12th Maths | Chapter 01 | Exercise 1.1 - All example sums one shot | Ist Mid-Term Revision

[00:00:00]Hello my dear 12.

[00:00:02]So I'm here with a quick revision session.

[00:00:08]So sentence chapter one max applications of matrices and determinance.

[00:00:23]So first example 1.1 if a is equal to a mat verify that a into adjint a is equal to adjint a into a is equal to determinant of a into i3 so adjint of so of Okay.

[00:01:00]And remember to give a transpose here.

[00:01:06]So 7 - 4 - 4 3. So 7 - 4 - 4 3. Next - 6 in the row and the column - 6 - 4 2 3 - 6 -4 2 3 So the two in the row in the column - 6 7 2 - 4 - 6 7 2 - 4 Next - 6 in the row in the column - 6 2 - 4 3 - 6 2 - 4 3 Next 7 in the column in the row 8 2 3 8 2 2 3 Next the minus 4. So in the row in the column 8 - 6 2 - 4 8 - 6 2 - 4 Next in the two in the row in the column - 6 2 7 - 4 - 6 2 7 - 4 Next in the - 4 row 8 - 6 - 4 8 2 - 6 - 4 Next in the three in the row in the column 8 - 6 - 6 7 8 - 6 - 6 7 Okay.

[00:02:46]See plus - plus - plus - plus - plus plus of 7 3 are 21 - - 4 into - 4 16 next minus of - 6 into 3 - 18 - - 4 into 2 - 8 next plus - 6 into -4 is + 24 - 7 2 are 14. Next - 6 into 3 minus - 6 into 3 is - 18 - - 4 into 2 - 8 already + 8. Next next plus of 8 3 are 86 24 - 2 2's are 4. Next minus of 8 into -4 is - 32 - 6 into 2 - 6 into 2 - 12 minus + 12 the transpose.

[00:03:54]Okay. So plus of - 6 into -4 is + 24 - 7 2 are 14. Next minus of 8 into -4 - 32 - - 6 into 2 - 12 already minus + 12 plus of 8 7 are 56 - - 6 into - 6 36 addition subtraction.

[00:04:20]Okay. 21 16 + 5. Okay. Next - 18 + 8us.

[00:04:30]Next 24 14 10 plus next - 18 + 8 - 10 alreadyus + 10 24 4 20 then - 32 + 12 is 20US already + 20 next in the third row 24 -4 10 plus 10 next - 32 + 12 - 20 already. Minus plus 20 56 36 20 plus sign. So plus 20 transpose 5 10. Next 10 20 10 20 Next 10 20 10 20 adjint of A into adjint of A into adjint of A. So another question 8 - 6 2 - 6 7 - 4 2 - 4 3 adjint 5 10 10 20 10 10 20 20 20 Okay.

[00:06:15]Yes. First row, first column. Okay.

[00:06:20]8 5 are 40 - 6 into 10 - 60 20 into 10 + 20.

[00:06:27]Next first row second column 8 into 10 80 - 6 into 20 - 120 2 into 20 + 40.

[00:06:36]First row third column 8 into 10 80 - 6 into 20 - 120 2 into 20 + 40.

[00:06:46]Second row first column - 6 into 5 - 30 7 into 10 + 70 - 4 into 10 - 40. Second row second column - 6 into 10 - 60 7 into 20 + 140 - 4 into 20 - 80. Second row third column - 6 into 10 - 60 7 into 20 is + 140 - 4 into 20 - 80 Next third row first column 2 into 5 10 - 5 4 into 10 - 40 3 into 10 + 30 Third row second column 2 into 10 20 - 4 into 20 - 80 3 into 20 + 60.

[00:07:35]Third row, third column. 2 into 10 20 - 4 into 20 - 80 3 into 20 is + 60 matrix.

[00:07:46]Okay.

[00:07:50]40 + 20 60 - 60.

[00:07:54]So 0 80 + 40 120 - 120 same 0 0 - 30 - 40 - 70 + 70 0 - 60 - 80 - 140 + 140 same. So 0 10 + 30 40 - 40 0 20 + 60 80 - 80 0 same adjint of first adjint of A into A of A into a PDF.

