Class 12 Maths NCERT Solutions for Chapter 4 Determinants Miscellaneous Exercise
![Class 12 Maths NCERT Solutions for Chapter 4 Determinants Miscellaneous Exercise Class 12 Maths NCERT Solutions for Chapter 4 Determinants Miscellaneous Exercise](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEga6QeWR42CWx9WIqR2ZV-x48gfXtE95iYJaBh4X1b2ZAQTWTvtK9Oq6qwgE663fW2E7Rg0pDWsGQSD0eRfGiUI_D29b6Qc_8j7O2Bh8tVLSwiBD4l8TM_p5uhTowtCvNmywOyNZemyZV8fg1Ich2B8oF48mQPbOtleXTYCb7JQT1VBHAPPf6mi6w6t/w662-h312-rw/cbse-class-12-maths-ncert-solutions-chapter-4--miscellaneous-exercise.jpg)
Determinants Miscellaneous Exercise Solutions
1. Prove that the determinant is independent of θ is independent of θ.
Solution
Δ =
= x(–x2 – 1) – sinθ(– x sinθ – cos θ) + cos θ(– sinθ + x cosθ)
= –x3 – x + xsin2 θ + sinθ cosθ – sinθ cosθ + x cos2θ
= –x3 – x + x(sin2θ + cos2θ)
= –x3 – x + x
= –x3 (Independent of θ)
Hence, Δ is independent of θ.
2. Without expanding the determinant, prove that
Solution
Hence, the given result is proved.
3. Evaluate ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_3a1XCIaL5DAOy91x2v60ereBzn8VtF0SJIgC--51_mrjwALeecPTFBlpU1jFi31kaXM3JBEhyZJA7yt4UEsX8_4kS3kbsJkdZUGFMruXJtWJgTlEKQm0wPPQAWLmWbJs9QINTyxipvfs21vVerTT5nP0_LuKqkZCrcoyvxEVAA7ZbDC4oFUbBq9M/w231-h65-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%204.JPG)
Solution Δ = ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_3a1XCIaL5DAOy91x2v60ereBzn8VtF0SJIgC--51_mrjwALeecPTFBlpU1jFi31kaXM3JBEhyZJA7yt4UEsX8_4kS3kbsJkdZUGFMruXJtWJgTlEKQm0wPPQAWLmWbJs9QINTyxipvfs21vVerTT5nP0_LuKqkZCrcoyvxEVAA7ZbDC4oFUbBq9M/w243-h69-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%204.JPG)
Expanding along C3 , we have :
∆ = − sin α(−sin α sin2β – cos2 β sinα) + cos α(cos α cos2 β + cos α sin2 β)
= sin2 α(sin2 β + cos2 β) + cos2 α(cos2 β + sin2 β)
= sin2 α(1) + cos2 α(1)
= 1
∆ = − sin α(−sin α sin2β – cos2 β sinα) + cos α(cos α cos2 β + cos α sin2 β)
= sin2 α(sin2 β + cos2 β) + cos2 α(cos2 β + sin2 β)
= sin2 α(1) + cos2 α(1)
= 1
4. If a, b and c are real numbers, and triangle =
Show that either a + b + c = 0 or a = b = c.
Solution
Expanding along R1 , we have :
Δ = 2 (a + b + c)(1)[(b - c)(c - b) - (b - a)(c - a)]
= 2(a + b + c)[-b2 - c2 + 2bc - bc + ba + ac - a2 ]
= 2(a + b + c)[ab + bc + ca - a2 - b2 - c2 ]
Δ = 2 (a + b + c)(1)[(b - c)(c - b) - (b - a)(c - a)]
= 2(a + b + c)[-b2 - c2 + 2bc - bc + ba + ac - a2 ]
= 2(a + b + c)[ab + bc + ca - a2 - b2 - c2 ]
It is given that Δ = 0.
(a + b + c)[ab + bc + ca - a2 - b2 - c2] = 0
⇒ Either a + b + c = 0, or ab + bc + ca - a2 - b2 - c2 = 0 .
