Question

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) =

Answer 1

1.

$$\begin{pmatrix} 1 & -4& \ |& b_1 \\ 1 & -3 & \ |& b_2 \\ 1 & 1 & \ |& b_3 \\ 1 & -5 & \ |& b_4 \\ 1 & 6 & \ |& b_5 \end{pmatrix}$$

$R_2=R_2-R_1$

$$\begin{pmatrix} 1 & -4& \ |& b_1 \\ 0 & 1 & \ |& -b_1+b_2 \\ 1 & 1 & \ |& b_3 \\ 1 & -5 & \ |& b_4 \\ 1 & 6 & \ |& b_5 \end{pmatrix}$$

$R_3=R_3-R_1$

$$\begin{pmatrix} 1 & -4& \ |& b_1 \\ 0 & 1 & \ |& -b_1+b_2 \\ 0 & 5 & \ |& -b_1+b_3 \\ 1 & -5 & \ |& b_4 \\ 1 & 6 & \ |& b_5 \end{pmatrix}$$

$R_4=R_4-R_1$

$$\begin{pmatrix} 1 & -4& \ |& b_1 \\ 0 & 1 & \ |& -b_1+b_2 \\ 0 & 5 & \ |& -b_1+b_3 \\ 0 & -1& \ |& -b_1+b_4 \\ 1 & 6 & \ |& b_5 \end{pmatrix}$$

$R_5=R_5-R_1$

$$\begin{pmatrix} 1 & -4& \ |& b_1 \\ 0 & 1 & \ |& -b_1+b_2 \\ 0 & 5 & \ |& -b_1+b_3 \\ 0 & -1& \ |& -b_1+b_4 \\ 0 & 10 & \ |& -b_1+b_5 \end{pmatrix}$$

$R_1=R_1+(4)R_2$

$$\begin{pmatrix} 1 & 0 & \ |& -3b_1+4b_2 \\ 0 & 1 & \ |& -b_1+b_2 \\ 0 & 5 & \ |& -b_1+b_3 \\ 0 & -1& \ |& -b_1+b_4 \\ 0 & 10 & \ |& -b_1+b_5 \end{pmatrix}$$

$R_3=R_3-(5)R_2$

$$\begin{pmatrix} 1 & 0 & \ |& -3b_1+4b_2 \\ 0 & 1 & \ |& -b_1+b_2 \\ 0 & 0 & \ |& 4b_1-5b_2+b_3 \\ 0 & -1& \ |& -b_1+b_4 \\ 0 & 10 & \ |& -b_1+b_5 \end{pmatrix}$$

$R_4=R_4+R_2$

$$\begin{pmatrix} 1 & 0 & \ |& -3b_1+4b_2 \\ 0 & 1 & \ |& -b_1+b_2 \\ 0 & 0 & \ |& 4b_1-5b_2+b_3 \\ 0 & 0 & \ |& -2b_1+b_2+b_4 \\ 0 & 10 & \ |& -b_1+b_5 \end{pmatrix}$$

$R_5=R_5-(10)R_2$

$$\begin{pmatrix} 1 & 0 & \ |& -3b_1+4b_2 \\ 0 & 1 & \ |& -b_1+b_2 \\ 0 & 0 & \ |& 4b_1-5b_2+b_3 \\ 0 & 0 & \ |& -2b_1+b_2+b_4 \\ 0 & 0 & \ |& 9b_1-10b_2+b_5 \end{pmatrix}$$

$$4b_1-5b_2+b_3=0$$ $$-2b_1+b_2+b_4=0$$ $$9b_1-10b_2+b_5=0$$

Let $b_4=r, b_5=s.$ Then

$$4b_1-5b_2+b_3=0$$ $$-2b_1+b_2+r=0$$ $$9b_1-10b_2+s=0$$

$$b_3=-4b_1+5b_2$$ $$b_2=2b_1-r$$ $$-11b_1+10r+s=0$$

$$b_1={10 \over 11}r+{1 \over 11}s ,$$ $$b_2={9 \over 11}r+{2 \over 11}s ,$$ $$b_3={5 \over 11}r+{6 \over 11}s ,$$ $$b_4=r ,$$ $$b_5=s ,$$

