Question 1

(i)

Accepted Answer

(i) 1, (ii) x^3-x^2+2. (i) Apply the same rule ad-bc, with a= cos θ, b=- sin θ, c= sin θ, d= cos θ: = ( cos θ)( cos θ)-(- sin θ)( sin θ). The minus sign in the entry - sin θ and the minus sign of the rule cancel, giving a plus: = cos^2θ+ sin^2θ. By the identity sin^2θ+ cos^2θ=1, = 1. The value is 1, and it is 1 for every θ — this matrix rotates the plane through θ, and a rotation does not change area.

Question 2

If A =[ ], then show that 2 A =4 A

Accepted Answer

The property being illustrated is: if A is a square matrix of order n and k is a scalar, then kA =k^n A. Here n=2 and k=2, so the predicted factor is 2^2=4. Verify both sides directly. Left side. Multiplying a matrix by a scalar multiplies every entry, so 2A = 2[ ] = [ ], 2A = (2)(4)-(4)(8) = 8-32 = -24. Right side. A = = (1)(2)-(2)(4) = 2-8 = -6, so 4 A = 4(-6) = -24. Since both sides equal -24, 2A = 4 A. The reason the factor is 4 and not 2: scaling by 2 scales both rows, and a determinant picks up the common factor of each row separately, once per row.

Question 3

If A =[ ], then show that 3 A =27 A

Accepted Answer

The property is kA =k^n A for a square matrix of order n; here n=3 and k=3, so the predicted factor is 3^3=27. Verify both sides. Right side. Expand A along the first column, which has only one non-zero entry: A = = 1 - 0 + 0 = 1(4-0) = 4. (Equivalently, A is upper triangular, so A is the product of the diagonal entries 1· 1· 4 = 4.) Hence 27 A = 27× 4 = 108. Left side. Every entry is tripled: 3A = [ ], 3A = 3 - 0 + 0 = 3(36-0) = 108. Both sides equal 108, so 3A = 27 A. The exponent is 3 because the factor 3 comes out of each of the three rows: 3× 3× 3 = 27.

Question 4

Evaluate the determinants (i) (ii) (iii) (iv)

Accepted Answer

(i) -12, (ii) 46, (iii) 0, (iv) 5. Each determinant is expanded by the cofactor rule A =Σ j a ij(-1)^i+jM ij, choosing the row or column with the most zeros so as to do the least work. (i) Row 2 is 0, 0, -1, so expand along R 2; only the entry a 23=-1 contributes: = (-1) (-1)^2+3. The sign factor (-1)^2+3=-1 is the step usually dropped. The minor is (3)(-5)-(-1)(3) = -15+3 = -12, so = (-1)(-1)(-12) = -12. (ii) Expand along R 1, remembering the alternating signs +, -, +: = 3 - (-4) + 5 = 3(1+6) + 4(1+4) + 5(3-2) = 21+20+5 = 46. (iii) Expand along R 1 (its first entry is 0 ): = 0 - 1 + 2 = -1 (0-6 ) + 2 (-3-0 ) = 6-6 = 0. This is no accident: the matrix is skew-symmetric ( a ij=-a ji ) of odd order, and every such determinant is 0. (iv) Expand along R 1: = 2 - (-1) + (-2) = 2(0-5) + 1(0+3) - 2(0-6) = -10+3+12 = 5. The values are: (i) -12, (ii) 46, (iii) 0, (iv) 5.

Question 5

If A =[ ], find A

Accepted Answer

0. Expand along the first row using A =a 11C 11+a 12C 12+a 13C 13, with the sign pattern +, -, +: A = 1 - 1 + (-2). Evaluate the three 2× 2 minors, watching the signs: = -9-(-12) = 3, = -18-(-15) = -3, = 8-5 = 3. Therefore A = 1(3) - 1(-3) + (-2)(3) = 3+3-6 = 0. So A = 0; the matrix is singular. The reason shows in the rows: 3R 1+R 2 = (3+2, 3+1, -6-3) = (5, 4, -9) = R 3, i.e. the third row is a combination of the first two, and a determinant with dependent rows vanishes.

Question 6

Find values of x, if (i) = (ii) =

Accepted Answer

(i) x= ± √ 3, (ii) x=2. In each part evaluate both determinants by the rule ad-bc and equate. (i) = (2)(1)-(4)(5) = 2-20 = -18, = (2x)(x)-(4)(6) = 2x^2-24. Equating, 2x^2-24 = -18 ⇒ 2x^2 = 6 ⇒ x^2 = 3. Because the equation is quadratic, both square roots are solutions — nothing in the problem rules out a negative x: x = ±√ 3. (ii) = 10-12 = -2, = (x)(5)-(3)(2x) = 5x-6x = -x. Equating, -x = -2 ⇒ x = 2. Hence (i) x=±√ 3 and (ii) x=2.

Question 7

If =, then x is equal to (A) 6 (B) ± 6 (C) -6 (D) 0

Accepted Answer

(B). Evaluate each side by ad-bc. Left side: = x· x - 2· 18 = x^2-36. Right side: = 36-36 = 0. So x^2-36 = 0 ⇒ x^2 = 36 ⇒ x = ± 6. The trap is to read off x=6 by matching the two arrays entry by entry; the condition is only on the value of the determinant, and x=-6 gives (-6)^2-36=0 just as well. Hence x = ± 6, which is option (B).

