Class 12 Maths NCERT Solutions for Chapter 4 Determinants Exercise 4.5
![Class 12 Maths NCERT Solutions for Chapter 4 Determinants Exercise 4.5 Class 12 Maths NCERT Solutions for Chapter 4 Determinants Exercise 4.5](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgCE6emrpNEKZOSoyVECpoZ9lQW1oikNb6iZKo2kkIFDJns5IjFU3LFLkQ9i0GblFIxPkzQz9puXXJHKrt7L3NBS0K9YpkEGDfs_ibN2s6EinVQAz9OMITY-z6nASnnJvLqz1-nFP1W3YRs84W8zqizqKp-VCqM-UNxeRpjpEgcGLbpYoinTelhspzg/w650-h307-rw/cbse-class-12-maths-ncert-solutions-chapter-4-exercise-4-5.jpg)
Determinants Exercise 4.5 Solutions
1. Find the adjoint of the matrix
Solution
Let A =
We have,
A11 = 4, A12 = -3, A21 = -2, A22 = 1
2. Find adjoint of each of the matrices.
Solution
3. Verify A (adj A) = (adj A) A = |A|
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEipyZMpyGUYj7pbLkIG4cptJO5TpboiUY4vEg7Qucw6sx7_xFJxcwJrVRz3GAeH135p5uaMcX20YPrruiFEr7Kj9Fm9YLx1vM5IPRXz4AF40h6mH9SpVAqolYb7LnveoPUBp26eW03ogsaw454W6Ql6QKfBiFy1rjavfKjwikRTN_DdM0DgveRZ0Agn/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2017.JPG)
Solution
4. Verify A (adj A) = (adj A) A = |A|
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh_CMG8DV9VBU8PfBeQT2UEoMfbpQZEMm5HLZw_dMT6EMm1z7b2KxBS8ffrv1-Cw_naIiJjd-uNeeUkLl5pNcMmmPtsVd6oTOVn7yCgtATMu1BbKByy0wl29FUcvspfR0Jqrf-0XFiPef-hymd0TGYCzPetLLwgfvwX9iXR0Z0oTaQuCvoZTtCQvjND/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2019.JPG)
Solution
5. Find the inverse of each of the matrices (if it exists).
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjQ9vFDEtPgUc3w-cMUu9MWd9CwWqj5mY0oZoZMGnKdptZqxyipvwy-d-3wto57LOeLr6f8rZn8iryIZL8x7ZqDXphzi6HwQW1AsBKV-f_TrZpeMbHjx31mNSqCwVZwP15UkWZt45uPphe5XD-2fcMbKngSAdgRNMmkCG-kzF2i5eiSHhkMBtbgmCCh/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2021.JPG)
Solution
Let A = ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjQ9vFDEtPgUc3w-cMUu9MWd9CwWqj5mY0oZoZMGnKdptZqxyipvwy-d-3wto57LOeLr6f8rZn8iryIZL8x7ZqDXphzi6HwQW1AsBKV-f_TrZpeMbHjx31mNSqCwVZwP15UkWZt45uPphe5XD-2fcMbKngSAdgRNMmkCG-kzF2i5eiSHhkMBtbgmCCh/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2021.JPG)
We have,
|A| = 6 + 8 = 14
Now,
A11 = 3, A12 = -4, A21 = 2, A22 = 2
We have,
|A| = 6 + 8 = 14
Now,
A11 = 3, A12 = -4, A21 = 2, A22 = 2
6. Find the inverse of each of the matrices (if it exists).
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgEF76gSr9ZUGpqqxP_mNHEzt4AuVyZfbbzuRn34Wp7ZqZTJFTOIXleQxOzzBOHMPSQ9Wd5ZXEy3fhdjZPpkFPDPZ3zdtwbwaBLvKeDqUqEo_HCMFPm9iRIg6wqEbMJFaTQaRSevg6MDl9xi3uizyixpXovRNt_H_-CTIgfU2fyWAeZRlTV7ONZdTjd/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2023.JPG)
Solution
Let A = ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgEF76gSr9ZUGpqqxP_mNHEzt4AuVyZfbbzuRn34Wp7ZqZTJFTOIXleQxOzzBOHMPSQ9Wd5ZXEy3fhdjZPpkFPDPZ3zdtwbwaBLvKeDqUqEo_HCMFPm9iRIg6wqEbMJFaTQaRSevg6MDl9xi3uizyixpXovRNt_H_-CTIgfU2fyWAeZRlTV7ONZdTjd/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2023.JPG)
