Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
To every square matrix A = [aij] we attach one number, its determinant, written |A|, det A or Δ. The vertical bars here do not mean 'modulus': a determinant can be negative. Only square matrices have determinants.
The determinant tells us whether a system of equations has a unique solution: for a₁x + b₁y = c₁, a₂x + b₂y = c₂ it is a₁b₂ − a₂b₁, and the solution is unique when this is not 0.
Example: 24−12 = 2(2) − 4(−1) = 4 + 4 = 8.
Expand along row 1 using second-order determinants, with alternating signs +, −, +:
Each small determinant is what is left after deleting the row and column of the entry in front of it.
The value is the same along any row or column, so expand along the one with the most zeros, and use the sign (−1)^(i+j): + when i + j is even, − when it is odd.
Example: 124−130410. Column 3 has two zeros, so expand along it: 4−1341 = 4(−1 − 12) = −52.
If A = kB where A and B are square matrices of order n, then |A| = kⁿ|B|. For a 3 × 3 matrix, |2A| = 8|A|. Do not confuse this with multiplying one row by k, which multiplies the determinant by k only once.
The determinant of a triangular matrix (all zeros above or below the diagonal) is the product of the diagonal entries. For example |1 2 3; 0 4 5; 0 0 6| = 24.
The product rule |AB| = |A||B| is in your textbook as Theorem 3. The facts on transposes and row interchange are standard results that you can use in the questions here, though the textbook exercises do not ask you to prove them.
The triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) has area
For a point P(x, y) to lie on the line through A and B, the triangle ABP must have zero area. So the line through (1, 3) and (0, 0) is 001131xy1 = 0, which gives y − 3x = 0, that is y = 3x.
Check that the answer passes through both given points.
The minor Mij of the element aij is the determinant left after deleting row i and column j. For order n ≥ 2, it has order n − 1.
The cofactor is Aij = (−1)^(i+j) Mij.
In 123456789 the element 6 is in row 2, column 3. Delete them: M₂₃ = 1278 = 8 − 14 = −6, and A₂₃ = (−1)⁵(−6) = 6.
That is, |A| is the sum of the products of the elements of any row (or column) with their own cofactors.
If the elements of one row (or column) are multiplied by the cofactors of the elements of another row (or column), the sum is 0. For example a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃ = 0, because it is the expansion of a determinant with two identical rows.
Check with 2−3560415−7: a₁₁A₃₁ + a₁₂A₃₂ + a₁₃A₃₃ = 2(−12) + (−3)(22) + 5(18) = −24 − 66 + 90 = 0.
The adjoint of a square matrix A is the transpose of the matrix of cofactors: adj A = [Aij]′.
For a 2 × 2 matrix there is a shortcut: swap the diagonal entries and change the signs of the other two. For example if A = 2314 then adj A = 4−3−12.
Reason: the sum of products of a row with its own cofactors is |A| (on the diagonal) and with another row's cofactors is 0 (everywhere else).
Example: A = 133143134. |A| = 1(16 − 9) − 3(4 − 3) + 3(3 − 4) = 1. The cofactors are A₁₁ = 7, A₁₂ = −1, A₁₃ = −1, A₂₁ = −3, A₂₂ = 1, A₂₃ = 0, A₃₁ = −3, A₃₂ = 0, A₃₃ = 1. So adj A = 7−3−3−110−101 and, since |A| = 1, A⁻¹ = adj A.
If A and B are non-singular of the same order, (AB)⁻¹ = B⁻¹A⁻¹. Also (A⁻¹)⁻¹ = A and |A⁻¹| = 1/|A|.
If A satisfies an equation such as A² − 4A + I = O and |A| ≠ 0, multiply by A⁻¹: A − 4I + A⁻¹ = O, so A⁻¹ = 4I − A. For A = 2312 this gives 2−3−12.
A system of three equations in x, y, z is written AX = B, where A is the matrix of coefficients, X = [x, y, z]ᵀ and B is the column of constants.
A⁻¹ exists, so X = A⁻¹B: the system is consistent with a unique solution. This is the matrix method.
Example: 2x + 5y = 1, 3x + 2y = 7. |A| = −11, A⁻¹ = -1112−5−32, and X = A⁻¹B = 3−1, so x = 3, y = −1.
Find (adj A)B.
A system is consistent if it has at least one solution and inconsistent if it has none.
Example: the sum of three numbers is 6; three times the third plus the second is 11; the first plus the third is twice the second. The equations are x + y + z = 6, y + 3z = 11, x − 2y + z = 0, and the solution is x = 1, y = 2, z = 3.
