Question 1 Find the dimension of each of the following vector spaces: (a) The vector space of all diagonal 𝑛 × 𝑛 n×n matrices. (b) The vector space of all symmetric 𝑛 × 𝑛 n×n matrices. (c) The vector space of all upper triangular 𝑛 × 𝑛 n×n matrices. Answer: (a) For diagonal matrices, only the diagonal entries can be nonzero. There are 𝑛 n diagonal entries, so the dimension is: Dimension = 𝑛 . Dimension=n. (b) For symmetric matrices, the elements satisfy 𝑎 𝑖 𝑗 = 𝑎 𝑗 𝑖 a ij =a ji . This results in 𝑛 ( 𝑛 + 1 ) / 2 n(n+1)/2 independent entries. Thus: Dimension = 𝑛 ( 𝑛 + 1 ) 2 . Dimension= 2 n(n+1) . (c) For upper triangular matrices, all entries below the diagonal are zero. This results in 𝑛 ( 𝑛 + 1 ) / 2 n(n+1)/2 independent entries. Thus: Dimension = 𝑛 ( 𝑛 + 1 ) 2 . Dimension= 2 n(n+1) . Question 2 Find a basis for the span of the given vectors: [ 0 1 − 2 ] , [ 3 1 − 0 ] , [ 2 1 5 ] . 0 1 −2 , 3 1 −0 , 2 1 5 . Answer: Represent the vectors as columns in a matrix: 𝐴 = [ 0 3 2 1 1 1 − 2 0 5 ] . A= 0 1 −2 3 1 0 2 1 5 . Perform row reduction to row echelon form: [ 0 3 2 1 1 1 − 2 0 5 ] → [ 1 0 0 0 1 0 0 0 1 ] . 0 1 −2 3 1 0 2 1 5 → 1 0 0 0 1 0 0 0 1 . The basis consists of the linearly independent columns corresponding to pivot columns: Basis = { [ 1 0 0 ] , [ 0 1 0 ] , [ 0 0 1 ] } . Basis= ⎩ ⎨ ⎧ 1 0 0 , 0 1 0 , 0 0 1 ⎭ ⎬ ⎫ . Question 3 Write the vector 𝑣 v in the form 𝑎 𝑖 + 𝑏 𝑗 ai+bj, given its magnitude ∣ 𝑣 ∣ = 8 ∣v∣=8 and angle 𝛼 = 4 5 ∘ α=45 ∘ . Answer: The vector can be expressed as: 𝑣 = ∣ 𝑣 ∣ ( cos ⁡ 𝛼 𝑖 + sin ⁡ 𝛼 𝑗 ) . v=∣v∣(cosαi+sinαj). Substituting ∣ 𝑣 ∣ = 8 ∣v∣=8 and 𝛼 = 4 5 ∘ α=45 ∘ : cos ⁡ ( 4 5 ∘ ) = sin ⁡ ( 4 5 ∘ ) = 2 2 . cos(45 ∘ )=sin(45 ∘ )= 2 2 . 𝑣 = 8 ( 2 2 𝑖 + 2 2 𝑗 ) = 4 2 𝑖 + 4 2 𝑗 . v=8( 2 2 i+ 2 2 j)=4 2 i+4 2 j. Question 4 Find the distance between the points ( 3 , − 2 ) (3,−2) and ( 4 , 3 ) (4,3). Answer: The distance formula is: 𝑑 = ( 𝑥 2 − 𝑥 1 ) 2 + ( 𝑦 2 − 𝑦 1 ) 2 . d= (x 2 −x 1 ) 2 +(y 2 −y 1 ) 2 . Substitute 𝑥 1 = 3 x 1 =3, 𝑦 1 = − 2 y 1 =−2, 𝑥 2 = 4 x 2 =4, 𝑦 2 = 3 y 2 =3: 𝑑 = ( 4 − 3 ) 2 + ( 3 − ( − 2 ) ) 2 = 1 2 + 5 2 = 1 + 25 = 26 . d= (4−3) 2 +(3−(−2)) 2 = 1 2 +5 2 = 1+25 = 26 . Question 5 Determine whether the planes 𝑥 + 2 𝑦 − 𝑧 = 2 x+2y−z=2 and 2 𝑥 − 2 𝑦 + 𝑧 = 1 2x−2y+z=1 are parallel, perpendicular, or neither. If neither, find the angle between them. Answer: Normal vectors of the planes: 𝑛 1 = ⟨ 1 , 2 , − 1 ⟩ , 𝑛 2 = ⟨ 2 , − 2 , 1 ⟩ . n 1 =⟨1,2,−1⟩,n 2 =⟨2,−2,1⟩. The planes are: Not parallel if 𝑛 1 n 1 and 𝑛 2 n 2 are not scalar multiples. Not perpendicular if 𝑛 1 ⋅ 𝑛 2 ≠ 0 n 1 ⋅n 2  =0. Angle 𝜃 θ between the planes is given by: cos ⁡ 𝜃 = 𝑛 1 ⋅ 𝑛 2 ∣ 𝑛 1 ∣ ∣ 𝑛 2 ∣ . cosθ= ∣n 1 ∣∣n 2 ∣ n 1 ⋅n 2 . Compute 𝑛 1 ⋅ 𝑛 2 n 1 ⋅n 2 : 𝑛 1 ⋅ 𝑛 2 = 1 ( 2 ) + 2 ( − 2 ) + ( − 1 ) ( 1 ) = 2 − 4 − 1 = − 3. n 1 ⋅n 2 =1(2)+2(−2)+(−1)(1)=2−4−1=−3. Magnitudes: ∣ 𝑛 1 ∣ = 1 2 + 2 2 + ( − 1 ) 2 = 6 , ∣ 𝑛 2 ∣ = 2 2 + ( − 2 ) 2 + 1 2 = 9 = 3. ∣n 1 ∣= 1 2 +2 2 +(−1) 2 = 6 ,∣n 2 ∣= 2 2 +(−2) 2 +1 2 = 9 =3. cos ⁡ 𝜃 = − 3 6 ⋅ 3 = − 1 6 . cosθ= 6 ⋅3 −3 = 6 −1 . 𝜃 = cos ⁡ − 1 ( − 1 6 ) ≈ 114. 1 ∘ . θ=cos −1 (− 6 1 )≈114.1 ∘ . Result: The planes are neither parallel nor perpendicular. The angle between them is approximately 114.1°.