We have,
|A| = -2 + 15= 13
Now,
A11 = 2, A12 = 3, A21 = -5, A22 = -1We have,
|A| = -2 + 15= 13
Now,
7. Find the inverse of each of the matrices (if it exists).
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgbahRRyfIO-dn6zP6JOj1tiO-gpVeIm6ckmfi94qqGijGiQ8fD_0rdZ0rvnBf_JKPPRb9vJRmLcdPtvudYkRA6MAzIsXW2NQvB4humsVITCPJqT46WnxOoJ5m5ALubq4BayvbmAeZLoUJywO_wjcIGafKW-KC6JKOSuEXUQzNCQzVoOTxHAS0e4BxT/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2025.JPG)
Solution
Let A = ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgbahRRyfIO-dn6zP6JOj1tiO-gpVeIm6ckmfi94qqGijGiQ8fD_0rdZ0rvnBf_JKPPRb9vJRmLcdPtvudYkRA6MAzIsXW2NQvB4humsVITCPJqT46WnxOoJ5m5ALubq4BayvbmAeZLoUJywO_wjcIGafKW-KC6JKOSuEXUQzNCQzVoOTxHAS0e4BxT/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2025.JPG)
We have,
|A| = 1(10 - 0) - 2(0 - 0) + 3(0 - 0) = 10
Now,
A11 = 10 - 0 = 10 , A12 = -(0 - 0) = 0, A13 = 0 - 0 = 0,
A21 = -(10 - 0) = -10, A22 = 5 - 0 = 5, A23 = -(0 - 0) = 0
A31 = 8 - 6= 2 , A32 = -(4 - 0) = -4, A33 = 2 - 0 = 2
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi6YV8GUPd8kypkSO4zs-4gA6_bgpmnOxW6540gSDKbLrMF7ztYphGUkQnec6_Kxh216YPMR2UFoXoGyEUne4-5oQiHYG5lExdpoMhaxY7uJF1L1D4JQJtAgk7Q6TUVqgGBwnGytBEpZPikC0QvW1eUy1rP3aLsl84NTlKRKTQYcsW-l_GdDvQv8fBf/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2026.JPG)
|A| = 1(10 - 0) - 2(0 - 0) + 3(0 - 0) = 10
Now,
A11 = 10 - 0 = 10 , A12 = -(0 - 0) = 0, A13 = 0 - 0 = 0,
A21 = -(10 - 0) = -10, A22 = 5 - 0 = 5, A23 = -(0 - 0) = 0
A31 = 8 - 6= 2 , A32 = -(4 - 0) = -4, A33 = 2 - 0 = 2
8. Find the inverse of each of the matrix (if it exists).
Solution
Let A = ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjAtmyJWxdqrFecJkACxjJCyZccIAdKj03FaAOqWw6AAL1qKviU8iOaTgW6fRt8W_I23fCFrFkDWl94Rf_632dOd3nzxIH4DTHC6WV7SNDMgdpBljs8TnjH3t0rtjWVXjoiJ3X8JHZ5IOs2jlafgG562eeSzpw32kkfPem1Al6s4XuAE9Q00j7hJCjp/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2027.JPG)