Question 8

Find area of the triangle with vertices at the point given in each of the following: (i) (1, 0), (6, 0), (4,3) (ii) (2, 7), (1, 1), (10,8) (iii) (-2, -3), (3, 2), (-1, -8)

Accepted Answer

(i) 15/2, (ii) 47/2, (iii) 15. The area of the triangle with vertices (x 1,y 1),(x 2,y 2),(x 3,y 3) is 12 Δ, where Δ=. The modulus sign is not decoration: the determinant itself can come out negative (it does in (iii)), while an area cannot. (i) Vertices (1,0),(6,0),(4,3). Expanding along R 1, Δ= = 1(0· 1-1· 3)-0(6· 1-1· 4)+1(6· 3-0· 4) = -3+0+18 = 15. Area = 12 15 = 15/2 sq units. (ii) Vertices (2,7),(1,1),(10,8). Δ= = 2(1-8)-7(1-10)+1(8-10) = -14+63-2 = 47. Area = 12 47 = 47/2 sq units. (iii) Vertices (-2,-3),(3,2),(-1,-8). Watch the signs of the negative coordinates: Δ= = -2 (2-(-8) )-(-3) (3-(-1) )+1 (3(-8)-2(-1) ), Δ = -2(10)+3(4)+(-24+2) = -20+12-22 = -30. Area = 12 -30 = 15 sq units. Answer: (i) 15/2 sq units, (ii) 47/2 sq units, (iii) 15 sq units.

Question 9

Show that points A (a, b+c), B (b, c+a), C (c, a+b) are collinear.

Accepted Answer

Three points are collinear precisely when the triangle they determine has zero area, i.e. when the area determinant vanishes. So it is enough to show Δ= = 0. Use the property that a determinant is unchanged if a multiple of one column is added to another; apply C 1 → C 1 + C 2: Δ =. Every entry of C 1 now carries the common factor a+b+c, which may be taken outside the determinant: Δ = (a+b+c). In the remaining determinant C 1 and C 3 are identical, and a determinant with two identical columns is zero. Hence Δ = (a+b+c)· 0 = 0. Since the area 12 Δ = 0, the points A (a,b+c), B (b,c+a), C (c,a+b) are collinear. (Note the conclusion holds for all a,b,c — no need to assume a+b+c ≠ 0, because the second factor is zero on its own.)

Question 10

Find values of k if area of triangle is 4 sq. units and vertices are (i) ( k, 0),(4,0),(0,2) (ii) (-2, 0), (0, 4), (0, k)

Accepted Answer

(i) 0, 8, (ii) 0, 8. With Δ the area determinant, area = 12 Δ = 4 gives Δ = 8. This is the step usually lost: the modulus splits into two equations, Δ = 8 and Δ = -8, and both must be solved — that is why each part has two answers. (i) Vertices (k,0),(4,0),(0,2). Expanding along R 1, Δ= = k(0· 1-1· 2)-0(4-0)+1(4· 2-0) = -2k+8. So 8-2k = 8, giving 8-2k = 8 ⇒ k = 0, or 8-2k = -8 ⇒ k = 8. (ii) Vertices (-2,0),(0,4),(0,k). Δ= = -2(4-k)-0(0-0)+1(0· k-4· 0) = 2k-8. So 2k-8 = 8, giving 2k-8 = 8 ⇒ k = 8, or 2k-8 = -8 ⇒ k = 0. Answer: k = 0 or k = 8 in both (i) and (ii).

Question 11

(i) Find equation of line joining (1,2) and (3,6) using determinants. (ii) Find equation of line joining (3,1) and (9,3) using determinants.

Accepted Answer

(i) y=2 x, (ii) x-3 y=0. A point (x,y) lies on the line through (x 1,y 1) and (x 2,y 2) exactly when the three points are collinear, i.e. when the area determinant is zero. So the equation of the line is = 0. (i) Through (1,2) and (3,6): = 0. Expanding along R 1, x(2· 1-1· 6)-y(1· 1-1· 3)+1(1· 6-2· 3) = 0, -4x+2y+0 = 0 ⇒ 2y = 4x. Equation of the line: y = 2x. (ii) Through (3,1) and (9,3): = 0. Expanding along R 1, x(1· 1-1· 3)-y(3· 1-1· 9)+1(3· 3-1· 9) = 0, -2x+6y+0 = 0 ⇒ x = 3y. Equation of the line: x - 3y = 0, i.e. x = 3y.

Question 12

If area of triangle is 35 sq units with vertices (2,-6),(5,4) and (k, 4). Then k is (A) 12 (B) -2 (C) -12, -2 (D) 12,-2

Accepted Answer

(D). Form the area determinant for (2,-6),(5,4),(k,4) and expand along R 1: Δ= = 2(4-4)-(-6)(5-k)+1(5· 4-4k) = 0+6(5-k)+(20-4k), Δ = 30-6k+20-4k = 50-10k. Area = 12 Δ = 35, so 50-10k = 70. Keeping both branches of the modulus: 50-10k = 70 ⇒ k = -2, or 50-10k = -70 ⇒ k = 12. Both values are admissible (neither makes the three points collinear), so k = 12 or k = -2. Answer: option (D) 12, -2.

