Linear Algebra Answers

1. What conditions must be satisfied by b1, b2, b3, b4, and b5 for the over determined linear system to be consistent?


x1 - 4x2 = b1

x1 - 3x2 = b2

x1 + x2 = b3

x1 - 5x2 = b4

x1 + 6x2 = b5


2. Find the rank and nullity of the matrix; then verify that the values obtained satisfy Formula (4) in the Dimension Theorem.


A = 3 -2 -5 7 -29

-2 0 2 -6 14

4 7 3 19 0


a. Rank (A) = ?

b. nullity (A) = ?

c. rank (A) + nullity (A) = ?


3. For purposes of this problem, let us define a "checkerboard matrix" to be a square matrix Upper A equals left-bracket a Subscript i j Baseline right-bracket such that


a Subscript i j Baseline equals Start Layout left-brace 1st Row 1st Column 1 if i plus j is even 2nd Row 1st Column 0 if i plus j is odd End Layout


Find the rank and nullity of the following checkerboard matrix.


The 9 times 9 checkerboard matrix.


a. rank (A) =

b. nullity (A) =

1. 1 3 6 0

A = -2 4 6 1

-3 1 0 1


(a) Find an equation relating nullity ⁢ left-parenthesis Upper A right-parenthesis and ⁢ nullity left-parenthesis Upper A Superscript Upper T Baseline right-parenthesis for the matrix A.


(b) Find an equation relating nullity ⁢ left-parenthesis Upper A right-parenthesis and ⁢ nullity left-parenthesis Upper A Superscript Upper T Baseline right-parenthesis for a general m times n matrix.


2. Are there values of r and s for which



Close and open Parenthesis 1 0 0

0 r-6 6

0 s-5 r+6

0 0 7


has rank 1 or 2? If so, find those values.

The matrix has ____ for


r = ?

s = ?

1. Find the largest possible value for the rank of A and the smallest possible value for the nullity of A.

A is 8 × 8


a. The largest possible value for the rank of A is ?

b. The smallest possible value for the nullity of A is ?


2. Find the largest possible value for the rank of A and the smallest possible value for the nullity of A.

A is 3 × 8


a. The largest possible value for the rank of A is ?

b. The smallest possible value for the nullity of A is ?

1. Find a basis for the subspace of R4 spanned by the given vectors.


(1,1,-5,-6), (2,0,2,-2), (3,-1,0,8).


2. Find a subset of the vectors that forms a basis for the space spanned by the vectors, then express each vector that is not in the basis as a linear combination of the basis vectors.


v1=(1,0,1,1), v2 = (-4,4,-1,4), v3 = (2,4,5,10), v4 = (-10,4,-7,-2)


3. Find a subset of the vectors that forms a basis for the space spanned by the vectors, then express each vector that is not in the basis as a linear combination of the basis vectors.


v1=(1,-1,5,2), v2 = (-2,3,1,0), v3 = (5,-6,14,6), v4 = (0,4,2,-3), v5 = (3,15,45,2)


4. Determine whether b is in the column space of A, and if so, express b as a linear combination of the column vectors of A.


A = 3 4 B = -1

6 -9 15

1. Find the vector form of the general solution of the given linear system Ax = b ; then use that result to find the vector form of the general solution of Ax = 0.


x subscript 1 + x subscript 2 + 2x subscript 3 = 6

x subscript 1 + x subscript 3 = -2

2x subscript 1 + x subscript 2 + 3x subscript = 4


2. Find a basis for the null space and row space of A.

1 -1 3

A = 5 -4 -2

7 -6 4


3. A matrix in row echelon form is given. By inspection, find bases for the row and column spaces of the matrix A.


1 0 2

A = 0 0 1

0 0 0

1. Express the product Ax as a linear combination of the column vectors of A.

Close and open parenthesis 4 1 Close and open Parenthesis 2

-2 3 3


2. Suppose that the solution set of the homogeneous system Ax = 0 is given by the formulas


x1 = − 5r + 6s, x2 = r − 3s, x3 = r, x4 = s


Find a vector form of the general solution of Ax = 0.


3. Suppose that x1 = − 3, x2 = 4, x3 = 1, x4 = − 2 is a solution of a non homogeneous linear system Ax = b, and that the solution set of the homogeneous system Ax = 0 is given by the formulas x1 = − 5r + 6s, x2 = r − 4s, x3 = r, x4 = s. Find a vector form of the general solution of Ax = b.

1. 1. Find a basis for the subspace of R Superscript 3 that is spanned by the vectors


v1 = (1, 0, 0), v2 = (1, 1, 0), v3 = (3, 1, 0), v4 = (0, -2, 0). Choose the correct letter.


A. v1 and v2 form a basis for span {v1, v2, v3, v4}.

B . v2 and v3 form a basis for span {v1, v2, v3, v4}.

C. v3 and v4 form a basis for span {v1, v2, v3, v4}.

D. v1 and v3 form a basis for span {v1, v2, v3, v4}.

E. v2 and v4 form a basis for span {v1, v2, v3, v4}.

F. v1 and v4 form a basis for span {v1, v2, v3, v4}.

1. Let Upper T Subscript Upper A Baseline colon Upper R Superscript 3 Baseline right-arrow Upper R Superscript 3 be multiplication by A and find the dimension of the subspace of Upper R Superscript 3 consisting of all vectors x for which Upper T Subscript Upper A Baseline left-parenthesis x right-parenthesis equals 0.


a. A= 1 1 0

1 0 3

1 0 3

The dimension is????


b. A = 1 4 0

1 4 0

1 4 0

The dimension is???


c. A = 3 0 0

-3 3 0

3 3 3

The dimension is????

1. Determine the dimension of and a basis for the solution space of the homogeneous linear system.


2x subscript 1 + x subscript 2 + 2x subscript 3 = 0

x subscript 1 + 6x subscript 3 = 0

x subscript 2 + x subscript 3 = 0


2. Find a basis for the given subspace of R Superscript 3, and state its dimension for the plane 2 x - 3 y + 5 z = 0.


3. Let v1 = (1,-1,3,-6) and v2 = (-3,4,-12,24).

Find standard basis vectors for R4 that can be added to the set {v1, v2} to produce a basis for R4.


A. v3 = (1,0,0,0) and v4 = (0,1,0,0)

B. v3 = (0,0,1,0) and v4 = (1,0,0,0)

C. v3 = (1,0,0,0) and v4 = (0,0,0,1)

D. v3 = (0,1,0,0) and v4 = (0,0,0,1)