Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

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Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Plus Two Maths Determinants Three Mark Questions and Answers

Question 1.
Using properties of determinants prove \(\left|\begin{array}{ccc}{x} & {y} & {x+y} \\{y} & {x+y} & {x} \\{x+y} & {x} & {y}\end{array}\right|\) = -2(x3 + y3).
Answer:
Plus Two Maths Determinants 3 Mark Questions and Answers 1
= 2(x + y)(-x2 + xy – y2) = -2(x3 + y3).

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 2.
If a, b, c are real numbers and \(\left|\begin{array}{lll}{b+c} & {c+a} & {a+b} \\{c+a} & {a+b} & {b+c} \\{a+b} & {b+c} & {c+a}\end{array}\right|\) = 0, Show that a = b = c.
Answer:
Plus Two Maths Determinants 3 Mark Questions and Answers 2
2(a + b + c) [(b – c) (c – b) – (b – a) (c – a)] =0 (a+b+c) = 0
(a + b + c) = 0 or (b – c) (c – b) = (b – a) (c – a)
(a + b + c) = 0 or a = b = c.

Question 3.
Solve using properties of determinants.
\(\left|\begin{array}{ccc}{2 x-1} & {x+7} & {x+4} \\{x} & {6} & {2} \\{x-1} & {x+1} & {3}
\end{array}\right|\) = 0
Answer:
Plus Two Maths Determinants 3 Mark Questions and Answers 3
⇒ (x – 1) (x2 + x – 6x + 6) = 0
⇒ (x – 1)(x2 – 5x + 6) = 0
⇒ (x – 1) (x – 3) (x – 2) = 0
⇒ x = 1, x = 3, x = 2.

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 4.
If \(\left|\begin{array}{cc}{3} & {x} \\{x} & {x}\end{array}\right|=\left|\begin{array}{cc}{-2} & {2} \\{4} & {1}\end{array}\right|\), find the value of x.
Answer:
\(\left|\begin{array}{cc}{3} & {x} \\{x} & {x}\end{array}\right|=\left|\begin{array}{cc}{-2} & {2} \\{4} & {1}\end{array}\right|\)
⇒ 3x – x2 = – 2 – 8
⇒ x2 – 3x – 10 = 0
⇒ x = 5, -2.

Question 5.
A = \(\left[\begin{array}{ccc}{1} & {-3} & {1} \\{2} & {0} & {4} \\{1} & {2} & {-2}
\end{array}\right]\)

  1. Calculate |A| (1)
  2. Find |adjA| {Hint: using the property A × adjA = |A|I} (1)
  3. Find |3A| (1)

Answer:
1. |A| = \(\left[\begin{array}{ccc}{1} & {-3} & {1} \\{2} & {0} & {4} \\{1} & {2} & {-2}
\end{array}\right]\) = – 28.

2. A × adjA = |A|I
Plus Two Maths Determinants 3 Mark Questions and Answers 4

3. |3A| = 27 × |A| = 27 × -28 = -756.

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 6.
Using properties of determinants prove the following.
Plus Two Maths Determinants 3 Mark Questions and Answers 5
Answer:
Plus Two Maths Determinants 3 Mark Questions and Answers 6
Plus Two Maths Determinants 3 Mark Questions and Answers 7

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants
Plus Two Maths Determinants 3 Mark Questions and Answers 8
= 2{-{-c){b{a – c)) – b(-c(c + a))}
= 2{c(ab – cb) + b(c2 + ac)}
= 2{abc – c2b + bc2 + abc)} = 4abc.