Now,
ab + bc + ca - a2 - b2 - c2 = 0
⇒ -2ab - 2bc - 2ca + 2a2 + 2b2 + 2c2 = 0
⇒ (a - b)2 + (b - c)2 + (c - a)2 = 0
⇒ (a - b)2 = (b - c)2 = (c - a)2 = 0 [(a-b)2, (b-c)2, (c-a)2 are non-negative]
⇒ (a - b) = (b - c) = (c - a) = 0
⇒ a = b = c
Hence, if Δ = 0 then either a + b + c = 0 or a = b = c.
(a + b + c)[ab + bc + ca - a2 - b2 - c2] = 0
⇒ Either a + b + c = 0, or ab + bc + ca - a2 - b2 - c2 = 0 .
Now,
ab + bc + ca - a2 - b2 - c2 = 0
⇒ -2ab - 2bc - 2ca + 2a2 + 2b2 + 2c2 = 0
⇒ (a - b)2 + (b - c)2 + (c - a)2 = 0
⇒ (a - b)2 = (b - c)2 = (c - a)2 = 0 [(a-b)2, (b-c)2, (c-a)2 are non-negative]
⇒ (a - b) = (b - c) = (c - a) = 0
⇒ a = b = c
Hence, if Δ = 0 then either a + b + c = 0 or a = b = c.
5. Solve the equations
= 0, a ≠ 0
Solution
Expanding along R1 , we have :
(3x + a)[1 × a2 ]= 0
⇒ a2 (3x + a) = 0
But a ≠ 0.
Therefore, we have :
3x + a = 0
⇒ x = -a/3
6. Prove that
= 4a2 b2 c2
Solution
Expanding along R3 , we have :
Δ = 2ab2 c[a(c - a) + a(a + c)]
= 2ab2 c[ac - a2 + a2 + ac]
= 2ab2 c(2ac)
= 4a2 b2 c2
Hence, the given result is proved.
7. If A ^(-1) =
find (AB)-1 .
Solution
We know that (AB)-1 = B-1 A-1 .
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjpCP9cyBAdS6qpjhAI1R0fTQRJnoxLPg22dT17IhbJVJ1PzGCNOQo4hqg0bzV7bs5tS7htILZ0nHhw1DxULDkEsPDe750GOEmhvZuFNoSNLaBOI2UwxDAVkySfsPAccijcFC-lYl5p0i-LbiwZfAH5slyzleaL1JOWaffMy7RrEYpFtzJI2nSm30fA/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%2012.JPG)
∴ |B| = 1 × 3 - 2× (-1) - 2(2)
= 3 + 2 - 4
= 5 - 4 = 1
Now, A11 = 3, A12 = 1, A13 = 2
A21 = 2, A22 = 1, A23 = 2
A31 = 6, A32 = 2, A33 = 5
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiXb65_XsD4xv_HFheSsMrEClFI3K1UMeMiVDMJWa2OkzVxWmRzvDepC_niKsGH-o8gxU0TO8cPIqoxvluD-mBZhYjex170zCbpQo41RTNVA8fiME1eqDc-shMTub9YW6StAfFBrxI1gju9UrsZcu6T2pC-A-Xj4BNjvrRmX2etn2v7fYlzhewdgDQ6/w318-h527-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%2013.JPG)
8. Let A = ∴ |B| = 1 × 3 - 2× (-1) - 2(2)
= 3 + 2 - 4
= 5 - 4 = 1
Now, A11 = 3, A12 = 1, A13 = 2
A21 = 2, A22 = 1, A23 = 2
A31 = 6, A32 = 2, A33 = 5
(i) [adj A]-1 = adj(A-1)
(ii) (A-1)-1 = A
Solution
A = ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgITsXXawm9M8dLV17sGNUL82enR2mtXV_M899S1QWpXzGEo-kC1wJ0UYH_tI-1wFCZn8ZKOmFO8scbSflf-6IEKy0anEIoXFvabUmyzQsr-9A1YmH3AnR6_AF5YkiQRhehYZLKSuiu9k4ggjmeYGOFeo9SXQhCSVCWJ1DeXBd7BuG_pRErQEgyq7KX/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%2014.JPG)
∴ |A| = 1 (15 - 1) + 2(-10 - 1) + 1(-2 - 3) = 14 - 22 - 5= -13
