Determinants

For a 2 × 2 determinant, use ad − bc.

2(−1) − 4(−5) = −2 + 20 = 18.

cos²θ − (−sin θ)(sin θ) = cos²θ + sin²θ = 1.

(x² − x + 1)(x + 1) − (x − 1)(x + 1)

= (x³ + 1) − (x² − 1) = x³ − x² + 2.

|A| = 1·2 − 2·4 = −6.

2A = [[2, 4], [8, 4]], so |2A| = 2·4 − 4·8 = −24 = 4(−6).

A is upper triangular, so |A| = 1·1·4 = 4.

For a 3 × 3 matrix, |3A| = 3³|A| = 27·4 = 108.

Expand along row 2, which has two zeros.

The cofactor of the −1 in position (2,3) has positive sign: (−1)[−(3·(−5) − (−1)·3)] = −12.

Expand along row 1: 3(1·1 − (−2)·3) − (−4)(1·1 − (−2)·2) + 5(1·3 − 1·2).

= 3·7 + 4·5 + 5·1 = 46.

Expand along row 1: 0 − 1(0 − 6) + 2(−3 − 0).

= 6 − 6 = 0.

Expand along row 1: 2(0 − 5) − (−1)(0 + 3) + (−2)(0 − 6).

= −10 + 3 + 12 = 5.

Expand along row 1: 1(−9 + 12) − 1(−18 + 15) + (−2)(8 − 5).

= 3 − (−3) − 6 = 0.

2·1 − 4·5 = (2x)(x) − 4·6.

−18 = 2x² − 24, so x² = 3 and x = ±√3.

2·5 − 3·4 = 5x − 3(2x).

−2 = −x, so x = 2.

x² − 36 = 6·6 − 2·18 = 0.

Thus x² = 36, so x = 6 or x = −6.

Use the coordinate-area formula: area = ½|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)|.

Area = ½|1(0−3)+6(3−0)+4(0−0)| = ½|−3+18| = 15/2.

Area = ½|2(1−8)+1(8−7)+10(7−1)|.

= ½|−14+1+60| = 47/2.

Area = ½|−2(2−(−8))+3(−8−(−3))+(−1)(−3−2)|.

= ½|−20−15+5| = 15.

For A: x+y = a+(b+c)=a+b+c.

For B and C the coordinate sums are also a+b+c. Thus all three points lie on the same line x+y=a+b+c.

Use the base on the x-axis: area = ½|4−k|·2 = |4−k|.

|4−k|=4, so 4−k=4 or 4−k=−4. Hence k=0 or k=8.

The last two vertices lie on x=0, so their vertical separation is |k−4|. The horizontal height from (−2,0) is 2.

Area = ½·|k−4|·2 = |k−4| = 4, so k=0 or k=8.

A point (x,y) on the line makes the three points collinear: |1 2 1; 3 6 1; x y 1|=0.

Expanding gives 4x−2y=0, so y=2x.

A point (x,y) on the line satisfies |3 1 1; 9 3 1; x y 1|=0.

Expanding gives 2x−6y=0, hence x−3y=0.

The horizontal base has length |k−5| and the height is |4−(−6)|=10.

35 = ½·10·|k−5|, so |k−5|=7. Thus k=12 or k=−2.

For a 2 × 2 determinant, each minor is the remaining entry. Apply the cofactor signs +, −; −, +.

M = [[3, 0], [−4, 2]] and C = [[3, 0], [4, 2]].

Delete the row and column of each entry to get its minor.

Apply signs +, −; −, + to obtain the cofactors.

Each diagonal entry has a 2 × 2 identity minor with determinant 1.

Each off-diagonal minor has a zero row or column, so its determinant is 0; applying signs leaves the same identity matrix.

For each entry, delete its row and column and evaluate the remaining 2 × 2 determinant.

Apply the checkerboard cofactor signs + − +; − + −; + − +.

The second-row cofactors are C₂₁ = 7, C₂₂ = −7 and C₂₃ = −7.

Δ = 2(7) + 0(−7) + 1(−7) = 7.

Expand along column 3 using signs +, −, +.

Δ = x(y²−z²) − y(x²−z²) + z(x²−y²) = (x−y)(y−z)(z−x).

