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 Depdendent Variable

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 Dependent Variable

 Number of inequalities to solve: 23456789
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# Systems of Linear Equations

1. True or False.

(a) The null space of a 4 × 6 real matrix is a subspace of R6.

(b) The column space of a 4 × 6 real matrix is a subspace of R6.

(c) The rank of a 6 × 4 real matrix is at most 4.

(d) The nullity of a 4 × 6 real matrix is at least 4.

(e) A vector in Rm is in the column space of an m × n real matrix A if the linear
system is solvable.

(f) Any vector in the range of an m × n real matrix A can be written as a linear
combination of the column vectors of A.

(g) Suppose that an m × n real matrix A is row reduced to another matrix B by row
reductions. Then the null space of A equals the null space of B.

(h) Suppose that an m × n real matrix A is row reduced to another matrix B by row
reductions. Then the range of A equals the range of B.

(i) Suppose that an m × n real matrix A is row reduced to another matrix B by row
reductions. Then rank(A) = rank(B) and nullity(A) = nullity(B).

(j) If the column vectors of a 6 × 4 real matrix A are linearly independent, then the null
space of A is { 0 }.

(k) If the column vectors of a 6 × 4 real matrix A are linearly independent, then the
column space of A is R6.

2. Let .

(a) Find the dimension and a basis of Span .

(b) Find the dimension and a basis of Span . Is in Span ? If yes, write it as
a scalar multiple of with a specific coefficient.

(c) Find the dimension and a basis of Span . Is in Span ? If yes, write
it as a linear combination of with specific coefficients.

(d) Find the dimension and a basis of Span . Is in Span ? If
yes, write it as a linear combination of with specific coefficients.

(e) Find the dimension and a basis of Span Is in Span ?
If yes, write it as a linear combination of with specific coefficients.

Hint: The row reduction of one single matrix provides the answers to all these questions.

3. A n × n square matrix is called a magic square if its n row sums, n column sums, and 2
diagonal sums are all equal. For instance, is a 3 × 3 magic square, since
8+1+6 = 3+5+7 = 4+9+2 = 8+3+4 = 1+5+9 = 6+7+2 = 8+5+2 = 6+5+4: is another one.
Let V be the vector space of all 3 × 3 real matrices, and let M be the set of all 3 × 3
magic squares with real entries.

(a) Show that M is a subspace of V .

(b) Find the dimension and a basis of M.

1. (a) Y (b) N (c) Y (d) N (e) Y (f) Y (g) Y (h) N (i) Y (j) Y (k) N

2. (a) dim = 3. Basis: .
(b) dim = 1. Basis: . Yes, .
(c) dim = 1. Basis: . No.
(d) dim = 2. Basis: . No.
(e) dim = 3. Basis: . Yes, .

3. (a) Verify by yourself. I'll skip here.

(b) dim(M) = 3. The following matrices form a basis of M: Remark 1: Of course you may have different choices. That's just fine, as long as your
basis consists of 3 different magic squares and they span M.

Remark 2: Solving this problem allows us to generate all 3 × 3 magic squares. For the
example matrices given in the problem we have the following: 