If you are given a matrix product that gives the identity, such as AB = I, then B = A⁻¹ and you can solve AX = C as X = BC without finding an inverse again. This is how Example 19 of the textbook solves a system.
| Result | Condition |
|---|---|
| |A| for 2 × 2: a₁₁a₂₂ − a₁₂a₂₁ | order 2 |
| Area = ½ |x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1| (absolute value) | vertices known |
| Aij = (−1)^(i+j) Mij | any square matrix |
| A (adj A) = |A| I | any square matrix |
| |adj A| = |A|ⁿ⁻¹ | order n |
| A⁻¹ = (1/|A|) adj A | |A| ≠ 0 |
| (AB)⁻¹ = B⁻¹A⁻¹ | A, B invertible |
| X = A⁻¹B | |A| ≠ 0 |
| |kA| = kⁿ|A| | order n |
Try each problem first, then open the solution to check your working.
Evaluate 3−254.
Use: ad − bc
= 3(4) − (−2)(5) = 12 + 10 = 22.
Answer: 22
Find x if x + 132x − 1 = 0.
(x + 1)(x − 1) − 3 × 2 = 0, so x² − 1 − 6 = 0 and x² = 7.
Answer: x = ±√7
Evaluate 203140502.
Use: Expand along the column with most zeros
Column 2 has two zeros, so only the entry 4 (row 2, column 2) remains. Its sign is (−1)²⁺² = +.
= 42352 = 4(4 − 15) = −44.
Answer: −44
A is a 3 × 3 matrix with |A| = −4. Find |2A|, |adj A| and |A⁻¹|.
Use: |kA| = kⁿ|A|; |adj A| = |A|ⁿ⁻¹
|2A| = 2³(−4) = −32.
|adj A| = |A|² = 16.
|A⁻¹| = 1/|A| = −1/4.
Answer: −32, 16 and −1/4
Find the area of the triangle with vertices (−1, 2), (3, 4) and (5, −2).
Use: Δ = ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
−1(4 + 2) + 3(−2 − 2) + 5(2 − 4) = −6 − 12 − 10 = −28.
Area = ½|−28| = 14 square units.
Answer: 14 square units
Find k if the points (1, 2), (3, 5) and (k, 8) are collinear.
Use: The determinant must be 0
121351k81 = 1(5 − 8) − 2(3 − k) + 1(24 − 5k) = −3 − 6 + 2k + 24 − 5k = 15 − 3k.
15 − 3k = 0 gives k = 5.
Answer: k = 5
For 2−1304152−2, find M₁₂ and A₁₂. Use cofactors along row 1 to evaluate the determinant.
M₁₂ = 015−2 = 0 − 5 = −5, and A₁₂ = (−1)³(−5) = 5.
M₁₁ = 412−2 = −8 − 2 = −10 so A₁₁ = −10. M₁₃ = 0452 = 0 − 20 = −20 so A₁₃ = −20.
Δ = 2(−10) + (−1)(5) + 3(−20) = −20 − 5 − 60 = −85.
Answer: M₁₂ = −5, A₁₂ = 5, Δ = −85
Find adj A for A = 3−124 and verify A (adj A) = |A| I.
Use: Swap the diagonal, change signs of the others
adj A = 41−23. |A| = 12 + 2 = 14.
A (adj A) = 140014 = 14 I = |A| I.
Answer: adj A = 41−23
Find the inverse of 3524.
|A| = 12 − 10 = 2 ≠ 0. adj A = 4−5−23.
A⁻¹ = 124−5−23 = 2−52−132.
Answer: 2−52−132
Find A⁻¹ for A = 120013201.
Use: Cofactors, transpose, divide
|A| = 1(1 − 0) − 2(0 − 6) + 0 = 1 + 12 = 13 ≠ 0.
Cofactors: A₁₁ = 1, A₁₂ = 6, A₁₃ = −2; A₂₁ = −2, A₂₂ = 1, A₂₃ = 4; A₃₁ = 6, A₃₂ = −3, A₃₃ = 1.
adj A (the transpose) = 1−2661−3−241.
A⁻¹ = 1131−2661−3−241.
Answer: 1131−2661−3−241
Show that A = 2412 has no inverse, and verify that (adj A)A = O.
|A| = 4 − 4 = 0, so A is singular and A⁻¹ does not exist.
adj A = 2−4−12, and (adj A)A = 0000 = O = |A| I.
Answer: |A| = 0; no inverse
For A = 1235, show that A² − 6A − I = O and use it to find A⁻¹.
Use: Multiply by A⁻¹
A² = 7121831. A² − 6A − I = 7121831 − 6121830 − 1001 = O.
|A| = 5 − 6 = −1 ≠ 0. Multiplying by A⁻¹: A − 6I − A⁻¹ = O, so A⁻¹ = A − 6I = −523−1.
