Question #161493

Solve by finding the basis over R for the solution space.

(A) X + 3y -3z=0

2x - 3y + z=0

3x -2y + 2z=0


(B) X + Y + Z + W=0

2X + 3Y - Z +W=0

3X + 4Y +2W=0



1
Expert's answer
2021-02-24T11:28:01-0500

(133231322)(xyz)=(000)\begin{pmatrix} 1 & 3 & -3 \\ 2 & -3 & 1 \\ 3 & -2 & 2 \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}


Performing row operations on the matrix, we have


(133231322)R31(3)R21(2)(13309701111)\begin{pmatrix} 1 & 3 & -3\\ 2&-3&1\\ 3&-2&2 \end{pmatrix} \to^{R_{21}(-2)}_{R_{31}(-3)} \begin{pmatrix} 1&3&-3\\ 0&-9&7\\ 0&-11&11 \end{pmatrix}


R2(19)(133017901111)R32(11)(133017900229)\to R_2(\frac{-1}{9}) \begin{pmatrix} 1&3&-3\\ 0&1&\frac{-7}{9}\\ 0&-11&11 \end{pmatrix} \to^{R_{32}(11)} \begin{pmatrix} 1&3&-3\\ 0&1&\frac{-7}{9}\\ 0&0&\frac{22}{9} \end{pmatrix}


R3(922)(1330179001)\to R_3(\frac{9}{22}) \begin{pmatrix} 1&3&-3\\ 0&1&\frac{-7}{9}\\ 0&0&1 \end{pmatrix}


x+3y3z=0y79z=0z=0x+3y-3z=0\\ y-\frac{7}{9}z = 0\\ z=0


Therefore x=0,y=0, z=0. This is a trivial solution hence it is not a basis

Question B

(111123113402)(xyzw)=(000)\begin{pmatrix} 1&1&1&1\\ 2&3&-1&1\\ 3&4&0&2 \end{pmatrix} \begin{pmatrix} x\\y\\z\\w \end{pmatrix} = \begin{pmatrix} 0\\0 \\ 0 \end{pmatrix}

The matrix of the coefficient is

(111123113402)R31(3)R21(2)(111101310131)R32(1)(111101310000)\begin{pmatrix} 1&1&1&1\\ 2&3&-1&1\\ 3&4&0&2 \end{pmatrix} \to^{R_{21}(-2)}_{R_{31}(-3)} \begin{pmatrix} 1&1&1&1\\ 0&1&-3&-1\\ 0&1&-3&-1\end{pmatrix} \to^{R_{32}(-1)} \begin{pmatrix} 1&1&1&1\\ 0&1&-3&-1\\ 0&0&0&0\end{pmatrix}


x+y+z+w=0y3zw=0x+y+z+w=0\\ y-3z-w=0

Let z=a,w=bz=a, w=b where a,bRa,b \in \mathbb{R}

So, we have

y3ab=0y=3a+bx=yzwx=(3a+b)abx=4a2bx=4a2b,y=3a+b,z=a,w=b(x,y,z,w)=a(4,3,1,0)+b(2,1,0,1)y-3a-b=0 \\ y = 3a+b \\ x = -y-z-w \\ x= -(3a+b) -a-b \\ x= -4a-2b\\ \therefore x = -4a-2b, y = 3a+b , z =a, w=b \\ (x,y,z,w) = a(-4,3,1,0)+b(-2,1,0,1)

So a basis for the system is {(-4,3,1,0),(-2,1,0,1)}


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