[00:08:54]Next adjint of 5 10 10 20 10 20int of a question 8 - 6 2 - 6 7 - 4 2 - 4 3 Okay multiply first row into first column 58 are 40 10 into - 6 - 60. 10 into 2 is + 20.

[00:09:27]Next first row second column 5 into - 6 - 30 10 into 7 + 70 10 into -4 is -40.

[00:09:37]Next first row third column uh 5 into 2 is 10. 10 into -4 is -40. 10 into 3 is + 30. The second row first column 10 into 8 80. 20 into - 6 - 120 20 into 2 is + 40. Second row second column par 10 into -6 - 60 20 into 7 is + 140 20 into -4 is - 80. Next second row third column 10 into 2 is 20. 20 into -4 is -8 20 into 3 is + 60.

[00:10:19]Next is third row first column 10 into 8 is 80. 20 into - 6 is - 120. 20 into 2 is + 40.

[00:10:29]Third row second column 10 into - 6 is -60. 20 into 7 is + 140. 20 into -4 is - 80. Third row third column 10 into 2 is 20. 20 into -4 is - 80. 20 into 3 is 60.

[00:10:46]So matrix symbol 40 + 20 60 - 60 - 30 - 40 - 70 + 70 0 So 10 + 30 40 - 40 80 + 40 120 - 120 - 60 - 80 - 140 + 140 20 + 60 80 - 80 0 80 + 40 120 - 120 - 60 - 80 - 140 + 140 20 + 60 80 - 80. So adjint of A into A. So I determinant of A into I3.

[00:11:35]So determinant of a so determinant symbol 8 - 6 2 - 6 7 - 4 2 - 4.

[00:11:47]So expanding along R1 8 into the column in the row 7 3 are 21us - - 4 into -4 16 plus - plus - into - + 6 in the row - 18 + 8 - 18 + 8 + 2 into - 6 into -4 is 24 - 7 2 14. So the next step 8 into 21 16 5. Okay. So 5 + 6 into - 18 + 8 - 10 + 2 into 24 14 10. Okay.

[00:12:33]So 8 5 are 40. 6 into 10 60 + uh 2 into 10 20. So 40 + 20 it is 60. Yeah. 60 - 60. So that determinant of A into I3 determinant of A into I3. So determinant of A. So 0 I3 the third equation into adjint of A I3 is the first equation adjint of A into A I3 second equation and determinant of A into I3 I3 is the third equation. So, so from equation 1, 2 and 3 we conclude that.

[00:13:45]So a into adjint of A equals to adjint of A into A equals to determinant of A into three easy sum. Okay. Second if A is equal to matrix 2x2 matrix A B C D is non singular find A inverse A inverse A inverse does not exist.

[00:14:24]So inverseal a inverse exists easy. Okay.

[00:14:42]If a is equal to a b is non singular non singular determinant not equal to zet inverse a inverse formula 1 by determinant of a into adjint of Determinant of matrix A C minus B C. Okay. Adjint of A. Okay. Or 2x2 matrix.

[00:15:31]First 2x2 matrix 2x2 matrix 2 3 4 5 2x2 matrix elements change fourus.

[00:16:07]Okay.

[00:16:16]elements.

[00:16:19]So d inverse formula a inverse is equal to 1 by determinant of a d minus b c 1 / a d minus b cint d minus b minus c a so a inverse answer.

[00:16:51]Okay. Third, find the inverse of the matrix mat.

[00:16:58]So 2x2x3 matrix.

[00:17:12]So first determinant. So determinant 2 into 3 3's are 9 - 2 minus of - 1 + 1 into - 5 into 3 -5 - - 3 into 1 - 3 so + 3 I know + 3 into the column - 10 - 9 so - of - + 9 so 2 into 9 - 2 7 + 1 into -15 + 3 -2 + 3 into - 10 + 9 - 1. So 2 7s are 14 1 into -2 -2 3 into -1 - 3. So -12 - 3 -5 + 14 - 15 + 14 -1. So determinant of so a inverse exist. Therefore a inverse exist.

[00:18:19]Okay.

[00:18:21]Okay. Determinant of adjint of A. a joint of transpose.

[00:18:39]Okay.

[00:18:50]3 1 2 3.