Answer: −523−1
Let A = 1201 and B = 2011. Verify (AB)⁻¹ = B⁻¹A⁻¹.
AB = 4211, |AB| = 4 − 2 = 2, and (AB)⁻¹ = 121−2−14 = 12−1−122.
A⁻¹ = 1−201 and B⁻¹ = 120−121. B⁻¹A⁻¹ = 12−1−122, the same.
Answer: verified
For A = 121012301, find |A| and verify |adj A| = |A|².
|A| = 1(1 − 0) − 2(0 − 6) + 1(0 − 3) = 1 + 12 − 3 = 10.
adj A = 1−236−2−2−361, and |adj A| = 100 = 10².
Answer: |A| = 10 and |adj A| = 100
Examine the consistency of x + y = 3, 2x + 2y = 7.
Use: |A| = 0, then (adj A)B
A = 1122, |A| = 2 − 2 = 0.
adj A = 2−1−21, B = 37, and (adj A)B = −11 ≠ O.
So the system has no solution: it is inconsistent.
Answer: inconsistent
Solve 3x + 2y = 8, x − y = 1 by the matrix method.
A = 321−1, |A| = −3 − 2 = −5 ≠ 0. A⁻¹ = 152515−35.
X = A⁻¹B = 152515−3581 = 21.
Answer: x = 2, y = 1
Solve x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2.
A = 1112−1112−1, |A| = 1(1 − 2) − 1(−2 − 1) + 1(4 + 1) = −1 + 3 + 5 = 7 ≠ 0.
A⁻¹ = −17372737−271757−17−37.
X = A⁻¹ 632 = 123.
Answer: x = 1, y = 2, z = 3
2 pens and 1 notebook cost ₹45, and 1 pen and 3 notebooks cost ₹85. Find the price of each using the matrix method.
Let a pen cost ₹x and a notebook ₹y: 2x + y = 45, x + 3y = 85.
A = 2113, |A| = 6 − 1 = 5. A⁻¹ = 35−15−1525.
X = A⁻¹B = 1025.
Answer: pen ₹10, notebook ₹25
The inverse of A = 1235 is −523−1. Use it to solve x + 2y = 4, 3x + 5y = 9.
Use: X = A⁻¹B
X = −523−149 = −20 + 1812 − 9 = −23.
Check: −2 + 2(3) = 4 and 3(−2) + 5(3) = 9.
Answer: x = −2, y = 3
For what values of k does 2x + ky = 3, 4x + 6y = 5 have a unique solution? Examine the case when it does not.
|A| = 12 − 4k. It is not 0 exactly when k ≠ 3, so the solution is unique for every k ≠ 3.
For k = 3: A = 2346, adj A = 6−3−42, B = 35, and (adj A)B = 3−2 ≠ O.
So the system is inconsistent when k = 3.
Answer: unique for k ≠ 3; inconsistent for k = 3
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24−5−1 = 2(−1) − 4(−5) = −2 + 20 = 18.
Answer: 18
(i) cos θ · cos θ − (−sin θ)(sin θ) = cos²θ + sin²θ = 1.
(ii) (x² − x + 1)(x + 1) − (x − 1)(x + 1) = (x³ + 1) − (x² − 1) = x³ − x² + 2.
(Here (x² − x + 1)(x + 1) = x³ + 1.)
Answer: (i) 1; (ii) x³ − x² + 2
|A| = 1·2 − 2·4 = −6. 2A = 2484 and |2A| = 2·4 − 4·8 = 8 − 32 = −24.
4|A| = 4(−6) = −24, so |2A| = 4|A|.
Answer: shown
A is upper triangular, so |A| = 1 × 1 × 4 = 4.
3A = 3030360012 is also triangular, so |3A| = 3 × 3 × 12 = 108.
27|A| = 27 × 4 = 108, so |3A| = 27|A|.
Answer: shown
(i) Expand along row 2, which has two zeros: only the entry −1 at (2, 3) remains. Its cofactor sign is (−1)²⁺³ = −1, so the determinant is (−1)(−1)3−13−5 = (−15 + 3) = −12.
(ii) 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) = 21 + 20 + 5 = 46.
(iii) Expand along row 1: 0 − 1((−1)(0) − (−3)(−2)) + 2((−1)(3) − 0·(−2)) = −1(0 − 6) + 2(−3) = 6 − 6 = 0.
(iv) Expand along row 1: 2(2·0 − (−1)(−5)) − (−1)(0·0 − (−1)(3)) + (−2)(0·(−5) − 2·3) = 2(−5) + 1(3) − 2(−6) = −10 + 3 + 12 = 5.