$r,s\in \Bbb{R}$

2.

$$A=\begin{pmatrix} 3 & -2 & -5 & 7 & -29 \\ -2 & 0 & 2 & -6 & 14 \\ 4 & 7 & 3 & 19 & 0 \\ \end{pmatrix}$$

a)

$R_1=R_1/3$

$$\begin{pmatrix} 1 & -2/3 & -5/3 & 7/3 & -29/3 \\ -2 & 0 & 2 & -6 & 14 \\ 4 & 7 & 3 & 19 & 0 \\ \end{pmatrix}$$

$R_2=R_2+(2)R_1$

$$\begin{pmatrix} 1 & -2/3 & -5/3 & 7/3 & -29/3 \\ 0 & -4/3 & -4/3 & -4/3 & -16/3 \\ 4 & 7 & 3 & 19 & 0 \\ \end{pmatrix}$$

$R_3=R_3-(4)R_1$

$$\begin{pmatrix} 1 & -2/3 & -5/3 & 7/3 & -29/3 \\ 0 & -4/3 & -4/3 & -4/3 & -16/3 \\ 0 & 29/3 & 29/3 & 29/3 & 116/3 \\ \end{pmatrix}$$

$R_2=(-{3 \over 4})R_2$

$$\begin{pmatrix} 1 & -2/3 & -5/3 & 7/3 & -29/3 \\ 0 & 1 & 1 & 1 & 4 \\ 0 & 29/3 & 29/3 & 29/3 & 116/3 \\ \end{pmatrix}$$

$R_1=R_1+({2 \over 3})R_2$

$$\begin{pmatrix} 1 & 0 & -1 & 3 & -7 \\ 0 & 1 & 1 & 1 & 4 \\ 0 & 29/3 & 29/3 & 29/3 & 116/3 \\ \end{pmatrix}$$

$R_3=R_3-({29 \over 3})R_2$

$$\begin{pmatrix} 1 & 0 & -1 & 3 & -7 \\ 0 & 1 & 1 & 1 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}$$

The rank of a matrix is the number of nonzero rows in the reduced matrix, so the rank is 2.

$$rank(A)=2$$

b)

Solve the matrix equation

$$\begin{pmatrix} 1 & 0 & -1 & 3 & -7 \\ 0 & 1 & 1 & 1 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

If we take $x_3=t, x_4=s, x_5=u,$ then $x_1=t-3s+7u, x_2=-t-s-4u,$

$x_3=t,x_4=s,x_5=u.$

Therefore

$$X=\begin{pmatrix} t-3s+7u \\ -t-s-4u \\ t \\ s \\ u \end{pmatrix}=$$ $$= \begin{pmatrix} 1 \\ -1 \\ 1 \\ 0 \\ 0 \end{pmatrix}t+ \begin{pmatrix} -3 \\ -1 \\ 0 \\ 1 \\ 0 \end{pmatrix}s+ \begin{pmatrix} 7 \\ -4 \\ 0 \\ 0 \\ 1 \end{pmatrix}u$$

This is a null space.

Thus the nullity of the matrix A is 3.

$$nullity(A)=3$$

c.

$$rank(A)+nullity(A)=2+3=5=n$$

$A$ is a matrix with $n=5$ columns.

3.

$$A=\begin{pmatrix} 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \end{pmatrix}$$

$R_3=R_3-R_1$

$$\begin{pmatrix} 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \end{pmatrix}$$

$R_5=R_5-R_1$

$$\begin{pmatrix} 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \end{pmatrix}$$

$R_7=R_7-R_1$

$$\begin{pmatrix} 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \end{pmatrix}$$

$R_9=R_9-R_1$

$$\begin{pmatrix} 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$

$R_4=R_4-R_2$

$$\begin{pmatrix} 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$

$R_6=R_6-R_2$

$$\begin{pmatrix} 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$

$R_8=R_8-R_2$

$$\begin{pmatrix} 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$

a.

The rank of a matrix is the number of nonzero rows in the reduced matrix, so the rank is 2.

$rank(A)=2$

b.

$$rank(A)+nullity(A)=n$$

$n=9$

$$nullity(A)=9-2=7$$

$nullity(A)=7$