A determinant equals the sum of the entries in one column multiplied by their cofactors from that same column.

Option (D) is the expansion down column 1.

For a 2 × 2 matrix [[a,b],[c,d]], adj A = [[d,−b],[−c,a]].

So adj A = [[4,−2],[−3,1]].

The cofactor matrix is [[3,−12,6],[1,5,2],[−11,−1,5]].

The adjoint is its transpose: [[3,1,−11],[−12,5,−1],[6,2,5]].

|A| = 2(−6) − 3(−4) = 0.

adj A = [[−6,−3],[4,2]]. Multiplication in either order gives [[0,0],[0,0]] = 0I.

The cofactor matrix is [[0,−11,0],[3,1,−1],[2,8,3]].

Therefore adj A = [[0,3,2],[−11,1,8],[0,−1,3]]. Multiplying either way gives [[11,0,0],[0,11,0],[0,0,11]].

|A| = 2·3 − (−2)·4 = 14 ≠ 0.

adj A = [[3,2],[−4,2]], so A⁻¹ = (1/14)adj A.

|A| = (−1)·2 − 5(−3) = 13 ≠ 0.

adj A = [[2,−5],[3,−1]], hence A⁻¹ = (1/13)adj A.

A is upper triangular, so |A| = 1·2·5 = 10.

The inverse shown multiplies A to give I₃; since the determinant is non-zero, it exists.

The matrix is lower triangular, so |A| = 1·3·(−1) = −3.

Using A⁻¹=(1/|A|)adj A gives the stated result.

Expansion gives |A| = −3.

A⁻¹=(1/−3)adj A, with adj A = [[−1,5,3],[−4,23,12],[1,−11,−6]].

The determinant is 1.

Therefore A⁻¹ = adj A = [[−2,0,1],[9,2,−3],[6,1,−2]].

The lower-right block squared is [[cos²α+sin²α, 0], [0, cos²α+sin²α]] = I₂.

The first diagonal entry is 1 and the other entries in its row and column are zero, so A²=I₃. Hence A⁻¹=A.

AB = [[67,87],[47,61]], so (AB)⁻¹ = (−1/2)[[61,−87],[−47,67]].

A⁻¹=[[5,−7],[−2,3]] and B⁻¹=(−1/2)[[9,−8],[−7,6]]. Multiplying B⁻¹A⁻¹ gives the same matrix.

A²=[[8,5],[−5,3]], so A²−5A+7I=[[0,0],[0,0]].

Multiply the identity by A⁻¹: A−5I+7A⁻¹=O. Thus A⁻¹=(1/7)(5I−A)=(1/7)[[2,−1],[1,3]].

A²=[[11,8],[4,3]].

Comparing entries in A²+aA+bI=O gives 11+3a+b=0 and 8+2a=0. Thus a=−4 and b=1.

A²=[[4,2,1],[−3,8,−14],[7,−3,14]] and A³=[[8,7,1],[−23,27,−69],[32,−13,58]].

Substitution gives A³−6A²+5A+11I=O. Multiply by A⁻¹ and rearrange: A⁻¹=(1/11)(−A²+6A−5I).

A²=[[6,−5,5],[−5,6,−5],[5,−5,6]] and A³=[[22,−21,21],[−21,22,−21],[21,−21,22]].

Substitution gives the zero matrix. Multiplying by A⁻¹ gives A⁻¹=(1/4)(A²−6A+9I).

For an n × n matrix, |adj A| = |A|ⁿ⁻¹.

Here n=3, so |adj A|=|A|².

Taking determinants in AA⁻¹=I gives det(A)det(A⁻¹)=1.

Therefore det(A⁻¹)=1/det(A).

Subtract twice the first equation from the second: −y = −1, so y = 1.

Then x + 2 = 2, so x = 0.

Add the equations: 3x = 9, so x = 3.

Substitute into x + y = 4: y = 1.

The left side of the second equation is twice the first equation’s left side.

Twice the first equation would have right side 10, but the second has right side 8.

The coefficient determinant is a.

If a ≠ 0, the coefficient matrix is invertible, so there is a unique solution. If a = 0, the third equation becomes 0 = 4, which is impossible.

