Linear-1 1

1a

Solve: 4x+6=0∣Z1111 is prime ⟹Z11 is a field4x+6=0∣+(−6)(−6)≡5(4x+6)+(−6)=54x+(6+(−6)=54x+0=54x=5∣∗(4−1)4−1≡34−1(4x)=5∗3(4−1∗4)x=4(1)x=4x=4

1b

Solve: x2=1∣Z8Let us try all possible values of x in Z802=012=122=432=142=052=162=472=1⟹[x=1x=3x=5x=7  over Z8

1c

Given: F is a field where 2 is an invertible (2≠0F)Prove: x2=1 over F has exactly two different solutionsx2=1x2−1=0(x+1)(x−1)=0[x=1x=−1{2≡0F⟹2+(−1)=0+(−1)⟹1=−1⟹x2=1 has exactly one distinct solution2≢0F⟹2+(−1)≠0+(−1)⟹1≠−1⟹x2=1 has exactly two distinct solutionsIf F=Z2, then:2≡0Z2⟹x2=1 has exactly one distinct solution

2

Prove: ∀a∈F:−(−a)=aProved in the first lecture: ∀a,b,c∈F:{a+b=0a+c=0⟹b=cb=−a{a+b=0b+−b=0⟹a=−b−b=−(−a)⟹a=−(−a)

3a

Solve: {3y+6z=−92x+4y+7z=−22x+5y+10z=−21 over R(036−9247−221510−21)→R3↔R1(1510−21247−22036−9)→R2=R2−2R1(1510−210−6−1320036−9)→R3=2R3+R2(1510−210−6−132000−12){x+5y+10z=−21−6y−13z=20−z=2−z=2⟹z=−2−6y−13z=−6y+26=20⟹−6y=−6⟹y=1x+5y+10z=x+5−20=−21⟹x=−6⟹{x=−6y=1z=−2

3b

Solve: {5x+3y+3z=07x+3y+7z=07x+9y=0 over Z11(533073707900)→R2=R2+4R1R1=9R1(155001507900)→R3=R3+4R1(155001500790)→R3=R3+4R2(155001500070){x+5y+5z=0y+5z=07z=07z=0⟹z=0y+5z=y=0x+5y+5z=x=0⟹{x=0y=0z=0

3c

Solve: {ix+2y+z+2w=3x+(1−2i)y+z=1−3i over C(i212311−2i101−3i)→R2=R2−R1R2=R2∗i(i21230i−1+i−2i)→R1=R1−2R2R2=R2∗(−i)(i0−1−2i2−4i1011+i2i1)→R1=R1∗(−i)(10−2+i−4−2i−i011+i2i1){x+0y+(−2+i)z+(−4−2i)w=−iy+(1+i)z+(2i)w=1⟹{x=(2−i)z+(4+2i)w−iy=(−1−i)z+(0−2i)w+1z∈C;w∈C

4a

Solve: {3y+6z=02x+4y+7z=0x+5y+10z=0 over RLet us use the transformations made in 3a:(0360247015100)→(1510?0−6−13?00−1?)The right side of the system is all zero over R, so basic transformations do not affect it:(151000−6−13000−10){x+5y+10z=0−6y−13z=0−z=0−z=0⟹z=0−6y−13z=−6y=0⟹y=0x+5y+10z=x=0⟹x=0{x=0y=0z=0

4b

Solve: {5x+3y+3z=57x+3y+7z=77x+9y=7 over Z11Let us use the transformations made in 3b:(533573777907)→(155?015?007?)Let us also note that the right side of the system is equal to the first column:(5..57..77..7)Equal matrices will stay equal after applying the same basic transformations⟹(577)→(100)⟹(533573777907)→(155101500070){x+5y+5z=1y+5z=07z=07z=0⟹z=0y+5z=y=0⟹y=0x+5y+5z=x⟹x=1{x=1y=0z=0

4c

Solve: {ix+2y+z+2w=0x+(1−2i)y+z=0 over CLet us use the transformations made in 3c:(i212011−2i100)→(10−2+i−4−2i?011+i2i?)The right side of the system is all zero over C, so basic transformations do not affect it:⟹(i212011−2i100)→(10−2+i−4−2i0011+i2i0){x+0y+(−2+i)z+(−4−2i)w=0y+(1+i)z+(2i)w=0⟹{x=(2−i)z+(4+2i)wy=(−1−i)z+(0−2i)wz∈C;w∈C