Now, A11 = 14, A12 = 11, A13 = -5
A21 = 11, A22 = 4, A23 = -3
A31 = -5, A32 = -3, A13 = -1
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjzGsf9q-amxAqcGs-HuKkKRlRBhXyTP7hMWHpsk9r6QRkF31hL2mTWrE5mmqskygvdTs7IW3PcDmVbO3srRCOfS2kSfho05p7qfdNhtAXpCqMJCZeHNDOn2LXvEOuJZJ-F0aZssvhn1fd5KkY6jzEZjGnclEnku8VF4TeGyv7WaMnUutr9fkaiYDfY/w373-h196-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%2015.JPG)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi3PyWhRNMnaa6nsgpTRUHnFI6RYZvYTBgvKGMr3ixNdg_TSrhhmzhBObc0fLvLVXK-pyzEEnGAhThbn2PN9DU3OctBMIKuaHbN9t4-lA7vBi5Egj2prLPuIVf6DZJeHcJlRd6BoVvGQxOx7MwMQY2Ql2-9QeFdR1_g09tvThlXb9Df1Flf8FnzDKxh/w452-h309-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%2017.JPG)
∴ |A| = 1 (15 - 1) + 2(-10 - 1) + 1(-2 - 3) = 14 - 22 - 5= -13
Now, A11 = 14, A12 = 11, A13 = -5
A21 = 11, A22 = 4, A23 = -3
A31 = -5, A32 = -3, A13 = -1
(i) |adj A| = 14(-4 - 9) - 11(-11 - 15) - 5(-33 + 20)
= 14(-13) - 11(-26) - 5(-13)
= -182 + 286 + 65 = 169
we have ,
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiXRwIOjIk5Ct7oW-Zd3982Nff382A2AO6HqLl1WTV_jxyHc7d_xXRLy4Pjyt9__-fntJjG3n8HEYgPDd7q0s71POtBETzE-_DBbT1XsdgqTBuIfJAHLDpoa_5Rc3V9YlL6_tipIv_k5wjtbcsd5W-sVkW8O2-MpOde3d_n00YhYCYlT2GOFuRX7uCr/w456-h606-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%2016.JPG)
= 14(-13) - 11(-26) - 5(-13)
= -182 + 286 + 65 = 169
we have ,
(ii) We have show that :
9. Evaluate ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2kDpWZK7ff6xY4rSnADr3FgSxPOZl-VFN5oks3dF9z9EexaWD-5NuIQZO0O7KdcO-6DdkA6i_awaX7paL9FIFsxUMv6PnAc4UZQnzJc7Iwghq9nIWcGFV9-NAGv3qmeWdMmwUAhcwg06Un3icdCM8hfKb3jB9MA3v2F-rddyU65OVN7vHD-A6vcfl/w164-h71-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%2018.JPG)
Solution
10. Evaluate ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhjvxAFvBSOTH4znZLHxLG5I8s8xg1WNdqtUc5DuBZwaaUMn3QdLBurMFW6krW1T2ykgj5tSPsaQJ4u2tk0DEB0zAitba4dtFxqnT074J87CV87nuhgG-XI-QfMOuUwbEicI4t2drAFlDRoEdcB-AF9LHwvDssGscsGMDpMGT2iWkIOsVb2lmhQKcc4/w130-h69-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%2022.JPG)
Solution
Expanding along C1 , we have :
Δ = 1(xy - 0) = xy
11. Using properties of determinants, prove that :
Solution
Expanding along R3 , we have :
Δ = (β - α) (γ - α)[- (γ - β)( - α - β - γ)]
= (β - α) (γ - α) (γ - β)(α + β + γ)
= (α - β) (β - γ) (γ - α)(α + β + γ)
Hence, the given result is proved.
12. Using properties of determinants, prove that :
= (1 + pxyz)(x - y)(y - z)(z -x), where p is any scalar.
