LINEAR ALGEBRA TEST #1 SOLUTIONS

SHOW YOUR WORK!!

SOLVE THESE SYSTEMS OF LINEAR EQUATIONS IF POSSIBLE:

A/

1)step 1: ¼(r2 – r1); and (r3 – 4r1)/5
step 2: (r1 + r2); (r3 – r2)
solution: x = 1, y = 3, z = 0

2) switch 2nd and 3rd eqs.
step 1: (r3 – 2r1); and (r4 – 3r1)
step 2: (r1 – 2r2); (r3 + r2); (r4 – r2)
step 3: (r4 – r3); –( r3 / 6)
solution:

3) step 1: (r2 + r1); (r3 – r1); (r4 – 2r1)
step 2: (r1 – 2r2); (r3 + r2); (r4 – r2)
system is inconsistent. No solutions.

(15)

B/

1) Det (A) = 0

2) A ^–1 doesn't exist since det (A) = 0, so A is not row equivalent to I ₃.

3) step 1: – (r2 – 2r1); (r3 + r1);
solution: x = 5t, y = 7t, z = t; t is an element of R

(9)

C/

1) A is invertible: Det A = 1 2) A – 1 = by reduction.		4) Multiply A – 1 (b) to get x = 14, y = 13, and z = – 12
3) There's only the unique solution

(11)

D/

1) PROOF

Since A, B, and C are invertible matrices of the same size, A ^–1 , B ^–1 and C ^–1 all exist.
Let's multiply ABC on the left by A ^–1
Now, A ^–1(ABC) = (A ^–1A)BC, which equals (I)BC or just BC.
Now, we do it again with B ^–1 to get B ^–1 (BC) = (I)C or just C.
One more time with C ^–1 and we've shown that (C ^–1 B ^–1A ^–1)(ABC) = I
So (C ^–1 B ^–1A ^–1) is the inverse of ABC which is (ABC) ^–1.

(5)

2) Since this matrix is its own inverse.

(3)

3) Find inverses for each of these elementary matrices.

(3)

4)

The elementary matrices E₁ and E₂ must a) switch 2 rows b) multiply a row by k.
The value of k depends on the order we choose for the row operations.

Solution 1: E₁ switches the rows of matrix A; and E₂ takes ½ of row 2 in B.

So,

Solution 2: E₁ doubles row1 of matrix A; and E₂ switches the rows of matrix B.

So,

(4)

TOTAL (50)

________________________________________

.

( Linear Algebra MathRoom Index )

(all content © MathRoom Learning Service; 2004 - ).