Answer: (i) −12; (ii) 46; (iii) 0; (iv) 5
Expand along row 1: |A| = 1(1(−9) − (−3)(4)) − 1(2(−9) − (−3)(5)) + (−2)(2·4 − 1·5) = 1(3) − 1(−3) − 2(3) = 3 + 3 − 6 = 0.
Answer: |A| = 0
(i) 2 − 20 = 2x² − 24, so −18 = 2x² − 24, so x² = 3 and x = ±√3.
(ii) 10 − 12 = 5x − 6x, so −2 = −x and x = 2.
Answer: (i) x = ±√3; (ii) x = 2
x² − 36 = 36 − 36 = 0, so x² = 36 and x = ±6.
Answer: (B) ±6
(i) Area = ½|101601431| = ½|1(0 − 3) − 0 + 1(18 − 0)| = ½|−3 + 18| = 15/2 square units.
(ii) Area = ½|2(1 − 8) + 1(8 − 7) + 10(7 − 1)| = ½|−14 + 1 + 60| = 47/2 square units.
(iii) Area = ½|−2(2 + 8) + 3(−8 + 3) + (−1)(−3 − 2)| = ½|−20 − 15 + 5| = ½(30) = 15 square units.
Answer: (i) 15/2; (ii) 47/2; (iii) 15 square units
Area = ½ab + c1bc + a1ca + b1.
Add column 2 to column 1 (C₁ → C₁ + C₂): every entry of the new first column is a + b + c.
= ½(a + b + c)1b + c11c + a11a + b1, and the first and third columns are identical, so the determinant is 0.
The area is 0, so A, B and C are collinear.
Answer: shown
(i) ½|k(0 − 2) − 0 + 1(8 − 0)| = ½|8 − 2k| = 4, so 8 − 2k = ±8, which gives k = 0 or k = 8.
(ii) ½|−2(4 − k) − 0 + 1(0)| = ½|2k − 8| = 4, so 2k − 8 = ±8, which gives k = 0 or k = 8.
Answer: (i) k = 0 or 8; (ii) k = 0 or 8
(i) For a point P(x, y) on the line, the area of the triangle with the two given points is 0: 121361xy1 = 0.
Expanding along row 1: 1(6 − y) − 2(3 − x) + 1(3y − 6x) = 6 − y − 6 + 2x + 3y − 6x = 2y − 4x = 0. So y = 2x.
(ii) 311931xy1 = 0. Expanding along row 1: 3(3 − y) − 1(9 − x) + 1(9y − 3x) = 9 − 3y − 9 + x + 9y − 3x = 6y − 2x = 0. So x − 3y = 0.
Answer: (i) y = 2x; (ii) x − 3y = 0
½|2−61541k41| = 35. Expanding along row 1: 2(4 − 4) + 6(5 − k) + 1(20 − 4k) = 50 − 10k.
So |50 − 10k| = 70, which gives 50 − 10k = 70 or 50 − 10k = −70, that is k = −2 or k = 12.
Answer: (D) 12, −2
(i) M₁₁ = 3, M₁₂ = 0, M₂₁ = −4, M₂₂ = 2. Cofactors: A₁₁ = (−1)²(3) = 3, A₁₂ = (−1)³(0) = 0, A₂₁ = (−1)³(−4) = 4, A₂₂ = (−1)⁴(2) = 2.
(ii) The determinant is acbd. M₁₁ = d, M₁₂ = b, M₂₁ = c, M₂₂ = a. Cofactors: A₁₁ = d, A₁₂ = −b, A₂₁ = −c, A₂₂ = a.
Answer: (i) M = 3, 0, −4, 2 and A = 3, 0, 4, 2; (ii) M = d, b, c, a and A = d, −b, −c, a
(i) M₁₁ = M₂₂ = M₃₃ = 1 and every other minor is 0 (each is a 2 × 2 determinant with a zero row or column). The cofactors are the same, since the signs only matter for the entries that are 0: A₁₁ = A₂₂ = A₃₃ = 1, all others 0.
(ii) Minors:
M11 = 11, M12 = 6, M13 = 3
M21 = -4, M22 = 2, M23 = 1
M31 = -20, M32 = -13, M33 = 5
Cofactors Aij = (−1)^(i+j) Mij: A11 = 11, A12 = -6, A13 = 3, A21 = 4, A22 = 2, A23 = -1, A31 = -20, A32 = 13, A33 = 5.
Answer: (i) A₁₁ = A₂₂ = A₃₃ = 1, all other minors and cofactors 0; (ii) see the minors and cofactors above
A₂₁ = (−1)³|3 8; 2 3| = −(9 − 16) = 7. A₂₂ = (−1)⁴|5 8; 1 3| = 15 − 8 = 7. A₂₃ = (−1)⁵|5 3; 1 2| = −(10 − 3) = −7.