Plus Two Maths Determinants Four Mark Questions and Answers

Question 1.
(i) If \(\left|\begin{array}{rrr}{1} & {-3} & {2} \\{4} & {-1} & {2} \\{3} & {5} & {2}\end{array}\right|\) = 40, then \(\left|\begin{array}{ccc}{1} & {4} & {3} \\{-3} & {-1} & {5} \\{2} & {2} & {2}\end{array}\right|\) = ?
(a)   0
(b)  – 40
(c)  40
(d)  2 (1)
(ii) \(\left|\begin{array}{rrr}{3} & {-3} & {2} \\{12} & {-1} & {2} \\{9} & {5} & {2}\end{array}\right|\) = ?
(a) 120
(b)  40
(c)  – 40
(d)  0 (1)
(iii) Show that ∆ = \(\left|\begin{array}{ccc}{-a^{2}} & {a b} & {a c} \\{b a} & {-b^{2}} & {b c} \\{a c} & {b c} & {-c^{2}}\end{array}\right|\) = 4a2b2c2 (2)
Answer:
(i) (c) 40

(ii) (a)120

(iii) ∆ = abc\(\left|\begin{array}{ccc}{-a} & {a} & {a} \\{b} & {-b} & {b} \\{c} & {c} & {-c}\end{array}\right|\) take a, b, c from C1, C2, C3
Plus Two Maths Determinants 4 Mark Questions and Answers 9

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 2.
Plus Two Maths Determinants 4 Mark Questions and Answers 10
Answer:
(i) \(\left|\begin{array}{ll}{2} & {4} \\{5} & {1}\end{array}\right|=\left|\begin{array}{ll}{2 x} & {4} \\{6} & {x}\end{array}\right|\) ⇒ -18 = 2x2 – 24.
⇒ 2x2 = 6 ⇒ x2 = 3 ⇒ x = \(\pm \sqrt{3}\).

Plus Two Maths Determinants 4 Mark Questions and Answers 11
Plus Two Maths Determinants 4 Mark Questions and Answers 12

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants
Question 3.
Prove that \(\left|\begin{array}{ccc}{(b+c)^{2}} & {a^{2}} & {a^{2}} \\{b^{2}} & {(c+a)^{2}} & {b^{2}} \\{c^{2}} & {c^{2}} & {(a+b)^{2}}\end{array}\right|\) = 2abc(a + b + c)3.
Answer:
Plus Two Maths Determinants 4 Mark Questions and Answers 13
Plus Two Maths Determinants 4 Mark Questions and Answers 14
= (a + b + c)2 × 2ab [(b + c) (c + a) – ab]
= (a + b + c)2 × 2ab [bc + ab + c2 + ac – ab)
= (a + b + c)2 × 2abc [a + b + c]
= 2abc (a + b + c)3.

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 4.
(i) Let the value of a determinant is ∆. Then the value of a determinant obtained by interchanging two rows is
(a) ∆
(b) -∆
(c) 0
(d) 1 (1)
(ii) Show that \(\left|\begin{array}{ccc}{a+b} & {b+c} & {c+a} \\{b+c} & {c+a} & {a+b} \\{c+a} & {a+b} & {b+c}\end{array}\right|=2\left|\begin{array}{lll}{a} & {b} & {c} \\{b} & {c} & {a} \\{c} & {a} & {b}\end{array}\right|\) (3)
Answer:
(i) (b) -∆

(ii) Operating C1 → C1 + C2 + C3, we have
Plus Two Maths Determinants 4 Mark Questions and Answers 15
Plus Two Maths Determinants 4 Mark Questions and Answers 16

Question 5.
Test the consistency 3x – y – 2z = 2, 2y – z = -1, 3x – 5y = 3.
Answer:
The given system of equations can be put in the matrix form, AX = B, where
Plus Two Maths Determinants 4 Mark Questions and Answers 17
|A| = 3(0 – 5) + 1(0 + 3) – 2(0 – 6) = 0
C11 = -5, C12 = -3, C21 = -6, C22 = 10, C23 = 6, C31 = 12, C32 = 5, C33 = 6
Plus Two Maths Determinants 4 Mark Questions and Answers 18
Plus Two Maths Determinants 4 Mark Questions and Answers 19
Therefore the system is inconsistent and has no solutions.

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 6.
Consider the system of equations 2x – 3y = 7 and 3x + 4y = 5

  1. Express the system in AX = B form. (1)
  2. Find adj A (2)
  3. Solve the system of equations. (1)

Answer:
1. |A| = \(\left|\begin{array}{cc}{2} & {-3} \\{3} & {4}\end{array}\right|\) = 8 + 9 = 17.

