Matrices

A has 3 rows and 4 columns, so its order is 3 × 4 and it has 12 entries.

a₁₃=19, a₂₁=35, a₃₃=−5, a₂₄=12 and a₂₃=5/2.

An m × n matrix has mn entries, so list the ordered factor pairs.

Write every positive factor pair in row-count × column-count order.

Use i,j = 1,2: a₁₁=2, a₁₂=9/2, a₂₁=9/2, a₂₂=8.

Substitute i,j = 1,2 into i/j.

For i,j = 1,2 the entries are 9/2, 25/2, 8 and 18, in row order.

Substitute i=1,2,3 and j=1,2,3,4 into the rule, one row at a time.

For each row, substitute its row number i and let j run from 1 to 4.

4=y, so y=4 3=z, so z=3 x=1

x+y=6 5+z=5, so z=0 xy=8 (t−2)(t−4)=t²−6t+8=0 So the pair (x,y) is (2,4) or (4,2) Check both: sum 6 and product 8

Equate the entries: x+y+z=9, x+z=5, y+z=7. Solving gives x=2, y=4, z=3.

a−b=−1 2a+c=5 2a−b=0 3c+d=13 (a−b)−(2a−b)=−1−0, so −a=−1 and a=1 2a−b=0 gives 2−b=0, so b=2 2a+c=5 gives 2+c=5, so c=3 3c+d=13 gives 9+d=13, so d=4

A square matrix has the same number of rows and columns.

3x+7=0 gives x=−7/3 2−3x=4 gives x=−2/3 −7/3 ≠ −2/3 Therefore the matrices cannot be equal; choose (B).

There are 9 positions and 2 choices for each: 2⁹ = 512.

A+B=[[4,1,−1],[9,2,7],[3,−1,4]] B−C=[[-1,−2,0],[4,−1,3],[1,2,0]] A+(B−C)=[[0,0,−3],[9,−1,5],[2,1,1]] (A+B)−C=[[0,0,−3],[9,−1,5],[2,1,1]]

3A=[[2,3,5],[1,2,4],[7,6,2]] 5B=[[2,3,5],[1,2,4],[7,6,2]] Therefore 3A−5B is the 3×3 zero matrix.

The top-left and bottom-right entries become cos²θ+sin²θ=1; the off-diagonal entries cancel to 0.

A+B = [[2+1, 4+3], [3+(−2), 2+5]] = [[3, 7], [1, 7]]

A−B = [[2−1, 4−3], [3−(−2), 2−5]] = [[1, 1], [5, −3]]

3A = [[6, 12], [9, 6]] 3A−C = [[6−(−2), 12−5], [9−3, 6−4]] = [[8, 7], [6, 2]]

Top-left: 2·1 + 4·(−2) = −6 Top-right: 2·3 + 4·5 = 26 Bottom-left: 3·1 + 2·(−2) = −1 Bottom-right: 3·3 + 2·5 = 19 AB = [[−6, 26], [−1, 19]]

Top-left: 1·2 + 3·3 = 11 Top-right: 1·4 + 3·2 = 10 Bottom-left: (−2)·2 + 5·3 = 11 Bottom-right: (−2)·4 + 5·2 = 2 BA = [[11, 10], [11, 2]]

Top row: a+a=2a and b+b=2b Bottom row: b+(−b)=0 and a+a=2a Sum = [[2a, 2b], [0, 2a]]

a²+b²+2ab=(a+b)² b²+c²+2bc=(b+c)² a²+c²−2ac=(a−c)² a²+b²−2ab=(a−b)²

Row 1: −1+12=11, 4+7=11, −6+6=0 Row 2: 8+8=16, 5+0=5, 16+5=21 Row 3: 2+3=5, 8+2=10, 5+4=9

Each diagonal entry: cos²x+sin²x=1 Each off-diagonal entry: sin²x+cos²x=1 The sum is the 2 × 2 matrix of all ones.