Question 13

(i) (ii)

Accepted Answer

(i) M 11=3, M 12=0, M 21=-4, M 22=2, A 11=3, A 12=0, A 21=4, A 22=2 (ii) M 11=d, M 12=b, M 21=c, M 22=a A 11=d, A 12=-b, A 21=-c, A 22=a. Definition. The minor M ij of the element a ij is the determinant of what is left after deleting the i th row and the j th column, and the cofactor is A ij=(-1)^i+jM ij. For a second order determinant, deleting one row and one column leaves a single entry, so each minor is just that entry; the whole work is in the sign (-1)^i+j, which is +,- in the first row and -,+ in the second. (i) For: Minors: deleting row 1, column 1 leaves 3; row 1, column 2 leaves 0; row 2, column 1 leaves -4; row 2, column 2 leaves 2. So M 11=3, M 12=0, M 21=-4, M 22=2. Cofactors, attaching (-1)^i+j: A 11=(-1)^2(3)=3, A 12=(-1)^3(0)=0, A 21=(-1)^3(-4)=4, A 22=(-1)^4(2)=2. Note that A 21=4, not -4: the sign is applied to the minor -4, and the two minus signs cancel. (ii) For: Minors: M 11=d, M 12=b, M 21=c, M 22=a. Cofactors: A 11=d, A 12=-b, A 21=-c, A 22=a. Check (each expansion must reproduce the determinant): in (i), a 11A 11+a 12A 12=2(3)+(-4)(0)=6, and indeed =6; in (ii), a A 11+c A 12=ad-bc.

Question 14

(i) (ii)

Accepted Answer

(i) M 11=1, M 12=0, M 13=0, M 21=0, M 22=1, M 23=0, M 31=0, M 32=0, M 33=1, A 11=1, A 12=0, A 13=0, A 21=0, A 22=1, A 23=0, A 31=0, A 32=0, A 33=1 (ii) M 11=11, M 12=6, M 13=3, M 21=-4, M 22=2, M 23=1, M 31=-20, M 32=-13, M 33=5 A 11=11, A 12=-6, A 13=3, A 21=4, A 22=2, A 23=-1, A 31=-20, A 32=13, A 33=5. Definition. M ij is the 2× 2 determinant obtained by deleting the i th row and j th column, and A ij=(-1)^i+jM ij. The sign board for a third order determinant is [ ]. (i) For: M 11= =1, M 12= =0, M 13= =0, M 21= =0, M 22= =1, M 23= =0, M 31= =0, M 32= =0, M 33= =1. Applying the signs (every off-diagonal minor is 0, so its sign is immaterial): A 11=1, A 12=0, A 13=0, A 21=0, A 22=1, A 23=0, A 31=0, A 32=0, A 33=1. (ii) For: Minors: M 11= =10+1=11, M 12= =6-0=6, M 13= =3-0=3, M 21= =0-4=-4, M 22= =2-0=2, M 23= =1, M 31= =0-20=-20, M 32= =-1-12=-13, M 33= =5-0=5. Cofactors A ij=(-1)^i+jM ij: A 11=11, A 12=-6, A 13=3, A 21=-(-4)=4, A 22=2, A 23=-1, A 31=-20, A 32=-(-13)=13, A 33=5. The places students slip are A 21 and A 32, where the minor is already negative and the sign change makes the cofactor positive. Check by expanding along the first row: 1(11)+0(-6)+4(3)=23, and expanding along the second column gives 0(-6)+5(2)+1(13)=23 as well.

Question 15

Using Cofactors of elements of second row, evaluate Δ=.

Accepted Answer

7. Rule used: expansion of a determinant along a row, Δ=a 21A 21+a 22A 22+a 23A 23, where A ij=(-1)^i+jM ij. Along the second row the signs are -,+,-. Here Δ=, so a 21=2, a 22=0, a 23=1. Cofactors of the second row: A 21=(-1)^2+1 =-(9-16)=7, A 22=(-1)^2+2 =15-8=7, A 23=(-1)^2+3 =-(10-3)=-7. Each minor is formed from the rows and columns that remain, so M 21 keeps rows 1 and 3 and columns 2 and 3, and so on. Hence Δ=a 21A 21+a 22A 22+a 23A 23=2(7)+0(7)+1(-7)=14+0-7=7. (Confirming along the first row: 5(0-2)-3(6-1)+8(4-0)=-10-15+32=7.) Δ=7.

Question 16

Using Cofactors of elements of third column, evaluate Δ=.