2. c11 = 4, c12 = -3, c21 = 3, c22 = 2,
Plus Two Maths Determinants 4 Mark Questions and Answers 20

3. The given equations can be expressed in the form AX = B,
Plus Two Maths Determinants 4 Mark Questions and Answers 21

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 7.
(i) If A and B are matrices of order 3 such that|A| = -1; |B| = 3, then |3AB| is
(a) -9
(b) -27
(c) -81
(d) 9 (1)
(ii) If A = \(\left[\begin{array}{cc}{1} & {\tan x} \\{-\tan x} & {1}\end{array}\right]\), Show that AT A-1 = \(\left[\begin{array}{cc}{\cos 2 x} & {-\sin 2 x} \\{\sin 2 x} & {\cos 2 x}\end{array}\right]\) (3)
Answer:
(i) (c) -81 (since |3AB| = 27|A||B|).

(ii) |A| = \(\left[\begin{array}{cc}{1} & {\tan x} \\{-\tan x} & {1}\end{array}\right]\) = sec2x ≠ 0, therefore A is invertible.
Plus Two Maths Determinants 4 Mark Questions and Answers 22

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 8.
Consider the determinant ∆ = \(\left|\begin{array}{ccc}{x} & {x^{2}} & {1+x^{3}} \\{y} & {y^{2}} & {1+y^{3}} \\{z} & {z^{2}} & {1+z^{3}}\end{array}\right|\), Where x, y, z, are different.
(i) Express the above determinant as sum of two determinants. (1)
(ii) Show that if ∆ = 0, then 1 + xyz = 0. (3)
Answer:
(i) Given,
∆ = \(\left|\begin{array}{ccc}{x} & {x^{2}} & {1+x^{3}} \\{y} & {y^{2}} & {1+y^{3}} \\{z} & {z^{2}} & {1+z^{3}}\end{array}\right|=\left|\begin{array}{ccc}{x} & {x^{2}} & {1} \\{y} & {y^{2}} & {1} \\{z} & {z^{2}} & {1}\end{array}\right|+\left|\begin{array}{ccc}{x} & {x^{2}} & {x^{3}} \\{y} & {y^{2}} & {y^{3}} \\{z} & {z^{2}} & {z^{3}}\end{array}\right|\)

Plus Two Maths Determinants 4 Mark Questions and Answers 23
Plus Two Maths Determinants 4 Mark Questions and Answers 24
Given, ∆ = 0 ⇒ (1 + xyz)(y – x)(z – x)(z – y) = 0 ⇒ 1 + xyz = 0
∵ x ≠ y ≠ z.

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 9.
(i) The value of the determinant \(\left|\begin{array}{cc}{\sin 10} & {-\cos 10} \\{\sin 80} & {\cos 80}\end{array}\right|\) is
(a) – 1
(b) 1
(c) 0
(d) – 2 (1)
(ii) Using properties of determinants, show that (3)
\(\left|\begin{array}{lll}{a} & {a^{2}} & {b+c} \\{b} & {b^{2}} & {c+a} \\{c} & {c^{2}} & {a+b}\end{array}\right| = (b – c) (c – a) (a – b) (a + b + c)\)
Answer:
(i) (b) Since,
sin 10 cos 80 + cos 10 sin 80 = sin (10 + 80) =sin 90 = 1.

(ii) Let C3 → C3 + C1
Plus Two Maths Determinants 4 Mark Questions and Answers 25
= (a + b + c)(b – a)(c – a)(c + a – b – a)
= (a + b + c)(b – a)(c – a)(c – b)
= (b – c)(c – a)(a – b)(a + b + c).