(1,1): a²+b² (1,2): −ab+ab=0 (2,1): −ab+ab=0 (2,2): b²+a² Product = (a²+b²)I₂

Row 1: 1·[2,3,4]=[2,3,4] Row 2: 2·[2,3,4]=[4,6,8] Row 3: 3·[2,3,4]=[6,9,12]

First row: 1·1+(−2)·2=−3; 1·2+(−2)·3=−4; 1·3+(−2)·1=1 Second row: 2·1+3·2=8; 2·2+3·3=13; 2·3+3·1=9

Row 1 gives: 2+0+12=14; −6+6+0=0; 10+12+20=42 Row 2 gives: 3+0+15=18; −9+8+0=−1; 15+16+25=56 Row 3 gives: 4+0+18=22; −12+10+0=−2; 20+20+30=70

Row 1: 2−1=1; 0+2=2; 2+1=3 Row 2: 3−2=1; 0+4=4; 3+2=5 Row 3: −1−1=−2; 0+2=2; −1+1=0

Top-left: 3·2+(−1)·1+3·3=14 Top-right: 3·(−3)+(−1)·0+3·1=−6 Bottom-left: (−1)·2+0·1+2·3=4 Bottom-right: (−1)·(−3)+0·0+2·1=5

2X=[[7+3, 0+0], [2+0, 5+3]]=[[10,0],[2,8]] 2Y=[[7−3, 0−0], [2−0, 5−3]]=[[4,0],[2,2]] X=[[5,0],[1,4]] Y=[[2,0],[1,1]]

3(2X+3Y)=6X+9Y=[[6,9],[12,0]] 2(3X+2Y)=6X+4Y=[[4,−4],[−2,10]] 5Y=[[2,13],[14,−10]] Y=[[2/5,13/5],[14/5,−2]] 3(3X+2Y)=9X+6Y=[[6,−6],[−3,15]] 2(2X+3Y)=4X+6Y=[[4,6],[8,0]] 5X=[[2,−12],[−11,15]] X=[[2/5,−12/5],[−11/5,3]]

2X=[[1,0],[−3,2]]−[[3,2],[1,4]] 2X=[[-2,−2],[−4,−2]] X=[[-1,−1],[−2,−1]]

2[[1,3],[0,x]]=[[2,6],[0,2x]] Sum = [[2+y,6],[1,2x+2]] 2+y=5, so y=3 2x+2=8, so 2x=6 and x=3 The other entries already agree: 6=6 and 1=1

3[[1,−1],[0,2]]=[[3,−3],[0,6]] 3[[3,5],[4,6]]=[[9,15],[12,18]] 2[[x,z],[y,t]]=[[9−3,15−(−3)],[12−0,18−6]]=[[6,18],[12,12]] [[x,z],[y,t]]=[[3,9],[6,6]] Read the positions: x=3, z=9, y=6, t=6

Top entry: 2x−y=10 Bottom entry: 3x+y=5 (2x−y)+(3x+y)=10+5, so 5x=15 and x=3 2(3)−y=10 gives 6−y=10, so y=−4 Check bottom entry: 3(3)+(−4)=5

Left side = [[3x,3y],[3z,3w]] Right side = [[x+4,6+x+y],[z+w−1,2w+3]] 3x=x+4 gives x=2 3y=6+x+y gives 2y=8, so y=4 3w=2w+3 gives w=3 3z=z+w−1 gives 2z=2, so z=1

(1,1): cos x cos y−sin x sin y=cos(x+y) (1,2): −cos x sin y−sin x cos y=−sin(x+y) (2,1): sin x cos y+cos x sin y=sin(x+y) (2,2): −sin x sin y+cos x cos y=cos(x+y) The third row and column leave the final diagonal entry 1 and all other third-row/column entries 0.

AB=[[5·2+1·3, 5·1+1·4],[6·2+7·3, 6·1+7·4]] AB=[[13,9],[33,34]] BA=[[2·5+1·6, 2·1+1·7],[3·5+4·6, 3·1+4·7]] BA=[[16,9],[39,31]] The top-left entries are 13 and 16, so AB≠BA.

AB=[[7,12,14],[0,1,1],[1,2,1]] BA=[[1,3,3],[1,3,1],[6,11,6]] Since 7≠1, AB≠BA.

A²=[[5,−1,2],[9,−2,5],[0,−1,−2]] −5A=[[-10,0,−5],[−10,−5,−15],[−5,5,0]] 6I=[[6,0,0],[0,6,0],[0,0,6]] A²−5A+6I=[[1,−1,−3],[−1,−1,−10],[−5,4,4]]

A²=[[5,0,8],[2,4,5],[8,0,13]] A³=[[21,0,34],[12,8,23],[34,0,55]] −6A²=[[-30,0,−48],[−12,−24,−30],[−48,0,−78]] 7A=[[7,0,14],[0,14,7],[14,0,21]] 2I=[[2,0,0],[0,2,0],[0,0,2]] Adding these to A³ gives [[0,0,0],[0,0,0],[0,0,0]]=O₃

A²=[[3·3+(−2)·4, 3·(−2)+(−2)·(−2)],[4·3+(−2)·4, 4·(−2)+(−2)·(−2)]] A²=[[1,−2],[4,−4]] A−2I=[[3−2,−2],[4,−2−2]]=[[1,−2],[4,−4]]=A² Therefore k=1.