[00:18:52]Okay. in the two in the row in the column 3 1 2 3. So the determinant 3 1 2 3 Next in the minus one in the column in the row - 5 1 - 3 3 - 5 in the row in the column - 5 1 - 3 3 1 - 3 3 Okay 3 - 3 2 - 5 3 - 3 Next in the - Okay. So the row and the column - 1 3 2 3. So -1 3 2 3 3. Next in the column row 2 3 - 3 3 2 3 - 3 3 Next in the row 2 - 1 - 3 2 -1 - 3 2 Next - 3. So the row -1 3 3 1.

[00:20:00]So in the two in the column in the row 2 3 - 5 1 2 3 - 5 1 Next in the three. So in the column 2 - 1 - 5 3 2 -1 - 5 and sign.

[00:20:19]So plus - plus plus - - plus + - - plus + - - plus determinance expand equal to uh plus of 3 3's are 9 - 2 7 - of - 5 + 3 - 15 uh + 3 - 12 next plus of - 10 - 5 into 2 - 10 + 9 so -1. Next second row minus of -3 - 6 so - 7 sorry -3 - 3 - 6 - 9 sorry next plus of 2 3's are 6 + 9 so 15 minus of 2 2's are 4 - 3 up one next plus of -1 - 9. So, - 10 - of 2 1 are 2 + 15 7 17 - 15 - 2. So, -7 plus of 2 3's are 6 - 5. So one transpose sorry 2 + 15 so + 17 okay sorry okay transpose 7us next minus of - 9 9 15 - - 10 -7 1. Okay. So joint of a determinant So in the formula inverse okay so next okay a inverse is equal to determinant of a 1 by determinant of a determinant of a minus so min -1 okay and into adjint of a adjint of a 7 9 - 10 12 15 - 17 -1 -1 1 /us soy 12 15 - 17 -1 -1us okay matrix number scalar multiplication.

[00:23:35]So number - 7 9 - 9 - 12 - 15 + 17 1 1 Okay inverse next okay next question if A is non-s singular matrix X determinant not equal to Z of odd order.

[00:24:10]Prove that determinant of adjint A is positive.

[00:24:15]Okay. So order of the matrix order of the matrix matrix A number 2 + 2k + 1.9 formula determinant of adjint A is equal to determinant of A to the power NUS. So order of the matrix.

[00:24:58]So determinant of adjint A is equal to determinant of a matrix. Okay.

[00:25:07]Order of matrix 2k + 2k + 1us cancel. So determinant of a ^ 2ket determinant of a power k the square.

[00:25:43]Therefore uh determinant of a 2k is always positive square.

[00:26:01]So determinant of a determinant of a determinant of so determinant of adjint A is always positive.

[00:26:24]Hence proved.

[00:26:32]Okay. Next.

[00:26:34]Find a ifal.

[00:26:37]So direct.

[00:26:47]Okay. So a is equal to plus or minus 1 / roo<unk> of adjint A into adjint of adjint A.

[00:27:00]So first root of determinant of so first determinant of adjint.

[00:27:12]So first matrix determinant of adjint.

[00:27:20]So 7 into 11 into 7 77 - 7 into 5 35 plus - plus follow - 7 of the row in the column - 7 - 77 - 7 of - 5 - 121 So 77 - 35 7 - 5 2 7 7 - 3 4 - 7 of 7 - 7 7 - 7 So add - 84 Okay - 7 of - 126 7 into 42 7 2 are 14 4 1 7 4 are 28 29 - into - plus 7 4 are 28 8 2 7 8 are 56 2 58US into minus plus 7 6 42. So 2 7 7 2 are 14 18 7 1 are 7 + 1 8. So 2 + 8 10 + 4 14 1. So 8 + 8 + 8 16 + 9 25 or 6 26 2 8 8 8 + 2 10 + 5 15 + 2 17.

[00:29:02]So root of adjint 1764 sorry 1764.

[00:29:23]Okay.

[00:29:28]So matrix or transpose.

[00:29:41]So number 117 minus 7 5 are 35.

[00:29:53]Next in the in the row in the column - 7 - 77.

[00:30:02]Next in the row in the column - 5 - 121 - 5 - 121. Next in theus row in the column 49 + 35. Okay. 49 + 35.