From 2y−z=−1, z=2y+1. From 3x−5y=3, 3x=3+5y.

Substitute into the first left side: (3+5y)−y−2(2y+1)=1, but the equation requires it to be 2.

Subtract the first equation from the third: y−2z=6.

Solving this with the first two equations gives z=−2, y=2 and x=3. Substitution checks all three equations.

A=[[5,2],[7,3]], B=[4,5]ᵀ and |A|=15−14=1.

A⁻¹=[[3,−2],[−7,5]], so [x,y]ᵀ=A⁻¹B=[2,−3]ᵀ.

A=[[2,−1],[3,4]], |A|=8+3=11.

A⁻¹=(1/11)[[4,1],[−3,2]], so [x,y]ᵀ=(1/11)[−5,12]ᵀ.

A=[[4,−3],[3,−5]], |A|=−20+9=−11.

A⁻¹=(−1/11)[[−5,3],[−3,4]], giving [x,y]ᵀ=[−6/11,−19/11]ᵀ.

A=[[5,2],[3,2]], |A|=10−6=4.

A⁻¹=(1/4)[[2,−2],[−3,5]], so [x,y]ᵀ=[−1,4]ᵀ.

Write AX=B with coefficient matrix [[2,1,1],[1,−2,−1],[0,3,−5]].

Solving gives [x,y,z]ᵀ=[1,1/2,−3/2]ᵀ. Substitution checks all three equations.

The coefficient matrix is [[1,−1,1],[2,1,−3],[1,1,1]].

Solving AX=B gives [x,y,z]ᵀ=[2,−1,1]ᵀ.

The coefficient matrix is [[2,3,3],[1,−2,1],[3,−1,−2]].

Solving AX=B gives [x,y,z]ᵀ=[1,2,−1]ᵀ.

The coefficient matrix is [[1,−1,2],[3,4,−5],[2,−1,3]].

Solving AX=B gives [x,y,z]ᵀ=[2,1,3]ᵀ.

|A|=−1 and adj A=[[0,−1,2],[2,−9,23],[1,−5,13]], so A⁻¹=−adj A.

For B=[11,−5,−3]ᵀ, X=A⁻¹B=[1,2,3]ᵀ.

Let prices be o,w,r. Then [[4,3,2],[2,4,6],[6,2,3]][o,w,r]ᵀ=[60,90,70]ᵀ.

Solving gives [o,w,r]ᵀ=[5,8,8]ᵀ.

Expand along row 1 and use sin²θ + cos²θ = 1.

The terms containing θ combine to x(sin²θ+cos²θ)−x, which is zero; the remaining term is −x³.

Expand along row 2.

The two non-zero contributions are sin²β and cos²β, whose sum is 1.

Use (AB)⁻¹ = B⁻¹A⁻¹. The determinant of B is 1, and B⁻¹=[[3,2,6],[1,1,2],[2,2,5]].

Multiplying B⁻¹ by the given A⁻¹ gives [[9,−3,5],[−2,1,0],[1,0,2]].

The determinant is −5, so A is invertible. Since adj A=|A|A⁻¹, (adj A)⁻¹=(1/|A|)A.

Also adj(A⁻¹)=|A⁻¹|(A⁻¹)⁻¹=(1/|A|)A. Finally, the inverse of A⁻¹ is A.

Expand and collect like terms: Δ = −2x³−2y³.

Factor the common −2 to get Δ=−2(x³+y³).

Subtract row 1 from rows 2 and 3; this does not change the determinant.

The determinant becomes |1 x y; 0 y 0; 0 0 x| = xy.

Set u=1/x, v=1/y and w=1/z. The system becomes 2u+3v+10w=4, 4u−6v+5w=1, 6u+9v−20w=2.

Solving gives u=1/2, v=1/3, w=1/5. Taking reciprocals gives x=2, y=3, z=5.

Multiply diag(x,y,z) by diag(1/x,1/y,1/z).

The product is I₃, so option (A) is the inverse.

Expanding gives det A=2(1+sin²θ).

Since 0≤sin²θ≤1, 2≤det A≤4, including both endpoints.