Δ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃ = 2(7) + 0(7) + 1(−7) = 14 − 7 = 7.
Answer: Δ = 7
A₁₃ = |1 y; 1 z| = z − y. A₂₃ = −|1 x; 1 z| = −(z − x) = x − z. A₃₃ = |1 x; 1 y| = y − x.
Δ = yz(z − y) + zx(x − z) + xy(y − x) = yz² − y²z + zx² − z²x + xy² − x²y.
This equals (x − y)(y − z)(z − x): for example at (x, y, z) = (1, 2, 4) both are 6.
Answer: Δ = (x − y)(y − z)(z − x)
Δ is the sum of the products of the entries of one row or column with their own cofactors. Option (D) uses column 1 with its cofactors A₁₁, A₂₁, A₃₁. The others pair entries with cofactors of a different row or column, and so give 0.
Answer: (D)
A₁₁ = 4, A₁₂ = −3, A₂₁ = −2, A₂₂ = 1.
adj A = transpose of the cofactor matrix = 4−2−31.
(For a 2 × 2 matrix: swap the diagonal entries and change the sign of the other two.)
Answer: 4−2−31
The cofactors are A11 = 3, A12 = -12, A13 = 6, A21 = 1, A22 = 5, A23 = 2, A31 = -11, A32 = -1, A33 = 5.
adj A is the transpose of the cofactor matrix: 31−11−125−1625.
Answer: 31−11−125−1625
|A| = 2(−6) − 3(−4) = −12 + 12 = 0, so |A| I = O.
adj A = −6−342.
A (adj A) = 0000 and (adj A) A = 0000.
Both equal O = |A| I, so the relation holds.
Answer: verified
|A| = 1(0 − 0) + 1(9 + 2) + 2(0 − 0) = 11.
The cofactors are A11 = 0, A12 = -11, A13 = 0, A21 = 3, A22 = 1, A23 = -1, A31 = 2, A32 = 8, A33 = 3, so adj A = 032−11180−13.
A (adj A) = 110001100011 = 11 I, and (adj A) A = 110001100011 = 11 I.
Both equal |A| I.
Answer: verified
|A| = 14 ≠ 0, so A⁻¹ exists.
The cofactors are A11 = 3, A12 = -4, A21 = 2, A22 = 2.
adj A = 32−42 (the transpose of the cofactor matrix).
A⁻¹ = (1/|A|) adj A = 31417−2717
Answer: 31417−2717
|A| = 13 ≠ 0, so A⁻¹ exists.
The cofactors are A11 = 2, A12 = 3, A21 = -5, A22 = -1.
adj A = 2−53−1 (the transpose of the cofactor matrix).
A⁻¹ = (1/|A|) adj A = 213−513313−113
Answer: 213−513313−113
|A| = 10 ≠ 0, so A⁻¹ exists.
The cofactors are A11 = 10, A12 = 0, A13 = 0, A21 = -10, A22 = 5, A23 = 0, A31 = 2, A32 = -4, A33 = 2.
adj A = 10−10205−4002 (the transpose of the cofactor matrix).
A⁻¹ = (1/|A|) adj A = 1−115012−250015
Answer: 1−115012−250015
|A| = -3 ≠ 0, so A⁻¹ exists.
The cofactors are A11 = -3, A12 = 3, A13 = -9, A21 = 0, A22 = -1, A23 = -2, A31 = 0, A32 = 0, A33 = 3.
adj A = −3003−10−9−23 (the transpose of the cofactor matrix).
A⁻¹ = (1/|A|) adj A = 100−1130323−1
Answer: 100−1130323−1
|A| = -3 ≠ 0, so A⁻¹ exists.
The cofactors are A11 = -1, A12 = -4, A13 = 1, A21 = 5, A22 = 23, A23 = -11, A31 = 3, A32 = 12, A33 = -6.
adj A = −153−423121−11−6 (the transpose of the cofactor matrix).
A⁻¹ = (1/|A|) adj A = 13−53−143−233−4−131132
Answer: 13−53−143−233−4−131132
|A| = -1 ≠ 0, so A⁻¹ exists.
The cofactors are A11 = 2, A12 = -9, A13 = -6, A21 = 0, A22 = -2, A23 = -1, A31 = -1, A32 = 3, A33 = 2.
adj A = 20−1−9−23−6−12 (the transpose of the cofactor matrix).
A⁻¹ = (1/|A|) adj A = −20192−361−2
Answer: −20192−361−2
|A| = 1(−cos²α − sin²α) = −1 ≠ 0, so A⁻¹ exists.