Accepted Answer

(x-y)(y-z)(z-x). Rule used: expansion along a column, Δ=a 13A 13+a 23A 23+a 33A 33, with A ij=(-1)^i+jM ij. Down the third column the signs are +,-,+. For Δ= the third column entries are a 13=yz, a 23=zx, a 33=xy. Cofactors of the third column (delete column 3 and the relevant row): A 13=(-1)^1+3 =z-y, A 23=(-1)^2+3 =-(z-x)=x-z, A 33=(-1)^3+3 =y-x. The middle one is where the sign is lost most often: the minor is z-x, and the factor (-1)^5 turns it into x-z. Therefore Δ=yz(z-y)+zx(x-z)+xy(y-x) =yz^2-y^2z+x^2z-xz^2+xy^2-x^2y. Now factorise. Group the terms containing x^2, then the rest: Δ=x^2(z-y)-x(z^2-y^2)+yz(z-y)=(z-y)[x^2-x(z+y)+yz], using z^2-y^2=(z-y)(z+y). The bracket factorises as x^2-x(y+z)+yz=(x-y)(x-z). Hence Δ=(z-y)(x-y)(x-z)=(x-y)(y-z)(z-x). Δ=(x-y)(y-z)(z-x). In particular Δ=0 whenever any two of x,y,z are equal, as it must be, since then two rows of the determinant coincide.

Question 17

If Δ= and A i j is Cofactors of a i j, then value of Δ is given by (A) a 11 A 31+a 12 A 32+a 13 A 33 (B) a 11 A 11+a 12 A 21+a 13 A 31 (C) a 21 A 11+a 22 A 12+a 23 A 13 (D) a 11 A 11+a 21 A 21+a 31 A 31

Accepted Answer

(D). Rule being tested: a determinant equals the sum of the products of the elements of ANY ONE row (or ANY ONE column) with their OWN corresponding cofactors, Δ=a i1A i1+a i2A i2+a i3A i3 (any fixed row i ), Δ=a 1jA 1j+a 2jA 2j+a 3jA 3j (any fixed column j ). If instead the elements of one row are multiplied by the cofactors of a DIFFERENT row, the sum is 0, not Δ. Test each option against this rule: (A) a 11A 31+a 12A 32+a 13A 33: elements of row 1 against the cofactors of row 3 — a cross combination, so this equals 0, not Δ. (B) a 11A 11+a 12A 21+a 13A 31: the elements come from row 1 while the cofactors A 11,A 21,A 31 come from column 1 — the indices do not match, so this is not an expansion at all. (C) a 21A 11+a 22A 12+a 23A 13: elements of row 2 against the cofactors of row 1 — again a cross combination, equal to 0. (D) a 11A 11+a 21A 21+a 31A 31: the elements a 11,a 21,a 31 are exactly the first column, each paired with its own cofactor. This is the expansion of Δ along the first column. Hence the correct option is (D).

Question 18

[ ]

Accepted Answer

. The adjoint of a square matrix is the transpose of its matrix of cofactors: adj A=[A ij]^T, where A ij=(-1)^i+jM ij. For A=[ ], deleting the row and column of each entry leaves a single number, so A 11=+4, A 12=-3, A 21=-2, A 22=+1. The cofactor matrix is [ ]; transposing it, adj A=[ ]. This is the standard 2× 2 pattern: interchange the two diagonal entries and reverse the signs of the other two.

Question 19

[ ]

Accepted Answer

adj A is the transpose of the cofactor matrix, with A ij=(-1)^i+jM ij and M ij the 2× 2 minor left after deleting row i and column j of A=[ ]. Row 1 cofactors: A 11=+ =3, A 12=- =-(2+10)=-12, A 13=+ =0+6=6. Row 2 cofactors: A 21=- =-(-1-0)=1, A 22=+ =1+4=5, A 23=- =-(0-2)=2. Row 3 cofactors: A 31=+ =-5-6=-11, A 32=- =-(5-4)=-1, A 33=+ =3+2=5. Now transpose the array of cofactors -- this is the step most often dropped, and it sends A 12=-12 into row 2, column 1: adj A=[ ].

Question 20

[ ]

Accepted Answer

For A=[ ], A =(2)(-6)-(3)(-4)=-12+12=0, so A is singular. The identity A( adj A)=( adj A)A= A I holds for every square matrix, singular or not, so it is still to be checked. By the 2× 2 rule (swap the diagonal entries, change the sign of the others), adj A=[ ]. Then A( adj A)=[ ][ ]=[ ]=[ ], ( adj A)A=[ ][ ]=[ ]=[ ]. Also A I=0· I=O. Hence A( adj A)=( adj A)A= A I, all three being the zero matrix -- which is exactly why a matrix with A =0 can have no inverse.

Question 21

[ ]

Accepted Answer

Let A=[ ]. Expand A along the second column, which carries two zeros: A =(-1)A 12+0· A 22+0· A 32, A 12=- =-(9+2)=-11, so A =(-1)(-11)=11. The remaining cofactors: A 11=+ =0, A 13=+ =0, A 21=- =-(-3)=3, A 22=+ =1, A 23=- =-(0+1)=-1, A 31=+ =2, A 32=- =-(-2-6)=8, A 33=+ =3. Transposing the cofactor matrix, adj A=[ ]. Now multiply, in both orders: A( adj A)=[ ]=[ ], ( adj A)A=[ ]=[ ]. Both products equal 11I= A I. Hence A( adj A)=( adj A)A= A I is verified.

Question 22

[ ]

Accepted Answer

1/14[ ]. A square matrix is invertible precisely when A ≠ 0, and then A^-1=1/( A ) adj A. For A=[ ], A =(2)(3)-(-2)(4)=6+8=14≠ 0, so the inverse exists. (Mind the sign: -(-2)(4)=+8, not -8.) Using the 2× 2 adjoint rule, adj A=[ ], and therefore A^-1=1/14[ ].