We have,
|A| = 1(-3 - 0) - 0 + 0 = -3
Now,
A11 = -3 - 0 = -3 , A12 = -(-3 - 0) = 3, A13 = 6 - 15 = -9,
A21 = -(0 - 0) = 0, A22 = -1 - 0 = -1, A23 = -(2 - 0) = -2
A31 = 0 - 0 = 0 , A32 = -(0 - 0) = 0, A33 = 3 - 0 = 3
We have,
|A| = 1(-3 - 0) - 0 + 0 = -3
Now,
A11 = -3 - 0 = -3 , A12 = -(-3 - 0) = 3, A13 = 6 - 15 = -9,
A21 = -(0 - 0) = 0, A22 = -1 - 0 = -1, A23 = -(2 - 0) = -2
A31 = 0 - 0 = 0 , A32 = -(0 - 0) = 0, A33 = 3 - 0 = 3
9. Find the inverse of each of the matrices (if it exists).
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEinRq5dy7_vESm7hZ7znE8d54M4H8Ywx6Bi1sRksaxAuvm3KF1pko3Gm2nSqh6PrSkRE1oDR843TV6yFQJzs-5Hw9WvhQeinXUxNac41oGJ6niajQoKMa05JEN1qtLGA-Z3HWcjURKSIiVVYkGC0myfYF7WXhqx92vY_RBHAEHo2ccWqnMAQfaga3tR/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2029.JPG)
Solution
Let A = ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEinRq5dy7_vESm7hZ7znE8d54M4H8Ywx6Bi1sRksaxAuvm3KF1pko3Gm2nSqh6PrSkRE1oDR843TV6yFQJzs-5Hw9WvhQeinXUxNac41oGJ6niajQoKMa05JEN1qtLGA-Z3HWcjURKSIiVVYkGC0myfYF7WXhqx92vY_RBHAEHo2ccWqnMAQfaga3tR/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2029.JPG)
We have,
|A| = 2(-1 - 0) - 1(4 - 0) + 3(8 - 7)
= 2(-1) - 1(4) + 3(1)
= -2 - 4 + 3
= - 3
Now,
A11 = -1 - 0 = -1 , A12 = -(4 - 0) = -4, A13 = 8 - 7 = 1
A21 = -(1 - 6) = 5, A22 = 2 + 21 = 23, A23 = -(4 + 7) = -11
A31 = 0 - 3 = 3 , A32 = -(0 - 12) = 12, A33 = -2 - 4 = -6
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhwI-ZCimYBk5vx7_XsLYEGF4uSIeWsX6Vi7MrDD3pnTqX19XGLovMjQOO1RPT9MFyPo_6PRvnMgnVK0YVnrleNy2Ukx2QZVRfVInzLkI_Vf7cALVc0iD3c0DXlCxfnbF--7kMXa3md1Mk31CjmUxhFMEIbIG8QHMVLNfpx9NbONdSWIL9-8aZ5ym3q/w335-h174-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2030.JPG)
We have,
|A| = 2(-1 - 0) - 1(4 - 0) + 3(8 - 7)
= 2(-1) - 1(4) + 3(1)
= -2 - 4 + 3
= - 3
Now,
A11 = -1 - 0 = -1 , A12 = -(4 - 0) = -4, A13 = 8 - 7 = 1
A21 = -(1 - 6) = 5, A22 = 2 + 21 = 23, A23 = -(4 + 7) = -11
A31 = 0 - 3 = 3 , A32 = -(0 - 12) = 12, A33 = -2 - 4 = -6
10. Find the inverse of each of the matrices (if it exists).
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiEZoazZxRWKe6eJyANlacWXJf3IgO24bXnjqhPju1Eu74xenHPZcR8fn7wTYlm0skqPpoUmUJrNKJMLvGn8S97XWW1r66p57axWv9KotG3mRIW_eBiAu2TIA9k_GanNvGTE2H5lcNrRPT9CgHSWfZtqrT3y6qZc6deZYNgsMa5q0zuYp-cNkLuS4MF/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2031.JPG)
Solution
Let A = ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiEZoazZxRWKe6eJyANlacWXJf3IgO24bXnjqhPju1Eu74xenHPZcR8fn7wTYlm0skqPpoUmUJrNKJMLvGn8S97XWW1r66p57axWv9KotG3mRIW_eBiAu2TIA9k_GanNvGTE2H5lcNrRPT9CgHSWfZtqrT3y6qZc6deZYNgsMa5q0zuYp-cNkLuS4MF/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2031.JPG)
By expanding along C1 , we have :
|A| = 1(8 - 6) - 0 + 3(3 - 4) = 2 - 3 = -1
Now,
A11 = 8 - 6 = 2 , A12 = -(0 + 9) = -9, A13 = 0 - 6 = -6
A21 = -(-4 + 4) = 0, A22 = 4 - 6 = -2, A23 = -(-2 + 3) = -1
A31 = 3 - 4 = -1 , A32 = -(-3 - 0) = 3, A33 = 2 - 0 = 2
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiDyS93H5zQstbED211bZHYIExpIrfkos00DofKgLnARNlwxD9L3Mgu9ovhy2OoqLquK7g0_-JMDKRlhEzr4DJjG57roaZ9cm5sf3778ID0_cJsoge2wDe4n2K3oe1q9jWq2H7bvnIxcN1IeZk3hxGb5C1vu6etuWlCeZYivp9I4dysen-58CqsZr_B/w451-h166-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2032.JPG)
By expanding along C1 , we have :