The cofactors: A₁₁ = −cos²α − sin²α = −1, A₁₂ = A₁₃ = A₂₁ = A₃₁ = 0, A₂₂ = −cos α, A₂₃ = −sin α, A₃₂ = −sin α, A₃₃ = cos α.
adj A = −1000−cos α−sin α0−sin αcos α.
A⁻¹ = (1/|A|) adj A = −adj A = 1000cos αsin α0sin α−cos α, which is A itself (A² = I).
Answer: A⁻¹ = A
AB = 67874761, and |AB| = 67·61 − 87·47 = 4087 − 4089 = −2.
adj(AB) = 61−87−4767, so (AB)⁻¹ = 1-261−87−4767 = −612872472−672.
|A| = 15 − 14 = 1, A⁻¹ = 5−7−23. |B| = 54 − 56 = −2, B⁻¹ = −92472−3.
B⁻¹A⁻¹ = −612872472−672, which equals (AB)⁻¹.
Answer: verified
A² = 85−53. A² − 5A + 7I = 85−53 − 155−510 + 7007 = 0000.
Multiply A² − 5A + 7I = O by A⁻¹: A − 5I + 7A⁻¹ = O, so A⁻¹ = (5I − A)/7 = 172−113 = 27−171737.
Answer: 27−171737
A² = 11843. A² + aA + bI = 11 + 3a + b8 + 2a4 + a3 + a + b = O.
From 8 + 2a = 0: a = −4. From 4 + a = 0: a = −4. Then 11 − 12 + b = 0 gives b = 1, and 3 − 4 + 1 = 0 also holds.
Answer: a = −4, b = 1
A² = 421−38−147−314, A³ = A²A = 871−2327−6932−1358.
A³ − 6A² + 5A + 11I = 871−2327−6932−1358 − 24126−1848−8442−1884 + 555510−1510−515 + 110001100011 = 000000000.
|A| = 1(6 − 3) − 1(3 + 6) + 1(−1 − 4) = −11 ≠ 0, so A⁻¹ exists. Multiplying the equation by A⁻¹: A² − 6A + 5I + 11A⁻¹ = O, so A⁻¹ = −111(A² − 6A + 5I).
A² − 6A + 5I = 3−4−5−914−531, so A⁻¹ = −311411511911−111−411511−311−111.
Answer: −311411511911−111−411511−311−111
A² = 6−55−56−55−56, A³ = 22−2121−2122−2121−2122.
A³ − 6A² + 9A − 4I = 22−2121−2122−2121−2122 − 36−3030−3036−3030−3036 + 18−99−918−99−918 − 400040004 = 000000000.
|A| = 2(4 − 1) + 1(−2 + 1) + 1(1 − 2) = 6 − 1 − 1 = 4 ≠ 0. Multiply by A⁻¹: A² − 6A + 9I − 4A⁻¹ = O, so A⁻¹ = 14(A² − 6A + 9I).
A² − 6A + 9I = 31−1131−113, so A⁻¹ = 3414−14143414−141434.
Answer: 3414−14143414−141434
A (adj A) = |A| I. Taking determinants: |A||adj A| = |A|³, and |A| ≠ 0, so |adj A| = |A|².
Answer: (B) |A|²
AA⁻¹ = I gives det(A) det(A⁻¹) = det(I) = 1, so det(A⁻¹) = 1/det(A).
Answer: (B) 1/det(A)
System: x + 2y = 2, 2x + 3y = 3. |A| = -1 ≠ 0, so A is non-singular and the system has a unique solution; it is consistent.
Solution: x = 0, y = 1.
Answer: consistent
System: 2x − y = 5, x + y = 4. |A| = 3 ≠ 0, so A is non-singular and the system has a unique solution; it is consistent.
Solution: x = 3, y = 1.
Answer: consistent
System: x + 3y = 5, 2x + 6y = 8. |A| = 0.
(adj A)B = 6−2 ≠ O, so the system has no solution. It is inconsistent.
(Check: x + 3y = 5 would give 2x + 6y = 10, which contradicts 2x + 6y = 8.)
Answer: inconsistent
A = 111232aa2a. Expanding along row 1: |A| = 1(6a − 2a) − 1(4a − 2a) + 1(2a − 3a) = 4a − 2a − a = a.
If a ≠ 0, |A| ≠ 0 and the system is consistent with a unique solution: from the first two equations, 2(first) − (second) gives y = 0, and then x = 2 − 4/a and z = 4/a − 1.
If a = 0, the third equation becomes 0 = 4, which cannot hold, so the system is inconsistent.
Answer: consistent for a ≠ 0 (with x = 2 − 4/a, y = 0, z = 4/a − 1); inconsistent for a = 0
System: 3x − y − 2z = 2, 2y − z = −1, 3x − 5y = 3. |A| = 0.