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 10.
(i) Choose the correct answer from the bracket. Consider a square matrix of order 3. Let C11, C12, C13 are cofactors of the elements a11, a12, a13 respectively, then a11C11 + a12C12 + a13C13 is (1)
(a) 0
(b) |A|
(c) 1
(d) none of these.
(ii) Verify A(adjA) = (adjA)A = |A|I for the matrix A = \(\left[\begin{array}{ll}{5} & {-2} \\{3} & {-2}\end{array}\right]\) that, where I = \(\left[\begin{array}{ll}{1} & {0} \\{0} & {1}\end{array}\right]\) (3)
Answer:
(i) (b) |A|

(ii) |A| = \(\left|\begin{array}{cc}{5} & {-2} \\{3} & {-2}\end{array}\right|\) = – 4
C11 = – 2, C12 = – 3, C21 = 2, C22 = 5
Plus Two Maths Determinants 4 Mark Questions and Answers 26
Hence A(adjA) = (adjA)A = |A|I.

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 11.
Consider the following system of equations x + 2y = 4,2x + 5y = 9

  1. If A = \(\left[\begin{array}{ll}{1} & {2} \\{2} & {5}\end{array}\right]\), find |A| (1)
  2. Express the above system of equations in the form AX = B (1)
  3. Find adj A, A-1 (1)
  4. Solve the system of equations. (1)

Answer:
1. |A| = \(\left[\begin{array}{ll}{1} & {2} \\{2} & {5}\end{array}\right]\) = 5 – 4 = 1

2. The given system of equation can be expressed in the form AX = B.
Plus Two Maths Determinants 4 Mark Questions and Answers 27

3. Cofactor matrix of A = \(\left[\begin{array}{cc}{5} & {-2} \\{-2} & {1}\end{array}\right]\)
Plus Two Maths Determinants 4 Mark Questions and Answers 28

4. We have,
Plus Two Maths Determinants 4 Mark Questions and Answers 29
x = 2, y = 1.

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 12.
Consider the point X(-2, -3), B(3, 2), C(-1, -8)

  1. Find the area of ∆ABC (2)
  2. Find third vertex of any other triangle with same area and base AB. (2)

Answer:
1. \(\frac{1}{2}\left|\begin{array}{ccc}{-2} & {-3} & {1} \\{3} & {2} & {1} \\{-1} & {-8} & {1}\end{array}\right|\)
\(\frac{1}{2}\) (- 2(2 + 8) + 3(3 + 1) + 1(- 24 + 2)) = – 15
Area of ∆ ABC = 15.

2. The base AB is fixed and the third point is variable. Therefore we can choose any x coordinate and find y coordinate or vice versa.
Plus Two Maths Determinants 4 Mark Questions and Answers 30
⇒ – 2(2 – y) + 3(3 – 1) + 1(3y – 2) = 30
⇒ – 4 + 2y + 6 + 3y – 2 = 30
⇒ 5y = 30 ⇒ y – 6
Therefore point is(1, 6).

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 13.
Find the inverse of the following
Plus Two Maths Determinants 4 Mark Questions and Answers 31
Answer:
(i) Let |A| = \(\left|\begin{array}{lll}{1} & {2} & {3} \\{0} & {2} & {4} \\{0} & {0} & {5}\end{array}\right|\) = 10
C11 = 10, C12 = 0, C13 = 0, C21 = – 10, C22 = 5, C23 = 0, C31 = – 2, C32 = – 4, C33 = 2
Plus Two Maths Determinants 4 Mark Questions and Answers 32

(ii) Let |A| = \(\left|\begin{array}{ccc}{1} & {0} & {0} \\{3} & {3} & {0} \\{5} & {2} & {-1}\end{array}\right|\) = -3
C11 = -3, C12 = 3, C13 = -9, C21 = 0, C22 = -1, C23 = -2, C31 = 0, C32 = 0, C33 = 3
Plus Two Maths Determinants 4 Mark Questions and Answers 33

(iii) Let |A| = \(\left|\begin{array}{ccc}{2} & {1} & {3} \\{4} & {-1} & {0} \\{-7} & {2} & {1}\end{array}\right|\)
= 2(-1 – 0) -1(4 – 0) + 3(8 – 7) = -3
C11 = -1, C12 = -4, C13 = 1, C21 = 5, C22 = 23, C23 = -11, C31 = 3, C32 = 12, C33 = -6

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants
Plus Two Maths Determinants 4 Mark Questions and Answers 34