Question 23

[ ]

Accepted Answer

1/13. Use A^-1=1/( A ) adj A, valid when A ≠ 0. For A=[ ], A =(-1)(2)-(5)(-3)=-2+15=13≠ 0, so A^-1 exists. Interchanging the diagonal entries and reversing the signs of the off-diagonal ones, adj A=[ ]. Hence A^-1=1/13[ ].

Question 24

[ ]

Accepted Answer

1/10[ ]. Let A=[ ]. Expanding along the first column (only one non-zero entry) and again along the first column of the resulting minor, A =1 =1· 2· 5=10≠ 0, the familiar fact that the determinant of a triangular matrix is the product of its diagonal entries. So A^-1 exists. Cofactors: A 11= =10, A 12=- =0, A 13= =0, A 21=- =-10, A 22= =5, A 23=- =0, A 31= =8-6=2, A 32=- =-4, A 33= =2. Transposing, adj A=[ ]. Therefore A^-1=1/10[ ]. (The inverse of an upper triangular matrix is again upper triangular, a useful check.)

Question 25

[ ]

Accepted Answer

. Let A=[ ]. Expanding along the first row, A =1 -0+0=1(-3-0)=-3≠ 0, so A^-1 exists. Cofactors: A 11= =-3, A 12=- =-(-3)=3, A 13= =6-15=-9, A 21=- =0, A 22= =-1, A 23=- =-2, A 31= =0, A 32=- =0, A 33= =3. Transposing, adj A=[ ]. Dividing by A =-3 (every sign flips -- the step to be careful with): A^-1=1/-3[ ]=1/3[ ]=[ ].

Question 26

[ ]

Accepted Answer

Let A=[ ]. Expanding along the first row, A =2 -1 +3 =2(-1)-1(4)+3(8-7)=-2-4+3=-3≠ 0, so A^-1 exists. Cofactors: A 11=-1, A 12=- =-4, A 13= =8-7=1, A 21=- =-(1-6)=5, A 22= =2+21=23, A 23=- =-(4+7)=-11, A 31= =0+3=3, A 32=- =-(0-12)=12, A 33= =-2-4=-6. Transposing, adj A=[ ]. Hence A^-1=1/-3[ ]=1/3[ ].

Question 27

[ ]

Accepted Answer

Let A=[ ]. Expanding along the first row, A =1 -(-1) +2 =(8-6)+(0+9)+2(0-6)=2+9-12=-1≠ 0, so A^-1 exists. Cofactors: A 11=2, A 12=- =-9, A 13= =-6, A 21=- =-(-4+4)=0, A 22= =-2, A 23=- =-(-2+3)=-1, A 31= =3-4=-1, A 32=- =3, A 33= =2. Transposing, adj A=[ ]. Since A =-1, dividing simply reverses every sign: A^-1=1/-1[ ]=[ ].

Question 28

[ ]

Accepted Answer

Let A=[ ]. Expanding along the first row and using sin^2α+ cos^2α=1, A =1 =- cos^2α- sin^2α=-1. This is ≠ 0 for every value of α, so the inverse always exists. Cofactors: A 11=-1, A 12=- =0, A 13= =0, A 21=- =0, A 22= =- cos α, A 23=- =- sin α, A 31=0, A 32=- =- sin α, A 33= = cos α. The cofactor matrix is symmetric here, so transposing changes nothing: adj A=[ ]. Dividing by A =-1, A^-1=[ ]=A. So A is its own inverse, i.e. A^2=I, for every α.

Question 29

Let A =[ ] and B =[ ]. Verify that ( AB )^-1= B^-1 A^-1.

Accepted Answer

Each inverse is found from A^-1=1/( A ) adj A. The two inverses. A =3(5)-7(2)=15-14=1, so A^-1=1/1[ ]=[ ]. B =6(9)-8(7)=54-56=-2, so B^-1=1/-2[ ]. Left-hand side. First AB=[ ][ ]=[ ]=[ ], AB =67(61)-87(47)=4087-4089=-2 (= A B, as expected ), (AB)^-1=1/-2[ ]. Right-hand side. Note the order: B^-1 comes first. B^-1A^-1=1/-2[ ][ ]=1/-2[ ]=1/-2[ ]. The two results agree, so (AB)^-1=B^-1A^-1 is verified. (Multiplying in the other order, A^-1B^-1, gives a different matrix -- the reversal of order is the whole content of the result.)

Question 30

If A =[ ], show that A^2-5 A +7 I = O. Hence find A^-1.

Accepted Answer

1/7[ ]. With A=[ ], A^2=[ ][ ]=[ ]=[ ]. Then A^2-5A+7I=[ ]-[ ]+[ ]=[ ]=[ ]=O. Hence A^-1. Rearrange the relation so that A is a factor -- do not divide by A, and do not assume in advance that A^-1 exists: 7I=5A-A^2=A(5I-A) ⇒ A[1/7(5I-A)]=I. So A is invertible and A^-1=1/7(5I-A)=1/7([ ]-[ ])=1/7[ ]. (Check: A =6+1=7 and adj A=[ ], giving the same answer.)

Question 31

For the matrix A =[ ], find the numbers a and b such that A^2+a A +b I = O.

