
26 25 10 9 8 7 #codes = 26*25*10*9*8*7.
l1 l2 d1 d2 d3 d4
(11) One of these is a permutation , and the other is a combination . The first one is the permuation because the commitee is ranked. The second is a combination because the commitee is not ranked. So the answers are:
(a)
(b)