(adj A)B = −5−3−6 ≠ O, so the system has no solution. It is inconsistent.
Answer: inconsistent
System: 5x − y + 4z = 5, 2x + 3y + 5z = 2, 5x − 2y + 6z = −1. |A| = 51 ≠ 0, so A is non-singular and the system has a unique solution; it is consistent.
Solution: x = 3, y = 2, z = -2.
Answer: consistent (x = 3, y = 2, z = −2)
Write the system as AX = B, where A = 5273, X = xy and B = 45.
|A| = 1 ≠ 0, so A is non-singular and the solution is unique, X = A⁻¹B.
A⁻¹ = 3−2−75.
X = A⁻¹B = 2−3, so x = 2, y = -3.
Answer: x = 2, y = −3
Write the system as AX = B, where A = 2−134, X = xy and B = −23.
|A| = 11 ≠ 0, so A is non-singular and the solution is unique, X = A⁻¹B.
A⁻¹ = 411111−311211.
X = A⁻¹B = −5111211, so x = -5/11, y = 12/11.
Answer: x = −5/11, y = 12/11
Write the system as AX = B, where A = 4−33−5, X = xy and B = 37.
|A| = -11 ≠ 0, so A is non-singular and the solution is unique, X = A⁻¹B.
A⁻¹ = 511−311311−411.
X = A⁻¹B = −611−1911, so x = -6/11, y = -19/11.
Answer: x = −6/11, y = −19/11
Write the system as AX = B, where A = 5232, X = xy and B = 35.
|A| = 4 ≠ 0, so A is non-singular and the solution is unique, X = A⁻¹B.
A⁻¹ = 12−12−3454.
X = A⁻¹B = −14, so x = -1, y = 4.
Answer: x = −1, y = 4
Write the system as AX = B, where A = 2111−2−103−5, X = xyz and B = 1329.
|A| = 34 ≠ 0, so A is non-singular and the solution is unique, X = A⁻¹B.
A⁻¹ = 1334417134534−517334334−317−534.
X = A⁻¹B = 112−32, so x = 1, y = 1/2, z = -3/2.
Answer: x = 1, y = 1/2, z = −3/2
Write the system as AX = B, where A = 1−1121−3111, X = xyz and B = 402.
|A| = 10 ≠ 0, so A is non-singular and the solution is unique, X = A⁻¹B.
A⁻¹ = 251515−12012110−15310.
X = A⁻¹B = 2−11, so x = 2, y = -1, z = 1.
Answer: x = 2, y = −1, z = 1
Write the system as AX = B, where A = 2331−213−1−2, X = xyz and B = 5−43.
|A| = 40 ≠ 0, so A is non-singular and the solution is unique, X = A⁻¹B.
A⁻¹ = 1834094018−1340140181140−740.
X = A⁻¹B = 12−1, so x = 1, y = 2, z = -1.
Answer: x = 1, y = 2, z = −1
Write the system as AX = B, where A = 1−1234−52−13, X = xyz and B = 7−512.
|A| = 4 ≠ 0, so A is non-singular and the solution is unique, X = A⁻¹B.
A⁻¹ = 7414−34−194−14114−114−1474.
X = A⁻¹B = 213, so x = 2, y = 1, z = 3.
Answer: x = 2, y = 1, z = 3
|A| = 2(−4 + 4) + 3(−6 + 4) + 5(3 − 2) = 0 − 6 + 5 = −1 ≠ 0.
The cofactors are A11 = 0, A12 = 2, A13 = 1, A21 = -1, A22 = -9, A23 = -5, A31 = 2, A32 = 23, A33 = 13, so adj A = 0−122−9231−513.
A⁻¹ = (1/|A|) adj A = −adj A = 01−2−29−23−15−13.
The system is AX = B with B = 11−5−3, so X = A⁻¹B = 123.
Answer: A⁻¹ = 01−2−29−23−15−13; x = 1, y = 2, z = 3
Let the prices per kg of onion, wheat and rice be x, y and z (in ₹). Then 4x + 3y + 2z = 60, 2x + 4y + 6z = 90, 6x + 2y + 3z = 70.
A = 432246623, X = xyz, B = 609070.
|A| = 4(12 − 12) − 3(6 − 36) + 2(4 − 24) = 0 + 90 − 40 = 50 ≠ 0.
A⁻¹ = 0−11015350−25−251515.
X = A⁻¹B = 588.
Answer: onion ₹5, wheat ₹8, rice ₹8 per kg
Expand along row 1: x(−x·x − 1·1) − sin θ(−sin θ·x − 1·cos θ) + cos θ(−sin θ·1 − (−x)cos θ).