(iv) Let |A| = \(\left|\begin{array}{ccc}{1} & {-1} & {2} \\{0} & {2} & {-3} \\{3} & {-2} & {4}\end{array}\right|\)
= 1(8 – 6) + 1(0 + 9) + 2(0 – 6) = -1
C11 = 2, C12 = -9, C13 = -6, C21 = 0, C22 = -2, C23 = -1, C31 = 3, C32 = 3, C33 = 2
Plus Two Maths Determinants 4 Mark Questions and Answers 35

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 14.
Consider the system of equations 5x + 2y = 4, 7x + 3y = 5. If A = \(\left[\begin{array}{ll}{5} & {2} \\{7} & {3}\end{array}\right]\), X = \(\left[\begin{array}{l}{\mathrm{r}} \\{y}\end{array}\right]\) and B = \(\left[\begin{array}{l}{4} \\{5}\end{array}\right]\)

  1. Find |A| (1)
  2. Find A-1 (2)
  3. Solve the above system of equations. (1)

Answer:
1. |A| = \(\left|\begin{array}{ll}{5} & {2} \\{7} & {3}\end{array}\right|\) = 15 – 14 = 1.

2. Given, A = \(\left[\begin{array}{ll}{5} & {2} \\{7} & {3}\end{array}\right]\)
Plus Two Maths Determinants 4 Mark Questions and Answers 36

3. X = A-1B
Plus Two Maths Determinants 4 Mark Questions and Answers 37
⇒ x = 2, y = -3.

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Plus Two Maths Determinants Six Mark Questions and Answers

Question 1.
(i) Let A be a square matrix of order ‘n’ then |KA| = …….. (1)
(ii) Find x if \(\left|\begin{array}{cc}{x} & {2} \\{18} & {x}\end{array}\right|=\left|\begin{array}{cc}{6} & {2} \\{18} & {6}\end{array}\right|\) (2)
(iii) Choose the correct answer from the bracket. The value of the determinant \(\left|\begin{array}{ccc}{0} & {p-q} & {p-r} \\{q-p} & {0} & {q-r} \\{r-p} & {r-q} & {0}
\end{array}\right|\) is ….. (1)
(iv) Consider \(\left|\begin{array}{ccc}{a} & {a+b} & {a+b+c} \\{2 a} & {3 a+2 b} & {4 a+3 b+2 c} \\{3 a} & {6 a+3 b} & {10 a+6 b+3 c}\end{array}\right|\) (2)
Answer:
(i) If A be a square matrix of order n, then |KA| = Kn|A|

(ii) \(\left|\begin{array}{cc}{x} & {2} \\{18} & {x}\end{array}\right|=\left|\begin{array}{cc}{6} & {2} \\{18} & {6}\end{array}\right|\) ⇒ x2 – 36 = 0
⇒ x2 = 36 ⇒ x = ±6.

(iii) (c) 0 (since the given determinant is the determinant of a third order skew symmetric matrix)

Plus Two Maths Determinants 6 Mark Questions and Answers 38
= a [7a2 + 3ab – 6a2 – 3ab] = a(a2) = a3

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 2.
(i) Let \(\left|\begin{array}{lll}{1} & {3} & {2} \\{2} & {0} & {1} \\{3} & {4} & {3}
\end{array}\right|\) = 3, then what is the value of \(\left|\begin{array}{lll}{1} & {3} & {2} \\{4} & {0} & {2} \\{3} & {4} & {3}\end{array}\right|\) = ? and\(\left|\begin{array}{lll}{6} & {7} & {6} \\{2} & {0} & {1} \\{3} & {4} & {3}\end{array}\right| \) = ? (2)
(Hint: Use the properties of determinants)
(ii) Using properties of determinants show that (4)
\(\left|\begin{array}{ccc}{1+a} & {1} & {1} \\{1} & {1+b} & {1} \\{1} & {1} & {1+c}\end{array}\right|=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Answer:
Plus Two Maths Determinants 6 Mark Questions and Answers 39

(ii) Taking ‘a’ from R1, ‘b‘ from R2,’C’ from R3
Plus Two Maths Determinants 6 Mark Questions and Answers 40
Plus Two Maths Determinants 6 Mark Questions and Answers 41