Accepted Answer

a=-4, b=1. With A=[ ], A^2=[ ][ ]=[ ]=[ ]. Substituting into A^2+aA+bI=O: [ ]+[ ]+[ ]=[ ]=[ ]. Two matrices are equal only if corresponding entries are equal, which gives four equations: 11+3a+b=0, 8+2a=0, 4+a=0, 3+a+b=0. The (1,2) and (2,1) entries both give a=-4 (consistent). Putting a=-4 into the (1,1) entry: 11-12+b=0 ⇒ b=1, and the (2,2) entry checks out: 3-4+1=0. a=-4, b=1 (These are - trace (A) and det (A) respectively, as the Cayley--Hamilton theorem predicts.)

Question 32

For the matrix A =[ ] Show that A^3-6 A^2+5 A +11 I = O. Hence, find A^-1.

Accepted Answer

A^-1=1/11[ ]. Let A=[ ]. Step 1: A^2. A^2=[ ]=[ ]. Step 2: A^3=A^2A. A^3=[ ]=[ ]. Step 3: combine. A^3-6A^2+5A+11I=[ ]=[ ]=O. Step 4: hence A^-1. Factor A out of the first three terms rather than "dividing by A ": 11I=-(A^3-6A^2+5A)=A(-A^2+6A-5I) ⇒ A[1/11(-A^2+6A-5I)]=I, so A is invertible with A^-1=1/11(-A^2+6A-5I). Now -A^2+6A-5I=[ ]=[ ], and therefore A^-1=1/11[ ].

Question 33

If A =[ ] Verify that A^3-6 A^2+9 A -4 I = O and hence find A^-1.

Accepted Answer

1/4[ ]. Let A=[ ]. Step 1: A^2. A^2=[ ]=[ ]. Step 2: A^3=A^2A. A^3=[ ]=[ ]. Step 3: verify the relation. A^3-6A^2+9A-4I=[ ]=O, since every diagonal entry is 22-36+18-4=0 and every off-diagonal entry is ±(21-30+9)=0. Step 4: hence A^-1. Rewrite the identity with A factored out: 4I=A^3-6A^2+9A=A(A^2-6A+9I) ⇒ A[1/4(A^2-6A+9I)]=I, so A is invertible and A^-1=1/4(A^2-6A+9I). Computing that bracket, A^2-6A+9I=[ ]=[ ], so A^-1=1/4[ ].

Question 34

Let A be a nonsingular square matrix of order 3 × 3. Then adj A is equal to (A) A (B) A ^2 (C) A ^3 (D) 3 A

Accepted Answer

B. Start from the theorem that holds for every square matrix of order n: A( adj A)= A I n. Here n=3. Take determinants of both sides and use XY = X Y together with kI 3 =k^3 (a scalar k multiplies each of the three rows): A adj A = A I 3 = A ^3. Because A is nonsingular, A ≠ 0, so we may cancel one factor of A: adj A = A ^2. (The nonsingularity is exactly what licenses the cancellation -- without it the step is invalid.) Hence the correct option is (B) A ^2. In general adj A = A ^ n-1.

Question 35

If A is an invertible matrix of order 2, then det ( A^-1) is equal to (A) det ( A ) (B) 1/( det ( A )) (C) 1 (D) 0

Accepted Answer

B. By the definition of the inverse, AA^-1=I. Take determinants of both sides and use the product rule XY = X Y with I =1: det (A)· det (A^-1)= det (I)=1. Since A is invertible, det (A)≠ 0, so dividing is legitimate: det (A^-1)=1/( det (A)). Hence the correct option is (B) 1/( det (A)). (The order 2 plays no part -- the result holds for an invertible matrix of any order.)

Question 36

gathered 2 x+y+z=1 x-2 y-z=3/2 3 y-5 z=9 gathered

Accepted Answer

x=1, y=1/2, z=-3/2. Write AX=B; the third equation has no x -term, so its first coefficient is 0. A=[ ], X=[ ], B=[ ] A =2(10+3)-1(-5-0)+1(3-0)=26+5+3=34≠ 0 So A^-1 exists. Cofactors: A 11=13, A 12=5, A 13=3, A 21=8, A 22=-10, A 23=-6, A 31=1, A 32=3, A 33=-5 Transposing the cofactor matrix, adj A=[ ], A^-1=1/34[ ] X=A^-1B=1/34[ ]=1/34[ ] x=1, y=1/2, z=-3/2

Question 37

gathered x-y+z=4 2 x+y-3 z=0 x+y+z=2 gathered

Accepted Answer

x=2, y=-1, z=1. Write AX=B and use X=A^-1B. A=[ ], X=[ ], B=[ ] A =1(1+3)-(-1)(2+3)+1(2-1)=4+5+1=10≠ 0 Cofactors: A 11=4, A 12=-5, A 13=1, A 21=2, A 22=0, A 23=-2, A 31=2, A 32=5, A 33=3 adj A=[ ], A^-1=1/10[ ] X=A^-1B=1/10[ ]=1/10[ ] x=2, y=-1, z=1

Question 38

If A =[ ], find A^-1. Using A^-1 solve the system of equations

Accepted Answer

Use A^-1=1/( A ) adj A, where adj A is the transpose of the matrix of cofactors. Expanding A along the first row: A =2 (-4+4)-(-3)(-6+4)+5 (3-2)=0-6+5=-1≠ 0 Cofactors of A: A 11=0, A 12=2, A 13=1 A 21=-1, A 22=-9, A 23=-5 A 31=2, A 32=23, A 33=13 Transposing the cofactor matrix, adj A=[ ] A^-1=1/-1[ ]=[ ] The coefficient matrix of the given system is exactly this A, so with B=[ ] the system is AX=B and X=A^-1B: X=[ ][ ]=[ ]=[ ] A^-1=[ ], x=1, y=2, z=3

Question 39

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹ 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹ 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹ 70. Find cost of each item per kg by matrix method.