Solution
13. Using properties of determinants, prove that :
= 3(a + b + c)(ab + bc + ca)
Solution
Expanding along C1 , we have :
Δ = (a + b + c) [(2b + a)(2c + a) - (a - b)(a - c)]
= (a + b +c)[4abc + 2ab + 2ac + a2 - a2 + ac + ba - bc]
= (a + b + c)(3ab + 3bc + 3ac)
= 3(a + b + c)(ab + bc + ca)
Hence, the given result is proved.
Δ = (a + b + c) [(2b + a)(2c + a) - (a - b)(a - c)]
= (a + b +c)[4abc + 2ab + 2ac + a2 - a2 + ac + ba - bc]
= (a + b + c)(3ab + 3bc + 3ac)
= 3(a + b + c)(ab + bc + ca)
Hence, the given result is proved.
14. Using properties of determinants, prove that :
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiLXJqbjPON7tIQ5yKgohuE7U2dlCxcCZlHN48dv19lMZCoVXX7dt8ToIChMJ9sQk1aG35lB5CHATHEOv_MLnTSY3GSJySlVrynuB3Y4wwjSB1OggIzM5PPm1bUh0OuyOXEesqc2ZApIohyHZnH8e0wbevE1NZ_93rLCkAKILJu48rzET7Ze7zBwgdT/w225-h71-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%2028.JPG)
Solution
2/x + 3/y + 10/z = 4
4/x - 6/y + 5/z = 1
6/x + 9/y - 20/x = 2
Solution
Let 1/x = p, 1/y = q, 1/z = r.
Then the given system of equations is as follows :
2p + 3q + 10r = 4
4p - 6q + 5r = 1
6p + 9q - 20r = 2
This system can be written in the form of AX = B, where
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjeP91xuNq1iwjhUYgT34MtuZgVfVdJm_bzFIVWjgGUFiXiesznS_Ifq2Je4l4sxrxynjZB4-PH0V3uf8kl4DseDHFoLyX3sHuNd0r19iG3jhLmZflguh51THsIQuLo1lbYGRFGNaD62i3jprpoprU_W7jRy4iIbq9JJjw4ne-7ksv9rP1TFaGFCnHC/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%2030.JPG)
Now,
|A| = 2(120 - 45) - 3 (-80 - 30) + 10(36 + 36)
= 150 + 330 + 720
=1200
Thus, A is non - singular. Therefore, its inverse exists.
Now,
A11 = 75, A12 = 110, A13 = 72
A21 = 150, A22 = -100, A23 = 0
A31 = 75, A32 = 30, A33 = -24
∴ A-1 = 1/|A| [adj A]
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhDMq5bfaBW5z8_UbKisrlQ0IYFUCFXa5-WjSoIz2kiAQX0aClZQAaw2ZqC8v5Fp9yqdvJJNdsHogrli0keLYk_k_D6seolW-ipkQ-wKfglV1hGI7PIlbl5WpG6MZwJu1vlQqOzwj42wZ9dnVezH6N41nd-nsqQhRAEvMTbZpIbGroEJYj1MAbIGIRU/w230-h359-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%2031.JPG)
∴ p = 1/2, q = 1/3, and r = 1/5
Hence, x = 2, y = 3, and z = 5.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjVsjtPfNt9m0YZvzohNSC376rpv_amxroTVx0KyjKY7fhKkMaDGLX8MHPhILBTjv4-sHfv9QKk-_rpv8qmMBoQPtQlTGQCGLgXxr6Gscae-C2VsdZMBrtRcdmQ4fQ_3R4py9w4U7DIbYIZvI2y4vp2rKamS6STco0r_cOxFe4VC6fd4JD5I_9Z8qQY/w452-h339-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%2033.JPG)
Here, all the elements of the first row (R1 ) are zero.
Hence, we have Δ = 0.