|A| = 1(8 - 6) - 0 + 3(3 - 4) = 2 - 3 = -1
Now,
A11 = 8 - 6 = 2 , A12 = -(0 + 9) = -9, A13 = 0 - 6 = -6
A21 = -(-4 + 4) = 0, A22 = 4 - 6 = -2, A23 = -(-2 + 3) = -1
A31 = 3 - 4 = -1 , A32 = -(-3 - 0) = 3, A33 = 2 - 0 = 2
11. Find the inverse of each of the matrices (if it exists).
Solution
Let A = ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEizjEUqs3BD90wwapPHayzVTMUE35s7DMz3yRUTUeOGLBtAadMEc6UEckk-O0LAw576774lsX3dGEK0C6Q-yOctRCtw-V50V8vGesAZiC8VclUb1Gt96xE9jf5oJYlXq9lJPGbqiRInI6jzNXy1WR8adk25uqagC9hkIp_Wm09w3lt4dWWYK__hmqCW/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2033.JPG)
We have,
|A| = 1(−cos2 α – sin2α) = −(cos2α + sin2 α) = − 1
Now,
A11 = −cos2 α – sin2α = -1 , A12 = 0, A13 = 0
A21 = 0, A22 = −cosα , A23 = – sinα
A31 = 0 , A32 = – sinα, A33 = cosα
We have,
|A| = 1(−cos2 α – sin2α) = −(cos2α + sin2 α) = − 1
Now,
A11 = −cos2 α – sin2α = -1 , A12 = 0, A13 = 0
A21 = 0, A22 = −cosα , A23 = – sinα
A31 = 0 , A32 = – sinα, A33 = cosα
12. Let A =
and B =
. Verify that (AB)-1 = B-1 A-1 .
Solution
Let A = ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5deLfKJqdDfAVm8QA2tRoE-pzx1uYqcxTq9COss56Fx3kPmrDjgXWwUnEVofxQtfebs3v6gHyCh2sRKmBeoV736yhsEjydP5P0LtOKZ_eT8VKl-oSVBhhIQeaKP2VEfcM6WM6_ISuGubOnvGJ-q-eJwGOsqrdi3ZaQ64QlWbVhQARAgcIISzxJ_6p/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2035.JPG)
We have,
|A| = 15 - 14 = 1
Now,
A11 = 5, A12 = - 2, A21 = -7, A22 = 3
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhYLq8mdZMZFQ8TsoKaVpLbRo7huN9Q36MphCneeuWRCY508cZpl9blGeCCwSUodOdTMrMRs8mI_bv_da4iTkN8dRkRceJyZedhN4jsqD3VQ8SYuQbXrS38U_EV25ELt9zQMEmXLA_uM-YwsqsNga-4-ePcwUJzwv2khzoFs5QYbN0sPKTafByBLLzI/w235-h113-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2037.JPG)
We have,
|A| = 15 - 14 = 1
Now,
A11 = 5, A12 = - 2, A21 = -7, A22 = 3
Now, Let B = ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhtdxF5U3xH4kEozc4AmdHu9m8bOZ7TJHDw0kIXlW-0vi0V3VZSlhKqS5WD238awLLn9JEJigMdppCyl220ffGXpMQS80gaYhBqdtJcUikM7GRpiH030YBYgTZ-q1zH98cMoatxGe_jFDqDq_SYkj1i7-tCYkPAk_WXn8l8qq_0Q5mAypeymmYwnwRE/s1600-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2036.JPG)
We have,
|B| = 54 - 56 = -2
We have,
|B| = 54 - 56 = -2
From (1) and (2), we have :
(AB)-1 = B-1 A-1
Hence, the given result is proved.
(AB)-1 = B-1 A-1
Hence, the given result is proved.
13. If A =
show that A2 - 5A + 7I = 0. Hence find A-1.
Solution
Hence, A2 - 5A + 7I = 0.
∴ A . A - 5A = -7I
⇒ A.A(A-1 ) - 5AA-1 = -7IA-1 [Post-multiplying by A-1 as |A| ≠ 0]
⇒ A(AA-1 ) - 5I = -7A-1
⇒ AI -5I = -7A-1
⇒ A-1 = - (1/7)[A - 5I]
⇒ A-1 = (1/7)[5I - A]
14. For the matrix A =
find the numbers a and b such that A2 + aA + bI = 0.
Solution
Now,
A2 + aA + bI = 0
⇒ (AA)A-1 - aAA-1 = bIA-1 = 0 [Post-multiplying by A-1 as |A| ≠ 0]
⇒ A(AA-1 ) - aI + b(IA-1 ) = 0
⇒ AI + aI + bA-1 = 0
⇒ A + aI = -bA-1
⇒ A-1 = -(1/b)[A + aI]
Now,
15. For the matrix A =
show that A3 - 6A2 + 5A + 11I = 0 . Hence, find A-1 .