Accepted Answer

cost of onions per kg =₹ 5 cost of wheat per kg =₹ 8 cost of rice per kg =₹ 8. Let the cost per kg be ₹x for onion, ₹y for wheat and ₹z for rice. Translating the three purchases: In matrix form AX=B: A=[ ], X=[ ], B=[ ] A =4(12-12)-3(6-36)+2(4-24)=0+90-40=50≠ 0 So A^-1 exists. Cofactors: A 11=0, A 12=30, A 13=-20, A 21=-5, A 22=0, A 23=10, A 31=10, A 32=-20, A 33=10 adj A=[ ], A^-1=1/50[ ] X=A^-1B=1/50[ ]=1/50[ ]=[ ] Onion costs ₹5 per kg, wheat ₹8 per kg and rice ₹8 per kg.

Question 40

Prove that the determinant is independent of θ.

Accepted Answer

Expand the determinant along the first row. Δ= =x - sin θ + cos θ Evaluating the three 2× 2 determinants, Δ=x(-x^2-1)- sin θ(-x sin θ- cos θ)+ cos θ(- sin θ+x cos θ) =-x^3-x+x sin^2θ+ sin θ cos θ- sin θ cos θ+x cos^2θ The step that decides the question: the two sin θ cos θ terms come out with opposite signs and cancel, and the two remaining θ -terms have the common factor x, so the Pythagorean identity sin^2θ+ cos^2θ=1 applies: Δ=-x^3-x+x( sin^2θ+ cos^2θ)=-x^3-x+x=-x^3 The value Δ=-x^3 contains no θ. Hence the determinant is independent of θ.

Question 41

Evaluate.

Accepted Answer

1. Δ= Expand along R 2; it carries a zero, so only two cofactors are needed. With C ij=(-1)^i+jM ij, Δ=(- sin β)C 21+( cos β)C 22+0· C 23 The minors are M 21= = cos^2α sin β+ sin^2α sin β= sin β( cos^2α+ sin^2α)= sin β M 22= = cos^2α cos β+ sin^2α cos β= cos β The sign is where this goes wrong: C 21=(-1)^2+1M 21=- sin β, while C 22=(-1)^2+2M 22=+ cos β. Hence Δ=(- sin β)(- sin β)+( cos β)( cos β)= sin^2β+ cos^2β Δ=1 The determinant equals 1 for every α and β.

Question 42

If A^-1=[ ] and B =[ ], find ( AB )^-1

Accepted Answer

[ ]. Use the reversal law for inverses of a product: ( AB )^-1= B^-1 A^-1 Note the order: B^-1 comes first. Since A^-1 is already given, only B^-1 has to be found; A itself is never needed. For B =[ ], expanding B along R 1, B =1(3· 1-0·(-2))-2((-1)· 1-0· 0)+(-2)((-1)(-2)-3· 0)=3+2-4=1 So B =1≠ 0 and B^-1 exists. The cofactors of B are B 11=3, B 12=1, B 13=2, B 21=2, B 22=1, B 23=2, B 31=6, B 32=2, B 33=5 The adjoint is the transpose of the cofactor matrix: adj B =[ ], B^-1=1/( B ) adj B =[ ] Now multiply, keeping the order B^-1 A^-1: ( AB )^-1=[ ][ ] Row 1: 9-30+30=9, -3+12-12=-3, 3-10+12=5. Row 2: 3-15+10=-2, -1+6-4=1, 1-5+4=0. Row 3: 6-30+25=1, -2+12-10=0, 2-10+10=2. ( AB )^-1=[ ]

Question 43

Let A =. Verify that (i) [adj A ]^-1=a d j ( A^-1) (ii) ( A^-1)^-1= A

Accepted Answer

Here A =[ ]. Expanding along R 1, A =1(15-1)-2(10-1)+1(2-3)=14-18-1=-5≠ 0 The cofactors are A 11=14, A 12=-9, A 13=-1, A 21=-9, A 22=4, A 23=1, A 31=-1, A 32=1, A 33=-1, so (taking the transpose) adj A =[ ], A^-1=1/( A ) adj A =-1/5[ ] (i) [ adj A ]^-1= adj ( A^-1). The key relation is M ( adj M )= M I, i.e. adj M = M M^-1. Left side: adj A = A A^-1=-5 A^-1, hence [ adj A ]^-1=(-5 A^-1)^-1=-1/5 A Right side: applying the same relation to A^-1, and using A^-1 =1/( A )=-1/5, adj ( A^-1)= A^-1 ( A^-1)^-1=-1/5 A Both sides equal -1/5 A =[ ] Check of the left side by direct multiplication: the first row of adj A times the first column of -1/5 A gives -14/5+18/5+1/5=1, and times the second and third columns gives -28/5+27/5+1/5=0 and -14/5+9/5+1=0; continuing likewise the product is I. Hence [ adj A ]^-1= adj ( A^-1), verified. (ii) ( A^-1)^-1= A. Apply the inverse formula to the matrix A^-1: ( A^-1)^-1=1/( A^-1 ) adj ( A^-1)=1/(-1/5)(-1/5 A )= A (Equivalently, A A^-1= A^-1 A = I already says A is the inverse of A^-1, and the inverse of a matrix is unique.) Verified.