[00:30:20]Next in the seven row in the column 7 5 are 35 minus 77 35 sorry sorry in the 11 in the row in the column 7 7 are 49. So 49 49 + 77 next in the seven. So in the row column 7 5 are 35 minus 7 11 are 77. Okay.

[00:31:00]Next in the 11 in the row in the column 7 7s are 49 plus 49 + 77 mix in the five. So 7 7s are 49 - 7 49 - 7. So the seven row 77 77 - 7 sorry minus one 7 + 7 sign conventions follow plus - plus - plus plus - plus - plus plus just the addition subtraction.

[00:31:46]So 77 - 35 42. So + 42. Next - 7 - 77 - 84us + 84 - 5 - 121 - 126 plus 49 + 35 84us - 84 49 + 77 126 35 - 77 is 42 - 42 already + 42 32 49 + 77 126 plus so 126 next 49 - 7 42us 42 7 + 7 84 so plus 84 transpose okay So transpose rows.

[00:32:51]So 42 84 - 126 - 84 126 42 126 - 42 84 transpose adjint of adjint So 42.

[00:33:17]So 42 42 42 into 42 42 into 2 84 42 into 3 6 126. So 42. So 42 common earth 42 one times 84 two times 126 three times 84 two times 126 three times 42 one time the three time one time two time so adjint of adjint of determinant of adjint So just so a= Okay.

[00:34:20]A is equal to 1 / roo<unk> of determinant of adjint. Determinant of adjint 1764 into adjint of adjint 42 into 1 - 2 3 2 3 - 1 - 3 1 2. So 1764 square.

[00:34:51]So 42 into 42 is 2 4 2 2's are 4 2 4 are 8 4 2's are 8 4 4's are 16. So 4 8 + 8 16 1. Yeah.

[00:35:03]Just so roo<unk> of 1764.

[00:35:10]So 42 42 by 42 plus for into 1 - 2 3 2 3 - 1 - 3 1 2 copy paste it. So 42 42 cancel. So plus or minus of 1 - 2 3 2 3 -1 - 3 1 Okay.

[00:35:41]So next Okay. Next.

[00:35:55]Okay. So, a inverse is equal to plus or minus 1 / roo<unk> of determinant of adjint A into adjint A root of A.

[00:36:14]So root of determinant determinant determinant of adjint is equal to -1 of 1 1 - 2 2's are 4 + - plus follow - 2 of so 1 - 4 1 - 4 + 2 of 2 - 2 so 2 - 2 so next -1 of 1 - 4 - 3 - 2 of 1 - 4 - 3 and + 2 of 2 - 2 0 - into - + 3 okay + 3 - plus + 6 + 0 2 into 0 0 So 3 + 6 9 determinant of adjint 3 + 6 Okay. Okay. So, averse plus or minus 1 / roo<unk> of roo<unk> of determinant of adjint. So root of three roo<unk> of 9 into adjinty.

[00:37:43]So -1 2 2 1 1 2 2 2 1 next step 9.

[00:37:54]So a inverse is equal to 1 by plus or minus plus or minus 1x 3 of -1 2 2 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 A inverse question. So next question.

[00:38:17]Okay. If A is symmetric. So 11 concepts symmetric Apose.

[00:38:31]Okay.

[00:38:34]Prove that adjint A is also symmetric.

[00:38:38]Aintmatos.

[00:38:46]So adjint of the whole transpose to prove so adjint of a whole transpose. So 1.9 of a the whole transpose adjint of a transpose.

[00:39:16]So adjint of a transpose of apostalint.

[00:39:43]So the transpose therefore adjint of a is equal to adjint of a the whole transpose. So adjint of A is symmetric.

[00:40:17]transpose of transpose.

[00:40:24]So adjint of so adjint of transpose adjint of apose so that pro so adjint is symmetric. Hence, hence proof next.

[00:40:58]Okay. Question. Verify the property. A transpose the whole inverse is equal to A inverse the whole transpose with a A matrix.

[00:41:11]So transpose.

[00:41:17]So first transpose a transpose is equal to 29.

[00:41:25]Next 1 7 1 transpose.

[00:41:35]Next meaning for 1 / determinant 1 a transpose or determinant and adjint of a transpose.