The correct answer is A.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiCLu2OTc9_ZPRp77pXJCbbpErvIwRG85qs5qfOJIXxVPbj2QxcIBzXjw1ZdlUDOSCR4GOtKtp0k6RRvq-lE2Pw6qf3s3xR1TUdxZyr1S6x5DvTr1QrvwnxBOqaYsUUEpG35KtgLdG0J6BXPLaHeJqMERTTAA14QQGG6J9YWm_BQeoj0ogcqhwwbylm/w134-h230-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%2035.JPG)
A31 = 0, A32 = 0, A33 = xy
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjxIRR7659xa9zY_zsNn-ZtF_vo19RCDarvVhgAtxXzqfNXLLmeXK0VtQpLE-58GBY3YvxBOGKkjiVKKPjlxpY2sVVeZsdqa2quT3N-woJAB_TEZoXDYQ59G5XuFi7aly9neaVHGRAq04JmFhIXc0ym3pVvyn8xlLEgXOVAnJsL8PF12IS-aP4PXm3a/w242-h403-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%2036.JPG)
The correct answer is A.
Then the given system of equations is as follows :
2p + 3q + 10r = 4
4p - 6q + 5r = 1
6p + 9q - 20r = 2
This system can be written in the form of AX = B, where
Now,
|A| = 2(120 - 45) - 3 (-80 - 30) + 10(36 + 36)
= 150 + 330 + 720
=1200
Thus, A is non - singular. Therefore, its inverse exists.
Now,
A11 = 75, A12 = 110, A13 = 72
A21 = 150, A22 = -100, A23 = 0
A31 = 75, A32 = 30, A33 = -24
∴ A-1 = 1/|A| [adj A]
∴ p = 1/2, q = 1/3, and r = 1/5
Hence, x = 2, y = 3, and z = 5.
16. Choose the correct answer.
If a, b, c, are in A.P., then the determinant
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiVcs3eY5FnRSdDati3SxhUALVqaBfI-n7OykuMtbIjMjUlbS_BYLotV_gHRDkQcQ6-Rbm2XNWS0o22UtYGqX5rmuw2PtxswEvZmjgVBC4SRrlb7GCHMAU-v6OetHNwI0jN_n10etlW9ce6zoFAQjAw6FIKX7E0LLzqhZTFS2BYiYGbZ-PSr3tXkjMo/w176-h74-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%2032.JPG)
A. 0
B. 1
C. x
D. 2x
If a, b, c, are in A.P., then the determinant
A. 0
B. 1
C. x
D. 2x
Solution
Here, all the elements of the first row (R1 ) are zero.
Hence, we have Δ = 0.
The correct answer is A.
18. If x, y, z are nonzero real numbers, then the inverse of matrix A =
is
Solution
A = ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgFEegT3CYc5osQFEv0xZ6iZnJdNv_9MtdhnEWtR9a2oHg8195Xi9tnyWIwu5aEt_DrMiceVvjizJ2pYnEjXhCWufXCC2rnYr9w0P1T9D4IoSagKOaqiPsfhw1hBbe-RO0xSTMmr3GWqeICw1jAfIhIv5M_RdAIS0F0yahPZbflwHPOH58it-q7npJw/w68-h66-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Miscellaneous%20Exercise%20img%2034.JPG)
∴ |A| = x(yz - 0) = xyz ≠ 0
Now, A11 = yz, A12 = 0, A13 = 0
A21 = 0, A22 = xz, A23 = 0 ∴ |A| = x(yz - 0) = xyz ≠ 0
Now, A11 = yz, A12 = 0, A13 = 0
A31 = 0, A32 = 0, A33 = xy
The correct answer is A.
19. Choose the correct answer .
Let A =
where 0 ≤ θ ≤ 2Ï€, then
A. Det (A) = 0
B. Det (A) ∈ (2, ∞)
C. Det (A) ∈ (2, 4)
D. Det (A)∈ [2, 4]
Solution
A =
∴ |A| = 1(1 + sin2 θ) - sinθ( - sinθ + sinθ) + 1 (sin2 θ + 1)
= 1 + sin2 θ + sin2 θ + 1
= 2 + 2sin2 θ
= 2(1 + sin2 θ)
Now, 0 ≤ θ ≤ 2Ï€
⇒ 0 ≤ sinθ ≤ 1
⇒ 0 ≤ sin2 θ ≤ 1
⇒ 1 ≤ 1 + sin2 θ ≤ 2
⇒ 2 ≤ 2(1 + sin2 θ) ≤ 2
∴ Det (A) ∊ [2, 4]
The correct answer is D.