NCERT Solutions Class 12- Mathematics Part I – Chapter 4 Determinants – Miscellaneous Exercises on Chapter 4
Last Updated :
01 May, 2024
Question 1. Prove that the determinant [Tex]\begin{vmatrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1\\ cos\theta & 1 & x \end{vmatrix} [/Tex]is independent of θ.
Solution:
A = [Tex]\begin{vmatrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1\\ cos\theta & 1 & x \end{vmatrix}[/Tex]
A = x(x2 – 1) – sinθ(-x sinθ – cosθ) + cosθ(-sinθ + x cosθ)
A = x3 – x + x sin2θ + sinθcosθ – sinθcosθ + x cos2θ
A = x3 – x + x(sin2θ + cos2θ)
A = x3 – x + x
A = x3(Independent of θ).
Hence, it is independent of θ
Question 2. Evaluate [Tex]\begin{vmatrix} cos\alpha cos\beta & cos\alpha sin\beta & -sin\alpha\\ -sin\beta & cos\beta & 0\\ sin\alpha cos\beta & sin\alpha sin\beta & cos\alpha \end{vmatrix}[/Tex]
Solution:
A = [Tex]\begin{vmatrix} cos\alpha cos\beta & cos\alpha sin\beta & -sin\alpha\\ -sin\beta & cos\beta & 0\\ sin\alpha cos\beta & sin\alpha sin\beta & cos\alpha \end{vmatrix}[/Tex]
Expanding along C3
A = -sinα(-sinα sin2β – cos2β sinα) + cosα(cosα cos2β + cosα sin2β)
A = sin2α(sin2β + cos2β) + cos2α(cos2β + sin2β)
A = sin2(1) + cos2(1)
A = 1
Question 3. If A-1 =[Tex]\begin{vmatrix} 3 & -1 & 1\\ -15 & 6 &-5\\ 5 & -2 & 2 \end{vmatrix} [/Tex]and B =[Tex]\begin{vmatrix} 1 & 2 & -2\\ -1 & 3 &0\\ 0 & -2 & 1 \end{vmatrix} [/Tex]. Find (AB)-1
Solution:
|B| = 1(3 – 0) + 1(2 – 4) = 1
B11 = 3 – 0 = 3
B12 = 1
B13 = 2 – 0 = 2
B21 = -(2 – 4) = 2
B22 = 1 – 0 = 1
B23 = 2
B31 = 0 + 6 = 6
B32 = -(0 – 2) = 2
B33 = 3 + 2 = 5
adj B = [Tex]\begin{bmatrix} 3 & 2 & 6\\ 1 & 1 & 2\\ 2 & 2 & 5 \end{bmatrix}[/Tex]
B-1 = (adj B)/|B|
B-1 = [Tex]\begin{bmatrix} 3 & 2 & 6\\ 1 & 1 & 2\\ 2 & 2 & 5 \end{bmatrix}[/Tex]
Now,
(AB)-1 = B-1A-1
(AB)-1 = [Tex]\begin{bmatrix} 3 & 2 & 6\\ 1 & 1 & 2\\ 1 & 2 &5 \end{bmatrix} \begin{bmatrix} 3 & -1 & 1\\ -15 & 6 & -5\\ 5 & -2 & 2 \end{bmatrix}[/Tex]
= [Tex]\begin{bmatrix} 9-30+30 & -3+12-12 & 3-10+12\\ 3-15+10 & -1+6-4 & 1-5+4\\ 6+12-10 & -2+12-10 & 2-10+10 \end{bmatrix}[/Tex]
(AB)-1 = [Tex]\begin{bmatrix} 9 & -3 & 5\\ -2 & 1 & 0\\ 8 & 0 & 2 \end{bmatrix}[/Tex]
Question 4. Let A =[Tex]\begin{bmatrix} 1 & -2 & 1\\ -2 & 3 & 1\\ 1 & 1 & 5 \end{bmatrix} [/Tex] verify that
(i) [adj A]-1 = adj(A-1)
(ii) (A-1)-1 = A
Solution:
A = [Tex]\begin{bmatrix} 1 & -2 & 1\\ -2 & 3 & 1\\ 1 & 1 & 5 \end{bmatrix}[/Tex]