I+A=[[1,−t],[t,1]] I−A=[[1,t],[−t,1]] Product=[[cos α+t sin α, −sin α+t cos α],[sin α−t cos α, t sin α+cos α]] cos α+t sin α=1 sin α−t cos α=t Thus the product is [[1,−t],[t,1]]=I+A.

Use the matrix system [[1,1],[5,7]][x,y]ᵀ=[30,000,100I]ᵀ. For ₹1,800 the amounts are ₹15,000 and ₹15,000; for ₹2,000 they are ₹5,000 and ₹25,000.

10×12=120; 8×12=96; 10×12=120 [120, 96, 120]·[80, 60, 40]ᵀ =120·80+96·60+120·40 =9,600+5,760+4,800=₹20,160

P(p×k)Y(3×k) requires k=3; then PY has order p×3 W(n×3)Y(3×k) is defined and WY has order n×k Required restrictions: k=3 and p=n Choose (A).

7X has order 2×n 5Z has order 2×p=2×n Therefore 7X−5Z has order 2×n.

The 3×1 column becomes a 1×3 row.

Turn each row into the matching column.

Rows become columns, keeping each entry in its new position.

Both sides equal [[-5, 6, -1], [3, 9, 4], [-2, 9, 2]].

Both sides equal [[3, 4, -3], [1, 5, -2], [8, 9, 0]].

First recover A=[[3, -1, 0], [4, 2, 1]]. Then both sides equal [[2, 5], [1, 4], [1, 4]].

First recover A=[[3, -1, 0], [4, 2, 1]]. Then both sides equal [[4, 3], [-3, 0], [-1, -2]].

Recover A by transposing Aᵀ, then add 2B and transpose.

Multiply A by B, transpose the result, then multiply Bᵀ by Aᵀ; both products match.

Multiply A by B, transpose the result, then multiply Bᵀ by Aᵀ; both products match.

Transpose A and multiply. The diagonal entries become sin²α+cos²α=1, while the off-diagonal entries cancel.

Transpose A and multiply. The diagonal entries become sin²α+cos²α=1, while the off-diagonal entries cancel.

The entries mirror across the main diagonal, so transposing leaves A unchanged.

After transposing, every off-diagonal entry changes sign and diagonal entries remain zero.

First Aᵀ=[[1, 6], [5, 7]]. Then A + Aᵀ=[[2, 11], [11, 14]]. Transposing gives [[2, 11], [11, 14]], confirming it is symmetric.

First Aᵀ=[[1, 6], [5, 7]]. Then A − Aᵀ=[[0, -1], [1, 0]]. Transposing gives [[0, 1], [-1, 0]], confirming it is skew-symmetric.

This matrix is skew-symmetric, so Aᵀ=−A. The sum cancels, while the difference is 2A.

[[3,5],[1,−1]]ᵀ=[[3,1],[5,−1]]

Aᵀ=A because the entries mirror across the diagonal.

Use A=½(A+Aᵀ)+½(A−Aᵀ), keeping exact halves.

Aᵀ=[[1,−1],[5,2]].

(AB−BA)ᵀ=BᵀAᵀ−AᵀBᵀ=BA−AB=−(AB−BA).

Adding A to its transpose gives 2cosα on both diagonal entries and zeros elsewhere. Set 2cosα=1.

An inverse must multiply with the original matrix to give the identity in both orders.

Transpose the difference and use Aᵀ=A and Bᵀ=B.

Transpose the product; its order reverses, then use the given property of A.

In AᵀA, diagonal entries are the squared lengths of A’s columns; the off-diagonal entries cancel.

Multiply the row by the 3×3 matrix first to get [6,2,4], then use it in the final product.

Square A, then subtract 5A and add 7 to the diagonal entries.

The row times the matrix is [x−2, −10, 2x−8]; multiplying by the final column gives x²−48=0.

Multiply each market’s sales row by the price or profit-per-unit column.

Let X be 2×2 and match entries in the product with the given matrix.

The diagonal entries of A² are α²+βγ. Setting them equal to 1 gives option C.

Symmetry gives Aᵀ=A, while skew-symmetry gives Aᵀ=−A. Therefore A=−A and every entry is zero.

Expand using A²=A and A³=A; then subtract 7A.