[00:41:55]So a transpose the whole inverse determinant of a transpose determinant of a transpose 2 are 14 - 1 9 so 14 - 9 5 so adjint of a transpose elements change.

[00:42:29]So a transpose the whole inverse 1 by determinant 1x 5 into 7 -1 - 9 2 so a transpose inverse first ls next hs r hs a inverse the whole transpose so First inverse.

[00:43:02]Okay. A matrix 2 91 7. So the matverse inverse determinant of 27 14 - 9 same determinant of a adjint of adjint of a same method 2x2 matrix 2.

[00:43:30]Okay. So determin 1 by determinant of a into adjint of a.

[00:43:45]So a inverse 1x 5 into adjint of 7 - 9 -1 2 A inverse the whole transpose A inverse the whole transpose. So a inverse 1 by 5 7 - 9 -1 2. So transpose so 1 / 5 7 - 9 - 1 second equation. Okay. A inverse the whole transpose. So first equation a transpose the whole inverse is equal to 1x 5 of 7 - 1 - 92. So second equation a inverse the whole transpose 1x 5 7 -1 - 9 from equations one and two we conclude that A transpose the whole inverse equal to A inverse the whole transpose.

[00:45:17]So hence verified.

[00:45:20]Hence verified.

[00:45:23]Okay.

[00:45:26]Okay. Next question. Verify a the whole inverse is equal to B inverse A inverse property invers with A and B the whole A= 0 - 3 1 4 B -2 -3 0 and - multiply just multiply. So first row first column 0 into -2 0 -3 into 0.

[00:46:12]Next first row second column 0 into -3 0 -3 into -3 + 3. Next uh second row first column 1 into -2 - 2 4 into 0 0 + 0.

[00:46:28]Second row second column 1 into -3 - 3 4 into -1 - 4. So 0 3 - 2 - Okay.

[00:46:49]So determinant of 0 - 3 into - 2 - 6 already - 6 = 6 adjint of a diagonal elements replacement okay - 7 0 other elements 3 + a the whole inverse easy.

[00:47:24]Okay.

[00:47:26]So a the whole inverse a the whole inverse. Okay. Determinant value 1 by 6 adjint of a copy - 7 - 3 2 0 A the whole inverse L so B inverse A inverse so B inverse inver okay so B inverse plus one by determinant of B into adjint of B matrix.

[00:48:11]So B matrix - 2 - 3 0 -1. So determinant of - 2 into -1 2 - - 3 into 0 0 so 2 - 0 adjint of B replacement so -1 - 2 sign change -3 + 3 0 so B inverse inverse 1 by 2 of -1 3 0 - 2 next a inverse okay a inverse so a inverse 1 by determinant of a into adjint of a so a 0 - 3 1 4 so determinant 0 as 0 1 by uh 0 - - 3 into 1 - 3 + 3 adjint replacement 4 0 sign change 3 -1 so a inverse A inverse 1 by 3 of 4 3 - 1 B inverse A inverse. So B inverse A inverse MIP.

[00:49:49]So B inverse A inverse is equal to B inverse 1 by 2 of 1 by 2 of -1 3 0 - 2 -1 3 0 - 2 A inverse 1x 3 of 4 3 - 1 0 So 1x 2 1x 3 multiply So 2 3 are 6. So 1 by 6 and -1 3 0 - 2 copy paste 4 3 - 1 0.

[00:50:30]Okay. Next step matrix multiplication first row first column 1x copy first row first column - 1 into 4 - 4 3 3 into -1 - 3 first row second column -1 into 3 - 3 3 into 0 + 0 into 4 second row first column 0 into 4 0 - 2 into -1 + 2 second row second column 0 into 3 0 - 2 into 0 0. So then I 1 by 6 of - 7 - 3 2 0 B inverse A inverse. So second equation.

[00:51:15]So from first and second equation first equation A the whole inverse 1x 6 of - 7 - 3 2 equation B inverse A inverse is equal to 1x 6 of - 7 - 3 2 0.

[00:51:36]Okay.

[00:51:38]From equations one and two we conclude that conclude first equation A the whole inverse. Second equation B inverse A inverse. So equal So hence verified.