|A| = 1(15 – 1) + 2(-10 – 1) + 1(-2 – 3) = 14 – 27 = -13
A11 = 14
A12 = 11
A13 = -5
A21 = 11
A22 = 4
A23 = -3
A31 = -5
A32 = -3
A33 = -1
adj A = [Tex]\begin{bmatrix} 14 & 11 & -5\\ 11 & 4 & -3\\ -5 & -3 & -1 \end{bmatrix}[/Tex]
Arrr-1 = (adj A)/|A|
= [Tex]-1/13\begin{bmatrix} 14 & 11 & -5\\ 11 & 4 & -3\\ -5 & -3 & -1 \end{bmatrix}[/Tex]
= [Tex]1/13\begin{bmatrix} -14 & -11 & 5\\ -11 & -4 & 3\\ 5 & 3 & 1 \end{bmatrix}[/Tex]
(i). |adj A| = 14(-4 – 9) – 11(-11 – 15) – 5(-33 + 20)
= 14(-13) – 11(-26) – 5(-13)
= -182 + 286 + 65 = 169
adj(adj A) = [Tex]\begin{bmatrix} -13 & 26 & -13\\ 26 & -39 & -13\\ -13 & -13 & -65 \end{bmatrix}[/Tex]
[adj A]-1 = (adj(adj A))/|adj A|
= [Tex]1/169\begin{bmatrix} -13 & 26 & -13\\ 26 & -39 & -13\\ -13 & -13 & -65 \end{bmatrix}[/Tex]
= [Tex]1/13\begin{bmatrix} -1 & 2 & -1\\ 2 & -3 & -1\\ -1 & -1 & -5 \end{bmatrix}[/Tex]
Now, A-1 = [Tex]1/13\begin{bmatrix} -14 & -11 & 5\\ -11 & -4 & 3\\ 5 & 3 & 1 \end{bmatrix}[/Tex]
= [Tex]\begin{bmatrix} \frac{-14}{13} & \frac{-11}{13} & \frac{5}{13}\\ \frac{-11}{13} & \frac{-4}{13} & \frac{3}{13}\\ \frac{5}{13} & \frac{3}{13} & \frac{1}{13} \end{bmatrix}[/Tex]
adj(A-1) = [Tex]\begin{bmatrix} \frac{-4}{169}-\frac{9}{169} & -(\frac{-11}{169}-\frac{15}{169}) & \frac{-33}{169}+\frac{20}{169}\\ -(\frac{-11}{169}-\frac{-15}{169}) & \frac{-14}{169}-\frac{25}{169} &-(\frac{-42}{169}+\frac{55}{169}) \\ \frac{-33}{169}+\frac{20}{169} & -(\frac{-42}{169}+\frac{55}{169}) & \frac{56}{169}-\frac{121}{169} \end{bmatrix}[/Tex]
= [Tex]1/169\begin{bmatrix} -13 & 26 & -13\\ 26 & -39 & -13\\ -13 & -13 & -65 \end{bmatrix}[/Tex]
= [Tex]1/13\begin{bmatrix} -1 & 2 & -1\\ 2 & -3 & -1\\ -1 & -1 & -5 \end{bmatrix}[/Tex]
Hence, [adj A]-1 = adj(A-1)
(ii). A-1 = [Tex]1/13\begin{bmatrix} -14 & -11 & 5\\ -11 & -4 & 3\\ 5 & 3 & 1 \end{bmatrix}[/Tex]
adj A-1 = [Tex]1/13\begin{bmatrix} -1 & 2 & -1\\ 2 & -3 & -1\\ -1 & -1 & -5 \end{bmatrix}[/Tex]
|A-1| = (1/13)3[-14 × (-13) +11 × (-26) + 5 × (-13)]
= (1/13)3 × (-169)
= -1/13
Now, (A-1)-1 = (adj A-1)/|A-1|
= [Tex]\frac{1}{(\frac{-1}{13})}*\frac{1}{13}\begin{bmatrix} -1 & 2 & -1\\ 2 & -3 & -1\\ -1 & -1 & -5 \end{bmatrix}[/Tex]
= [Tex]\begin{bmatrix} 1 & -2 & 1\\ -2 & 3 & 1\\ 1 & 1 & 5 \end{bmatrix}[/Tex]
= A
Hence, it is proved that (A-1)-1 = A
Question 5. Evaluate [Tex]\begin{vmatrix} x & y & x+y\\ y & x+y & x\\ x+y & x &y \end{vmatrix}[/Tex]