5

For the system of equations: {(10+2a)x+(−5+5a)y+5z=a(−5+a)y+5z=a(15+3a)x+(−10+8a)y+(3+a)z=−2+2a over RFor all values of parameter a define whether system has 0, 1 or ∞ solutions(10+2a−5+5a5a0−5+a5a15+3a−10+8a3+a−2+2a)→R3=2R3−3R1(10+2a−5+5a5a0−5+a5a0−5+a−9+2a−4+a)→R3=R3−R2(10+2a−5+5a5a0−5+a5a00−14+2a−4)→R1=12R1R1=R1−R2(5+a2a000−5+a5a00−14+2a−4){(5+a)x+(2a)y+0z=0(−5+a)y+5z=a(−14+2a)z=−4(−14+2a)z=−4⟹(−7+a)z=−2{∅:a=7z=−2−7+a:a≠7a≠7:(−5+a)y+5z=(−5+a)y+−10−7+a=a⟹(−5+a)y=a2−7a+10−7+a=(a−2)(a−5)a−7⟹{y∈R:a=5y=a−2a−7:a≠5{(5+a)x+(2a)y=10x+10y=0⟹x=−y:a=5(5+a)x+(2a)y=10y=0⟹y=0⟹a−2=0:a=5⟹∅:a=−5(5+a)x+2a(y)=0⟹(5+a)x=−2a(a−2)a−7⟹x=−2a(a−2)(a−7)(a+5):|a|≠5⟹{∅:a=7⟹0 solutions∅:a=−5⟹0 solutionsx=−y;z=1:a=5⟹∞ solutionsx=−2a(a−2)(a−7)(a+5);y=a−2a−7;z=−2a−7:a∉{−5,5,7}⟹1 solution

6i

For the system of equations: {x+y+az=1x+ay+z=1ax+y+z=1For all values of parameter a∈R define whether system has 0, 1 or ∞ solutions(11a11a11a111)→R3=R3−a∗R1R2=R2−R1(11a10a−11−a001−a1−a21−a)→R3=R3+R2(11a10a−11−a0002−a−a21−a){x+y+az=1(a−1)y+(1−a)z=0(1−a)(a+2)z=(1−a)(1−a)(a+2)z=(1−a)⟹{∅:a=−2z∈R:a=1z=1a+2:a≠1a≠−2:(a−1)y+(1−a)z=0⟹(a−1)(y−z)=0{y∈R:a=1y=z:a≠1x+y+az=1⟹{x+y+z=1⟹x=1−y−z:a=1x+a+1a+2=1⟹x=1a+2:a≠1⟹{∅:a=−2⟹0 solutionsx=1−y−z;y∈R;z∈R:a=1⟹∞ solutionsx=y=z=1a+2:a∉{−2,1}⟹1 solution

6ii

For the system of equations: {x+y+az=1x+ay+z=1ax+y+z=1For all values of parameter a∈Z3 define whether system has 0, 1 or ∞ solutions(11a11a11a111)→R3=R3+2aR1R2=R2+2R1(11a10a+21+2a001+2a1+2a21+2a)→R3=R3+R2(11a10a+21+2a0002(a2+a+1)1+2a){x+y+az=1(a+2)y+(1+2a)z=02(a2+a+1)z=1+2aLet a=0:{2z=1⟹z=22y+z=0⟹2y+2=0⟹y=−1Z3=2x+y=1⟹x+2=1⟹x=−1Z3=2Let a=1:{2(0)z=0⟹0=0⟹z∈Z30y+0z=0⟹0=0⟹y∈Z3x+y+z=1⟹x=1+(−y)+(−z)As there are 32=9 combinations of y and z in Z3, there are 9 solutions in this caseLet a=2:{2(1)z=2⟹2z=2⟹z=1y+2z=0⟹y=1x+y+2z=1⟹x+1+2=1⟹x=1⟹{x=y=z=1:a=0⟹1 solutionx=y=z=2:a=2⟹1 solutionx=1+(−y)+(−z);y∈Z3;z∈Z3:a=1⟹9 solutions

7a

Determine if the following matrices are row equivalent: A=(123456789),B=(10211−232−2)A≡(123456789)→R3=R3−7R1R2=R2−4R1(1230−3−60−6−12)→R2=−13R2R3=R3−2R2(123012000)→R1=R1−2R2(10−1012000)B≡(10211−232−2)→R3=R3−3R1R2=R2−R1(10201−402−8)→R3=R3−2R2(10201−4000)Both matrices are canonical, but the third column of the matrices is not equal⟹matrices A,B are not row equivalent

7b

Determine if the following matrices are row equivalent: A=(101222345),B=(1234567810)A≡(101222345)→R3=R3−3R1R2=12R2−R1(101010042)→R1=R1−R3R3=12R3−2R2(100010001)B≡(1234567810)→R3=R3−7R1R2=R2−4R1(1230−3−60−6−11)→R3=R3+6R2R2=−13R2(123012001)→R1=R1−2R2−3R3R2=R2−2R3(100010001)Both matrices are canonical and equal to I3⟹matrices A,B are row equivalent