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants
Question 3.
If A = \(\left[\begin{array}{ccc}{2} & {-3} & {5} \\{3} & {2} & {-4} \\{1} & {1} & {-2}\end{array}\right]\)

  1. Find |A| (1)
  2. Find adj.A. (2)
  3. Solve 2x – 3y + 5z = 11, 3x + 2y – 4z = -5, x + y – 2z = -3 (3)

Answer:
1. A = \(\left[\begin{array}{ccc}{2} & {-3} & {5} \\{3} & {2} & {-4} \\{1} & {1} & {-2}\end{array}\right]\)
|A| = 2 × 0 + 3x – 2 + 5 = -1.

2. Co.factor A
Plus Two Maths Determinants 6 Mark Questions and Answers 42

3. Given
Plus Two Maths Determinants 6 Mark Questions and Answers 43
i.e; AX = B ⇒ X = A-1 B
Plus Two Maths Determinants 6 Mark Questions and Answers 44

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 4.
Let A = \(\left[\begin{array}{ccc}{1} & {-1} & {1} \\{2} & {1} & {-3} \\{1} & {1} & {1}\end{array}\right]\)

  1. Is A singular? (1)
  2. Find adj A. (2)
  3. Obtain A-1 (1)
  4. Using A-1 solve the system of equations x – y + z = 4, 2x + y – 3z = 0, x + y + z = 2 (2)

Answer:
1. A = \(\left[\begin{array}{ccc}{1} & {-1} & {1} \\{2} & {1} & {-3} \\{1} & {1} & {1}\end{array}\right]\)
⇒ |A| = 4 + 5 + 1 = 10 ≠ 0
A is non singular matrix.

2. Cofactor A
Plus Two Maths Determinants 6 Mark Questions and Answers 45

3. A-1 = \(\frac{1}{10}\) \(\left[\begin{array}{ccc}{4} & {2} & {2} \\{-5} & {0} & {5} \\{1} & {-2} {3}\end{array}\right]\)

4. Given, AX = B ⇒ X = A-1 B
Plus Two Maths Determinants 6 Mark Questions and Answers 46
⇒ x = 2, y = -1, z = 1.

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 5.
Solve the following system of linear equations.

  1. x + y + z = 3, y – z = 0, 2x – y = 1 (6)
  2. 5x – 6y + 4z = 15 , 7x + 4y – 3z = 19, 2x + y + 6z = 46 (6)
  3. x + 2y + 5z = 10, x – y – z = -2, 2x + 3_y-2 = -11 (6)

Answer:
1. Let AX = B
Where A = \(\left[\begin{array}{ccc}{1} & {1} & {1} \\{0} & {1} & {-1} \\{2} & {-1} & {0}\end{array}\right], X=\left[\begin{array}{c}{x} \\{y} \\{z}\end{array}\right], B=\left[\begin{array}{l}{3} \\{0} \\{1}\end{array}\right]\)
|A| = 1(0 – 1) – 1(0 + 2) + 1(0 – 2) = -5
C11 = -1, C12 = -2, C13 = -2, C21 = -1, C22 = 3, C23 = 3, C31 = -2, C32 = 1, C33 = 1
Plus Two Maths Determinants 6 Mark Questions and Answers 47

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

2. Let AX = B,
Where A = \(\left[\begin{array}{ccc}{5} & {-6} & {4} \\{7} & {4} & {-3} \\{2} & {1} & {6}\end{array}\right], X=\left[\begin{array}{c}{x} \\{y} \\{z}\end{array}\right],B=\left[\begin{array}{c}{15} \\{19} \\{46}\end{array}\right]\)
|A| = 5(24 + 3) + 6(42 + 6) + 4(7 – 8) = 419
C11 = 27, C12 = -48, C13 = -1, C21 = -1, C22 = 22, C23 = -17, C31 = 2, C32 = 43, C33 = 62
Plus Two Maths Determinants 6 Mark Questions and Answers 48
Plus Two Maths Determinants 6 Mark Questions and Answers 49