Solution:
A = [Tex]\begin{vmatrix} x & y & x+y\\ y & x+y & x\\ x+y & x &y \end{vmatrix}[/Tex]
Applying R1 -> R1+R2+R3
A = [Tex]\begin{vmatrix} 2(x+y) & 2(x+y) & 2(x+y)\\ y & x+y & x\\ x+y & x &y \end{vmatrix}[/Tex]
= 2(x+y)[Tex]\begin{vmatrix} 1 & 1 & 1\\ y & x+y & x\\ x+y & x &y \end{vmatrix}[/Tex]
Applying C2-> C2 – C1 and C3-> C3 – C1
A = 2(x + y)[Tex]\begin{vmatrix} 1 & 0 & 0\\ y & x & x-y\\ x+y & -y &-x \end{vmatrix}[/Tex]
Expanding along R1
A = 2(x + y)[-x2 + y(x – y)]
= -2(x + y)(x2 + y2 – yx)
A = -2(x3 + y3)
Question 6. Evaluate [Tex]\begin{vmatrix} 1 & x & y\\ 1 & x+y & y\\ 1 & x &x+y \end{vmatrix}[/Tex]
Solution:
A = [Tex]\begin{vmatrix} 1 & x & y\\ 1 & x+y & y\\ 1 & x &x+y \end{vmatrix}[/Tex]
Applying R2->R2 – R1 and R3->R3 – R1
A = [Tex]\begin{vmatrix} 1 & x & y\\ 0 & y & 0\\ 0 & 0 &x \end{vmatrix}[/Tex]
Expanding along C1
A = 1(xy – 0)
A = xy
Question 7. Solve the system of the following questions:
2/x + 3/y + 10/z = 4
4/x – 6/y + 5/z = 1
6/x + 9/y – 20/z = 2
Solution:
Assume 1/x = p ; 1/y = q; 1/z = r
then. the above equations will be like
2p + 3Q + 10r = 4
4p – 6q + 5r = 1
6p + 9q – 20r = 2
This can be written in the form of AX=B
where,
A = [Tex]\begin{bmatrix} 2 & 3 & 10\\ 4 & -6 & 5\\ 6 & 9 & -20 \end{bmatrix}[/Tex]
X = [Tex]\begin{bmatrix} p\\ q\\ r \end{bmatrix}[/Tex]
B = [Tex]\begin{bmatrix} 4\\ 1\\ 2 \end{bmatrix}[/Tex]
We have,
|A| = 2(120 – 45) – 3(-80 – 30) + 10(36 + 36)
|A| = 150 + 330 + 720
|A| = 1200 ≠0
Hence A is invertible matrix.
A11 = 75
A12 = 110
A13 = 72
A21 = 150
A22 = -100
A23 = 0
A31 = 75
A32 = 30
A33 = -24
A-1 = (adj A)/|A|
A-1 = [Tex]\frac{1}{1200}\begin{bmatrix} 75 & 150 & 75\\ 110 & -100 & 30\\ 72 & 0 & -24 \end{bmatrix}[/Tex]
Now,
X = A-1B
[Tex]\begin{bmatrix} p\\ q\\ r \end{bmatrix} [/Tex]= [Tex]\frac{1}{1200}\begin{bmatrix} 75 & 150 & 75\\ 110 & -100 & 30\\ 72 & 0 & -24 \end{bmatrix} \begin{bmatrix} 4\\ 1\\ 2 \end{bmatrix}[/Tex]
[Tex]\begin{bmatrix} p\\ q\\ r \end{bmatrix} [/Tex]= [Tex]\frac{1}{1200}\begin{bmatrix} 300+150+150\\ 440-100+60\\ 288+0-48 \end{bmatrix}[/Tex]
= [Tex]\frac{1}{1200}\begin{bmatrix} 600\\ 400\\ 240 \end{bmatrix}[/Tex]
= [Tex]\begin{bmatrix} \frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{5} \end{bmatrix}[/Tex]
From above p = 1/2; q = 1/3 ; r = 1/5
So, x = 2; y = 3; z = 5
Question 8. Choose the correct answer.
