Just Solving Questions

Q) Suppose L:R3to R3 is a linear operator and L[1,0,0]=[-2,1,0], L=[0,1,0]=[-3,-2,1], L[0,0,1]=[0,-1,3]  find basis Find L[-3,2,4]. give a formula for L[x,y,z] for all [x,y,z] belongs to R3 Given that L L L is a linear operator and L [ 1 , 0 , 0 ] = [ − 2 , 1 , 0 ] L[1, 0, 0] = [-2, 1, 0] L[1,0,0]=[−2,1,0], L [ 0 , 1 , 0 ] = [ − 3 , − 2 , 1 ] L[0, 1, 0] = [-3, -2, 1] L[0,1,0]=[−3,−2,1], and L [ 0 , 0 , 1 ] = [ 0 , − 1 , 3 ] L[0, 0, 1] = [0, -1, 3] L[0,0,1]=[0,−1,3], we can write: [ − 3 , 2 , 4 ] = − 3 [ 1 , 0 , 0 ] + 2 [ 0 , 1 , 0 ] + 4 [ 0 , 0 , 1 ] = − 3 L [ 1 , 0 , 0 ] + 2 L [ 0 , 1 , 0 ] + 4 L [ 0 , 0 , 1 ] \begin{align*} [-3, 2, 4] &= -3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1] \\ &= -3L[1, 0, 0] + 2L[0, 1, 0] + 4L[0, 0, 1] \end{align*} [−3,2,4]​=−3[1,0,0]+2[0,1,0]+4[0,0,1]=−3L[1,0,0]+2L[0,1,0]+4L[0,0,1]​ Using the given values for L [ 1 , 0 , 0 ] L[1, 0, 0] L[1,0,0], L [ 0 , 1 , 0 ] L[0, 1, 0] L[0,1,0], and L [ 0 , 0 , 1 ] L[0, 0, 1] L[0,0,1], we can compute: − 3 L [ 1 , 0 ...

Q) Suppose L:R3-> R3 is a linear operator  and L([1,0,0])=[-3,2,4], l=([0,1,0]) = [5,-1,3] and L([0,0,1])=[-4,0,-2]find L([6,2,-7]) and find L([x,y,z]) for any [x,y,z] belongs to R3. To find L ( [ 6 , 2 , − 7 ] ) L([6,2,-7]) L([6,2,−7]), we'll use linearity: Given: L ( [ 1 , 0 , 0 ] ) = [ − 3 , 2 , 4 ] L([1,0,0]) = [-3,2,4] L([1,0,0])=[−3,2,4] L ( [ 0 , 1 , 0 ] ) = [ 5 , − 1 , 3 ] L([0,1,0]) = [5,-1,3] L([0,1,0])=[5,−1,3] L ( [ 0 , 0 , 1 ] ) = [ − 4 , 0 , − 2 ] L([0,0,1]) = [-4,0,-2] L([0,0,1])=[−4,0,−2] We have [ 6 , 2 , − 7 ] = 6 [ 1 , 0 , 0 ] + 2 [ 0 , 1 , 0 ] − 7 [ 0 , 0 , 1 ] [6,2,-7] = 6[1,0,0] + 2[0,1,0] - 7[0,0,1] [6,2,−7]=6[1,0,0]+2[0,1,0]−7[0,0,1]. Since L L L is linear, we can apply it to each term separately and then add the results: L ( [ 6 , 2 , − 7 ] ) = 6 L ( [ 1 , 0 , 0 ] ) + 2 L ( [ 0 , 1 , 0 ] ) − 7 L ( [ 0 , 0 , 1 ] ) L([6,2,-7]) = 6L([1,0,0]) + 2L([0,1,0]) - 7L([0,0,1]) L([6,2,−7])=6L([1,0,0])+2L([0,1,0])−7L([0,0,1]) = 6 [ − 3 , 2 , 4 ] + 2 [ 5 , − 1 , 3 ] −...

Q) Use Gauss Jordan Method To Solve - x+2y+z=8; 2x+3y+2z=14; 3x+2y+2z=13 Ans.

Q) Prove that if (x+y).(x-y)=0 then ||x||=||y|| where x,y are vectors in |Rn Ans.  To prove that if ( 𝑥 + 𝑦 ) ⋅ ( 𝑥 − 𝑦 ) = 0 ( x + y ) ⋅ ( x − y ) = 0 , then ∥ 𝑥 ∥ = ∥ 𝑦 ∥ ∥ x ∥ = ∥ y ∥ where 𝑥 x and 𝑦 y are vectors in 𝑅 𝑛 R n , let's use the given dot product equation and properties of vectors: Given: ( 𝑥 + 𝑦 ) ⋅ ( 𝑥 − 𝑦 ) = 0 ( x + y ) ⋅ ( x − y ) = 0 Expanding the dot product: ( 𝑥 + 𝑦 ) ⋅ ( 𝑥 − 𝑦 ) = 𝑥 ⋅ 𝑥 − 𝑥 ⋅ 𝑦 + 𝑦 ⋅ 𝑥 − 𝑦 ⋅ 𝑦 ( x + y ) ⋅ ( x − y ) = x ⋅ x − x ⋅ y + y ⋅ x − y ⋅ y = ∥ 𝑥 ∥ 2 − ∥ 𝑦 ∥ 2 = ∥ x ∥ 2 − ∥ y ∥ 2 Given that ( 𝑥 + 𝑦 ) ⋅ ( 𝑥 − 𝑦 ) = 0 ( x + y ) ⋅ ( x − y ) = 0 , we have: ∥ 𝑥 ∥ 2 − ∥ 𝑦 ∥ 2 = 0 ∥ x ∥ 2 − ∥ y ∥ 2 = 0 This implies: ∥ 𝑥 ∥ 2 = ∥ 𝑦 ∥ 2 ∥ x ∥ 2 = ∥ y ∥ 2 Taking the square root of both sides (assuming ∥ 𝑥 ∥ ∥ x ∥ and ∥ 𝑦 ∥ ∥ y ∥ are non-negative): ∥ 𝑥 ∥ = ∥ 𝑦 ∥ ∥ x ∥ = ∥ y ∥ Thus, we've proven that if ( 𝑥 + 𝑦 ) ⋅ ( 𝑥 − 𝑦 ) = 0 ( x + y ) ⋅ ( x − y ) = 0 , then ∥ 𝑥 ∥ = ∥ 𝑦 ∥ ∥ x ∥ = ∥ y ∥ for...