3. Let AX = B
\(\text { Where } A=\left[\begin{array}{ccc}{1} & {2} & {5} \\{1} & {-1} & {-1} \\{2} & {3} & {-1}
\end{array}\right], X=\left[\begin{array}{c}{x} \\{y} \\{z}\end{array}\right], B=\left[\begin{array}{c}{10} \\{-2} \\{-11}\end{array}\right]\)
|A| = 1(4) – 2(1) + 5(5) = 27
C11 = 4, C12 = -1, C13 = 5, C21 = 17, C22 = -11, C23 = 1, C31 = 3, C32 = 6, C33 = -3
Plus Two Maths Determinants 6 Mark Questions and Answers 50
⇒ x = -1, y = -2, z = 3.

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 6.
If f(x) = \(\left[\begin{array}{ccc}{\cos x} & {-\sin x} & {0} \\{\sin x} & {\cos x} & {0} \\{0} & {0} & {1}\end{array}\right]\)
(i) Find f(-x) (2)
(ii) Find (f(x)]-1 (2)
(iii) Is |f(x)]-1 = f(-x)? (2)
Answer:
Plus Two Maths Determinants 6 Mark Questions and Answers 51

(ii) |f(x)| = \(\left[\begin{array}{ccc}{\cos x} & {-\sin x} & {0} \\{\sin x} & {\cos x} & {0} \\{0} & {0} & {1}\end{array}\right]\) = cos x (cos x) + sin x (sin x) = 1 ≠ 0
Therefore , [f(x)]-1 exists.
The cofactors are as follows.
C11 = cos x, C12 = -sin x, C13 = 0, C21 = sin x, C22 = cos x, C23 = 0, C31 = 0, C32 = 0, C33 = 1
Plus Two Maths Determinants 6 Mark Questions and Answers 52
Since, |f(x)|= 1

(iii) Yes. From (1) and (2) we have,
[f(x)]-1 =f(-x).

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 7.
(i) Choose the correct answer from the bracket. If A = \(\left[\begin{array}{cc}{2} & {3} \\{1} & {-2}\end{array}\right]\) and A-1 = kA, then the value of ‘k’ is
(a) 7
(b) -7
(c) \(\frac{1}{7}\)
(d)\(-\frac{1}{7}\) (1)
(ii) If A = \(\left[\begin{array}{ccc}{1} & {-1} & {1} \\{2} & {-1} & {0} \\{1} & {0} & {0}
\end{array}\right]\),
(a) A2 (2)
(b) Show that A2 = A-1 (3)
Answer:
Plus Two Maths Determinants 6 Mark Questions and Answers 53
Plus Two Maths Determinants 6 Mark Questions and Answers 54
C11 = 0, C12 = 0, C13 = 1, C21 = 0, C22 = -1, C23 = -1, C31 = 1, C32 = 2, C33 = 1
Plus Two Maths Determinants 6 Mark Questions and Answers 55

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 8.
‘Arjun’ purchased 3 pens, 2 purses, and 1 instrument box and pays Rs. 410. From the same Shop ‘Deeraj’ purchases 2 pens, 1 purse, and 2 instrument boxes and pays Rs.290, while ‘Sindhu’ purchases 2pens, 2 purses, 2 instrument boxes and pays Rs. 440.

  1. Translate the equation into system of linear equations. (2)
  2. The cost of one pen, one purse and one instrument box using matrix method. (4)

Answer:
1. Let The price of one pen is Rs.x, one purse is Rs.y and one instrument box be Rs.z
3x + 2y + z = 410; 2x + y + 2z =290; 2x + 2y + 2z = 440(1) 2 mts.

2. The system can be represented by the matrix equation AX = B
Plus Two Maths Determinants 6 Mark Questions and Answers 56
Plus Two Maths Determinants 6 Mark Questions and Answers 57
C11 = -2, C12 = 0, C13 = 2, C21 = -2, C22 = 4, C23 = -2, C31 = 3, C32 = -4, C33 = -1
Plus Two Maths Determinants 6 Mark Questions and Answers 58
Hence the cost one pen is Rs.20, one purse is Rs. 150 and one instrument box is Rs. 50.

