Question

4. (10+10pts.) Consider the homogeneous system 21 +22+ (3 - 2a).x3 = 0 2x1 + 12 + 7.03 - 14 = 0 -22 + 20.73 +2.04 = 0 21 +22 + 4x3 = 0 where a is a real constant. a. Find the value of a for which the dimension of the solution space of the system is 1. b. Find a basis of the solution space of the system for the value of a found in part (a).

Answer 1

The given system of equation is :

r1 +82 + (3 - 2a).r3 = 0 2.01 + r2 + 773 - 74 = 0 - 22 + 2a.r3 +2r4 = 0 21+x2 + 4.3 = 0

a) We have to find the value of $\small \alpha$ for which the dimension of solution space is 1 .

Now dimension of solution space of above system is equal to ( 4 - rank of coefficient matrix ) .

So we need to check for which value of $\small \alpha$ , rank of coefficienct matrix is 3 .

So For that we have to bring down the coefficient matrix in row - echelon form and then see for what value of $\small \alpha$ , the coefficient matrix has rank 3 .

So the coefficient matrix is given as :

1 (3 - 2a) 1 7 1 2 0 -1 1 1 0 -1 2 0 2a 4

We have to now reduce this matrix in row echelon form .

For that we perform elementary row operations .

R2 + R2 – 2R1 R4 → R4 R4 - R1

1 0 -1 0 -1 0 0 (3 - 20) 1 + lar 2. 1 +2ar 0 -1 2 0

R3 R3 - R2

$\small \begin{bmatrix} 1 & 1 & (3-2 \alpha ) & 0 \\ 0 & -1 & 1+4\alpha &-1 \\ 0 & 0 & -1-2\alpha & 3 \\ 0 &0 & 1+2\alpha & 0 \end{bmatrix}$

$\small R_4 \rightarrow R_4+R_3$

$\small \begin{bmatrix} 1 & 1 & (3-2 \alpha ) & 0 \\ 0 & -1 & 1+4\alpha &-1 \\ 0 & 0 & -1-2\alpha & 3 \\ 0 &0 & 0 & 3 \end{bmatrix}$

Now as we can see , there are elements in each pivot position . But in 3rd pivot ( 3rd row , 3rd column ) , we can

see that the element  $\small -1-2 \alpha$ can be made 0 . And if this element is made zero , we will have only 3 pivot elements and the rank of the matrix will reduce to 3 . And therefore then the corresponding solution space will have dimension 1 .

Now $\small -1-2 \alpha = 0 \implies \alpha = \frac{-1}{2}$

So we have for  $\small \alpha = \frac{-1}{2}$ , the coefficient matrix becomes

$\small \begin{bmatrix} 1 & 1 & (3-2 (\frac{-1}{2} )) & 0 \\ 0 & -1 & 1+4(\frac{-1}{2} ) &-1 \\ 0 & 0 & -1-2(\frac{-1}{2} ) & 3 \\ 0 &0 & 0 & 3 \end{bmatrix}=\begin{bmatrix} 1 & 1 & 4& 0 \\ 0 & -1 & -1 &-1 \\ 0 & 0 & 0 & 3 \\ 0 &0 & 0 & 3 \end{bmatrix}$

And then performing row operation   $\small R_4 \rightarrow R_4-R_3$

$\small \begin{bmatrix} 1 & 1 & 4& 0 \\ 0 & -1 & -1 &-1 \\ 0 & 0 & 0 & 3 \\ 0 &0 & 0 & 0 \end{bmatrix}$

So this matrix has 1 -zero row and other 3 rows are linearly independent (due to presence of pivot elements ) .

So we conclude for   $\small \alpha = \frac{-1}{2}$ , the coefficient matrix has rank 3 .

And therefore for  $\small \alpha = \frac{-1}{2}$ , the dimension of solution space becomes 1.

b)

Now we have to find basis of solution space .

For that we first find the solution using the reduced row echelon form of coefficient matrix above .

So for   $\small \alpha = \frac{-1}{2}$ in matrix form the system of given equations becomes .

$\small \begin{bmatrix} 1 & 1 & 4& 0 \\ 0 & -1 & -1 &-1 \\ 0 & 0 & 0 & 3 \\ 0 &0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2 \\x_3 \\ x_4 \end{bmatrix}=\begin{bmatrix} 0\\0 \\ 0 \\ 0 \end{bmatrix}$

writing in equation form .

$\small .\\ x_1+x_2+4x_3 =0\\ -x_2-x_3-x_4=0 \\ 3x_4=0$

From 3rd equation we get , $\small 3x_4=0 \implies x_4=0$

So putting it in second equation we get ,

$\small -x_2-x_3-0=0 \implies x_2=-x_3$

And finally using first equation , we get

$\small x_1+(-x_3)+4x_3=0 \implies x_1+3x_3=0 \implies x_1=-3x_3$

So we get solution as

$\small x_1=-3x_3 \ \ , \ \ x_2=-x_3 \ \ , \ \ x_3=x_3 \ \ , \ \ x_4=0$

Solution space is :

$\small (x_1,x_2,x_3,x_4)=(-3x_3,-x_3,x_3,0)=(-3,-1,1,0)x_3$

So basis of the solution space is  $\small \{(-3,-1,1,0) \}$