Question 44

Evaluate

Accepted Answer

-2(x^3+y^3). Δ= Every row has the same sum 2(x+y), so use the property that adding columns to a column leaves the determinant unchanged. Apply C 1→ C 1+C 2+C 3: Δ= =2(x+y) (a common factor of a column may be taken outside the determinant). Now R 2→ R 2-R 1 and R 3→ R 3-R 1: Δ=2(x+y) Expanding along C 1, Δ=2(x+y) =2(x+y)[-x^2+y(x-y)] =2(x+y)(-x^2+xy-y^2)=-2(x+y)(x^2-xy+y^2) Recognising (x+y)(x^2-xy+y^2)=x^3+y^3, Δ=-2(x^3+y^3)

Question 45

Evaluate

Accepted Answer

x y. Δ= The first column is constant, so subtracting R 1 from the other rows (a row operation that does not change the value of a determinant) clears it. Apply R 2→ R 2-R 1 and R 3→ R 3-R 1: Δ= This is upper triangular, or expand along C 1: Δ=1· =1·(yx-0) Δ=xy

Question 46

Solve the system of equations

Accepted Answer

x=2, y=3, z=5. The equations are not linear in x,y,z, but they are linear in the reciprocals. Since no denominator may vanish, x,y,z≠ 0; put a=1/x, b=1/y, c=1/z The system becomes 2a+3b+10c=4, 4a-6b+5c=1, 6a+9b-20c=2 that is A X= B with A =[ ], X=[ ], B =[ ] Expanding along R 1, A =2(120-45)-3(-80-30)+10(36+36)=150+330+720=1200≠ 0 so the system has the unique solution X= A^-1 B. The cofactors of A are A 11=75, A 12=110, A 13=72, A 21=150, A 22=-100, A 23=0, A 31=75, A 32=30, A 33=-24 Taking the transpose, adj A =[ ], A^-1=1/1200 adj A Hence [ ]=1/1200[ ][ ]=1/1200[ ]=1/1200[ ] So a=1/2, b=1/3, c=1/5. The step most often forgotten is the last one: these are the reciprocals, so substitute back. x=1/a=2, y=1/b=3, z=1/c=5 Check: 2/2+3/3+10/5=1+1+2=4; 4/2-6/3+5/5=2-2+1=1; 6/2+9/3-20/5=3+3-4=2. All three hold, and x,y,z≠ 0 as required.

Question 47

If x, y, z are nonzero real numbers, then the inverse of matrix A =[ ] is (A) [ ] (B) x y z[ ] (C) 1/(x y z)[ ] (D) 1/(x y z)[ ]

Accepted Answer

A. A is diagonal, so expanding gives A =xyz, which is non-zero because x,y,z are non-zero. Hence A^-1 exists. Rather than testing options, construct the inverse from the definition A A^-1= I. Try D =[ ]. Multiplying two diagonal matrices multiplies corresponding diagonal entries: A D =[ ]=[ ]= I and likewise D A = I. Since the inverse is unique, A^-1= D. (The same result follows from A^-1=1/( A ) adj A: here adj A =[ ], and dividing by xyz gives diag (x^-1,y^-1,z^-1).) Option (B) is xyz times this and option (D) is 1/xyz I; neither is diagonal-reciprocal, so both fail. The correct answer is (A).

Question 48

Let A =[ ], where 0 ≤ θ ≤ 2 π. Then (A) Det ( A )=0 (B) Det ( A ) ∈(2, ∞) (C) Det ( A ) ∈(2,4) (D) Det ( A ) ∈[2,4]

Accepted Answer

D. Expand Det ( A ) along R 1. Det ( A )=1 - sin θ +1 =1(1+ sin^2θ)- sin θ(- sin θ+ sin θ)+1( sin^2θ+1) The middle 2× 2 determinant is exactly 0, so Det ( A )=(1+ sin^2θ)+(1+ sin^2θ)=2+2 sin^2θ Now use the range. For 0≤θ≤ 2π, sin θ takes every value in [-1,1], so sin^2θ takes every value in [0,1]: 0≤ sin^2θ≤ 1 ⇒ 2≤ 2+2 sin^2θ≤ 4 Both endpoints are actually attained — θ=0 gives 2 and θ=(π)/2 gives 4 — so the interval is closed, which rules out the open intervals in (B) and (C), and Det ( A )≥ 2 0 rules out (A). Det ( A )∈[2,4] The correct answer is (D).

EXERCISE 4.1 1–8 (part 1 of 8)

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Exercise 7

Exercise 8