Solution
A3 - 6A2 + 5A + 11I = 0
⇒ (AAA)A-1 - 6(AA)A-3 + 5AA-1 + 11IA-1 = 0 [Post-multiplying by A-1 as |A| ≠ 0]
⇒ (AAA)A-1 - 6(AA)A-3 + 5AA-1 + 11IA-1 = 0 [Post-multiplying by A-1 as |A| ≠ 0]
⇒ AA(AA-1 ) - 6A(AA-1 ) + 5(AA-1 ) = 11(IA-1 )
⇒ A2 - 6A + 5I = -11A-1
⇒ A + aI = -bA-1
⇒ A-1 = -(1/11)[A2 - 6A + 5I] ...(1)
Now,
A2 - 6A + 5I
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgMIs6FZbudE31Dx2jzhFeU282Jgq9IRFPYi3FNknOI9clIdXwGwCnF0DXgJyvb23t_RJIYGBPGBuErK--YEnE2KmSVGAL9CfWG79CBF5iZPbOX6qcn0-XwhYiLSKiGY5PKxc-pTocuRPnmalAzL9piLexKgJTzrFuTeAXa-3sMZfZH8StZu4v2sBNM/w453-h440-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2047.JPG)
⇒ A-1 = -(1/11)[A2 - 6A + 5I] ...(1)
Now,
A2 - 6A + 5I
16. If A =
verify that A3 - 6A2 + 9A - 4I = 0 and hence find A-1.
Solution
∴ A3 - 6A2 + 9A + 4I = 0
Now,
A3 - 6A2 + 9A + 4I = 0
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhF89mHwdRt6nHbUlUZc2HWGkqL3ksy_3hGs7i5A_OnkvK8gWKbV-95-rLz56jQWORrbvz7MVLGYQlnMn8eeP_KBZLltyEPm2CLaWPoigWBdfu6sVUH-awzQCeRp161RRFnNi-ISJeB5oDFcd1BExn-ylxtkG-q43hqaePGjZttK6YnoOMve41UTgGP/w317-h345-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2051.JPG)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgct-vlyZslyU47eIjdmgNDoqmbQ7FkO29sW7QTh4C6Qiyjdju5blkypDNkHANzkgG34H5PQYBuwI3CK_GyRqyLT_WX3GBeKzDwdTP3CQQKj5JbwtO_GAN54C7X7HmHNdbYgk_K-XN6Er3IZdqNScJQb-0j4hM6oel1cH5Kp-qYWesdS2A077LXi5qJ/w470-h440-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2052.JPG)
Now,
A3 - 6A2 + 9A + 4I = 0
⇒ (AAA)A-1 - 6(AA)A-3 + 9AA-1 + 4IA-1 = 0 [Post-multiplying by A-1 as |A| ≠ 0]
⇒ AA(AA-1 ) - 6A(AA-1 ) + 9(AA-1 ) = 4(IA-1 )
⇒ AAI - 6AI + 9I = 4A-1
⇒ AAI - 6AI + 9I = 4A-1
⇒ A2 - 6A + 9I = 4A-1
⇒ A + aI = -bA-1
⇒ A-1 = (1/4)[A2 - 6A + 9I] ...(1)
A2 - 6A + 9I
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjbuStzA2j5SYRM1eaiNJnUzyNV7HjQW7SjIL-fqBKqLMl8WK0wvAVx8VhhARLDJl4IMl4gFEwrGXyd9xfP8HhBQvZzcH6PsbJ4jWKEFd8E2LQ0fUoMmsjJqUYLEz9X0HHKDaHRow1_cT4FCO2sKWFoQ2i5ewAhiTancY-ImSWfFqvE2m3af_tFKiEP/w473-h389-rw/NCERT%20Solution%20Class%2012%20Chapter%204%20-%20Determinants%20Exercise%204.4%20img%2050.JPG)
⇒ A-1 = (1/4)[A2 - 6A + 9I] ...(1)
A2 - 6A + 9I
17. 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|
(A) |A |
(B) |A|2
(C) |A|3
(D) 3|A|
Solution
We know that,
18. 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
(A) det (A)
(B) 1/det (A)
(C) 1
(D) 0
Solution
Since A is an invertible matrix, A-1 exists and A-1 = (1/|A|) adj A.