If x, y, z are non-zero real numbers, then the inverse of matrix A = [Tex]\begin{bmatrix} x & 0 & 0\\ 0 & y &0 \\ 0 & 0 & z \end{bmatrix} [/Tex] is
(A) [Tex]\begin{bmatrix} x^{-1} & 0 & 0\\ 0 & y^{-1} &0 \\ 0 & 0 & z^{-1} \end{bmatrix}[/Tex]
(B) xyz[Tex]\begin{bmatrix} x^{-1} & 0 & 0\\ 0 & y^{-1} &0 \\ 0 & 0 & z^{-1} \end{bmatrix}[/Tex]
(C) [Tex]\frac{1}{xyz}\begin{bmatrix} x & 0 & 0\\ 0 & y &0 \\ 0 & 0 & z \end{bmatrix}[/Tex]
(D) [Tex]\frac{1}{xyz}\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}[/Tex]
Solution:
A = [Tex]\begin{bmatrix} x & 0 & 0\\ 0 & y &0 \\ 0 & 0 & z \end{bmatrix}[/Tex]
|A| = x(yz – 0) = xyz ≠0
Hence, the matrix is invertible
Now,
A11 = yz
A12 = 0
A13 = 0
A21 = 0
A22 = xz
A23 = 0
A31 = 0
A32 = 0
A33 = xy
adj A = [Tex]\begin{bmatrix} yz & 0 & 0\\ 0 & xz &0 \\ 0 & 0 & xy \end{bmatrix}[/Tex]
A-1 = (adj A)/|A|
A-1 = [Tex]\frac{1}{xyz}\begin{bmatrix} yz & 0 & 0\\ 0 & xz &0 \\ 0 & 0 & xy \end{bmatrix}[/Tex]
A-1 = [Tex]\begin{bmatrix} \frac{yz}{xyz} & 0 & 0\\ 0 & \frac{xz}{xyz} &0 \\ 0 & 0 & \frac{xy}{xyz} \end{bmatrix}[/Tex]
A-1 = [Tex]\begin{bmatrix} \frac{1}{x} & 0 & 0\\ 0 & \frac{1}{y} &0 \\ 0 & 0 & \frac{1}{z} \end{bmatrix}[/Tex]
A-1 = [Tex]\begin{bmatrix} x^{-1} & 0 & 0\\ 0 & y^{-1} &0 \\ 0 & 0 & z^{-1} \end{bmatrix}[/Tex]
Hence, the correct answer is A.
Question 9. Choose the correct answer
Let A = [Tex]\begin{bmatrix} 1 & sin\theta & 1\\ -sin\theta & 1 & sin\theta\\ -1 & -sin\theta & 1 \end{bmatrix} [/Tex], where 0 ≤ θ ≤ 2π, then
(A) Det(A) = 0 (B) Det(A) ∈ (2, ∞)
(C) Det(A) ∈ (2, 4) (D) Det(A) ∈ [2, 4]
Solution:
A = [Tex]\begin{bmatrix} 1 & sin\theta & 1\\ -sin\theta & 1 & sin\theta\\ -1 & -sin\theta & 1 \end{bmatrix}[/Tex]
|A| = 1(1 + sin2θ) – sinθ(-sinθ + sinθ) + 1(sin2θ + 1)
|A| = 1 + sin2θ + sin2θ + 1
= 2 + 2 sin2θ
= 2(1 + sin2θ)
Now 0 ≤ θ ≤ 2π
So, 0 ≤ sinθ ≤ 1
0 ≤ sin2θ ≤ 1
0 + 1 ≤ 1 + sin2θ ≤ 1 + 1
2 ≤ 2(1 + sin2θ) ≤ 4
Det(A) ∈ [2, 4]
Hence, the correct answer is D.
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