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 9.
If A = \(\left[\begin{array}{ccc}{2} & {-3} & {5} \\{3} & {2} & {-4} \\{1} & {1} & {-2}\end{array}\right]\)

  1. Find A-1 (3)
  2. Using it solve the system of equations 2x – 3y + 5z = 16, 3x + 2y – 4z = -4, x + y – 2z = -3 (3)

Answer:
1. A = \(\left[\begin{array}{ccc}{2} & {-3} & {5} \\{3} & {2} & {-4} \\{1} & {1} & {-2}\end{array}\right]\)
⇒ |A| = 0 + 3x – 2 + 5 = -1
Plus Two Maths Determinants 6 Mark Questions and Answers 59

2. Given AX = B
⇒ X = A-1B
Plus Two Maths Determinants 6 Mark Questions and Answers 60
⇒ x = 2, y = 1, z = 3.

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 10.
Consider the following system of equations x + y + 3z = 5, x + 3y – 3z = 1, -2x – 4y – 4z = -10
(i) Convert the given system in the form AX = B (1)
(ii) Find A-1 (3)
(iii) Hence solve the system of equations. (2)
Answer:
Plus Two Maths Determinants 6 Mark Questions and Answers 61

(ii) i.e; AX = B, ⇒ X = A-1 B ⇒ |A| = -24 + 10 + 6 = -8
Plus Two Maths Determinants 6 Mark Questions and Answers 62

(iii) X = A-1B
Plus Two Maths Determinants 6 Mark Questions and Answers 63
= \(-\frac{1}{8}\) \(\left[\begin{array}{l}{-8} \\{-8} \\{-8}\end{array}\right]\)
⇒ x = 1, y = 1, z = 1.

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 11.
Solve the following system by equations by matrix method x + 2y + 5z = 10; x – y – z = -2; 2x + 3y – z = -11.
Answer:
x + 2y + 5z = 10; x – y – z = -2; 2x + 3y – z = 11
Plus Two Maths Determinants 6 Mark Questions and Answers 64
⇒ x = -1, y = -2, z = 3.

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 12.
If A = \(\left[\begin{array}{ccc}{3} & {-2} & {3} \\{2} & {1} & {-1} \\{4} & {-3} & {2}\end{array}\right]\)

  1. Find |A| (1)
  2. Find A-1 (3)
  3. Solve the linear equations 3x – 2y + 3z = 8; 2x + y – z = 1; 4x – 3y + 2z = 4 (2)

Answer:
1. |A| = \(\left[\begin{array}{ccc}{3} & {-2} & {3} \\{2} & {1} & {-1} \\{4} & {-3} & {2}\end{array}\right]\)
= 3(2 – 3) + 2(4 + 4) + 3(- 6 – 4) = -17.

2. |A| ≠ 0, hence its inverse exists.
A-1 = \(\frac{1}{|A|}\)adj A
C11 = -1, C12 = -8, C13 = -10, C21 = -5, C22 = -6, C23 = 1, C31 = -1, C32 = 9, C33 = 7
Plus Two Maths Determinants 6 Mark Questions and Answers 65

3. The given system of linear equations is of the form
Plus Two Maths Determinants 6 Mark Questions and Answers 66
∴ We have, x = 1, y = 2, z = 3.

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Question 13.
if \(\left[\begin{array}{cc}{2} & {5} \\{-3} & {7}\end{array}\right] \times A=\left[\begin{array}{cc}{17} & {-1} \\{47} & {-13}\end{array}\right]\) then
(i) Find the 2 × 2 matrix A. (3)
(ii) Find A2. (1)
(iii) Show that A2 + 5A – 6I = 0, where I is the identity matrix of order 2. (2)
Answer:
Plus Two Maths Determinants 6 Mark Questions and Answers 67
Plus Two Maths Determinants 6 Mark Questions and Answers 68
Plus Two Maths Determinants 6 Mark Questions and Answers 69
Plus Two Maths Determinants 6 Mark Questions and Answers 70

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