= x(−x² − 1) + sin θ(x sin θ + cos θ) + cos θ(−sin θ + x cos θ)
= −x³ − x + x sin²θ + sin θ cos θ − sin θ cos θ + x cos²θ = −x³ − x + x(sin²θ + cos²θ) = −x³.
The value −x³ does not contain θ.
Answer: The determinant equals −x³, independent of θ
Expand along row 2 (which has a zero): −(−sin β)·(cos α sin β·cos α − (−sin α)(sin α sin β)) + cos β·(cos α cos β·cos α − (−sin α)(sin α cos β)) − 0.
The first bracket: cos²α sin β + sin²α sin β = sin β. The second bracket: cos²α cos β + sin²α cos β = cos β.
So the value is sin β · sin β + cos β · cos β = 1.
Answer: 1
(AB)⁻¹ = B⁻¹A⁻¹.
|B| = 1(3 − 0) − 2(−1 − 0) − 2(2 − 0) = 3 + 2 − 4 = 1 ≠ 0. The cofactors of B are A11 = 3, A12 = 1, A13 = 2, A21 = 2, A22 = 1, A23 = 2, A31 = 6, A32 = 2, A33 = 5, so B⁻¹ = adj B = 326112225.
(AB)⁻¹ = B⁻¹A⁻¹ = 3261122253−11−156−55−22 = 9−35−210102.
Answer: 9−35−210102
|A| = 1(15 − 1) − 2(10 − 1) + 1(2 − 3) = 14 − 18 − 1 = −5. The cofactors are A11 = 14, A12 = -9, A13 = -1, A21 = -9, A22 = 4, A23 = 1, A31 = -1, A32 = 1, A33 = -1, so adj A = 14−9−1−941−11−1 and A⁻¹ = −145951595−45−1515−1515.
(i) |adj A| = |A|² = 25 ≠ 0, so adj A is invertible. Since A(adj A) = |A| I, we get [adj A]⁻¹ = A/|A| = −15−25−15−25−35−15−15−15−1.
Now adj(A⁻¹) has the cofactors of A⁻¹; computing them gives −15−25−15−25−35−15−15−15−1, which is the same matrix. So [adj A]⁻¹ = adj(A⁻¹).
(ii) A⁻¹ is invertible since |A⁻¹| = −1/5 ≠ 0, and (A⁻¹)⁻¹ = 121231115 = A.
Answer: verified
Add rows 2 and 3 to row 1 (R₁ → R₁ + R₂ + R₃): every entry of row 1 becomes 2(x + y).
= 2(x + y)111yx + yxx + yxy. Now C₂ → C₂ − C₁ and C₃ → C₃ − C₁:
= 2(x + y)100yxx − yx + y−y−x = 2(x + y)[x(−x) − (x − y)(−y)] = 2(x + y)(−x² + xy − y²).
Since (x + y)(x² − xy + y²) = x³ + y³, the value is −2(x³ + y³).
Answer: −2(x³ + y³)
Subtract row 1 from rows 2 and 3 (R₂ → R₂ − R₁, R₃ → R₃ − R₁):
1xy0y000x.
This is upper triangular, so the value is 1 · y · x = xy.
Answer: xy
Put u = 1/x, v = 1/y, w = 1/z. The system becomes 2u + 3v + 10w = 4, 4u − 6v + 5w = 1, 6u + 9v − 20w = 2, that is AX = B with A = 23104−6569−20, X = uvw, B = 412.
|A| = 2(120 − 45) − 3(−80 − 30) + 10(36 + 36) = 150 + 330 + 720 = 1200 ≠ 0, so A is non-singular.
Solving X = A⁻¹B gives u = 1/2, v = 1/3, w = 1/5. (Check: 1 + 1 + 2 = 4, 2 − 2 + 1 = 1, 3 + 3 − 4 = 2.)
So x = 2, y = 3, z = 5.
Answer: x = 2, y = 3, z = 5
|A| = xyz ≠ 0. The cofactors are A₁₁ = yz, A₂₂ = xz, A₃₃ = xy and all others 0, so adj A = yz000xz000xy.
A⁻¹ = (1/xyz) adj A = 1/x0001/y0001/z = x⁻¹000y⁻¹000z⁻¹.
Answer: (A)
Expand along row 1: Det(A) = 1(1 + sin²θ) − sin θ(−sin θ + sin θ) + 1(sin²θ + 1) = 2 + 2 sin²θ.
Since 0 ≤ sin²θ ≤ 1, we get 2 ≤ Det(A) ≤ 4. Both ends are reached (sin θ = 0 gives 2, sin θ = ±1 gives 4), so Det(A) ∈ [2, 4].
Answer: (D) Det(A) ∈ [2, 4]