[00:52:05]Okay. Next question.

[00:52:08]If a= mat find x and y such that in the equation x and y.

[00:52:20]So first square a into a upon 4 325 into 4 3 2 5. Okay. So first row first column 4 4's are 16 3 2's are 6.

[00:52:46]First row, second column, four fours are sorry, four 3's are 12, 3 fives are 15.

[00:52:52]Second row, first column 2 4's are 8 + 5 2's are 10. Second row second column 2 3's are 6 + 5 5 are 25. So 16 + 6 22 12 + 15 is 27 8 + 10 is 18 6 + 25 is 31 a square. So in the equation equation given that a² + x a + y i2 is equal to z.

[00:53:31]So A square 22 27 18 31 + X into A a question 4 3 2 5 + Y into I2 I2 identity matrix.

[00:53:54]So 1 0 0 1 = 0 in the step. Okay.

[00:54:06]So in the a square I just copy paste 22 27 18 and 31 plus x 4x 3x 2x 5x y into 1 y anything into y= 0.

[00:54:33]So equations.

[00:54:38]So 22 + 4x + y. Okay. 22 first element 22 + 4x + y.

[00:54:50]Second elements 27 + 3x 0. So next 18 + 2x + 0. So 18 + 2x 31 + 5 x + y 31 + 5 x + yal 0.

[00:55:11]So 0.

[00:55:29]So 27 + 3x 0. So 27 + 3x = 0. So 3x 27 + 27 position - 27. So 3 1 * 27 9. So x - 9. Okay. XUS 9 22 + 4x + y = 0 - 9 22 + 4 into - 9 + y = 0. So y - 22us 36 + 36. So y 6 2 4 3 2 1. Sous + 14. Okay.

[00:56:31]So, so hence find a square + x a + y i2 = 0 x - 9 y + 14. So substitute a² - 9 a + 14 I2= post multiply post multiplying by a inverse a inverse post multiply on both sides.

[00:57:41]Okay. So last a inverse. So a² a inverse - 9 a a inverse + 14 I2 into a inverse is equal to 0. So just 2 - 1 just add 2.

[00:58:13]So - 1 - 1. So anything into I2 I2 14 into anything 1 into I2 into a inverse a inverse equal to Z.

[00:58:39]So 14 A inverse 9 I2US - 9 I2 plus a minus A inverse is equal to 1 by 14 of 9 I2 A I 1 by 14 of 9 into I2 I 1 0 0 1 Next A 4 3 2 5 Okay 1 by 14 of 9 0 0 9us the 4 3 2 5 Okay so 1 by 14 of just subract 9 - 4 5 0 - 3 - 3 0 - 2 - 2 9 - 5 4 So inverse 1x 14 of 5 - 3 - 2 4 So answer 10.

[00:59:55]Okay. So next question.

[00:59:58]Okay. Question. Prove that cos theta sin theta sin theta is a a transpose is equal to identity matrix. So cos theta minus sin theta sin theta cos theta a transpose wait a second.

[01:00:36]So a transpose so theta sin theta cos theta sin theta sin theta cos theta sin theta cos theta mat okay so a apose is equal to a question cos theta minus sin theta sin sin theta cos theta. Next a transpose cos theta minus sin theta sin theta cos theta okayip.

[01:01:20]So first row first column cos theta cos theta cos² theta minus sin theta minus sin theta plus sin² theta next first row second column cos theta sin theta is sin theta cos theta minus sin theta cos theta so minus sin theta cos theta next second row first column sin theta into cos theta sin theta cos theta cos theta minus sin theta and I minus sin theta cos theta next second row second column so sin theta into sin theta sin² theta cos theta into cos theta plus cos² theta matrix yeah sin theta cos theta sin theta same theta theta US and since sin square theta + cos² theta is equal to 1 sin square + square 1.

[01:02:37]So I2. So since a Apose Ape Apose, so A is an orthogonal matrix.

[01:03:00]Easier. Okay. Next.

[01:03:04]Next question.

[01:03:06]If a equal is orthogonal find a b and hence find a inverse find okay and hence find inverse.

[01:03:37]Okay.

[01:03:42]Soon matrix Apose 1 by 7 Apose 1 by 7 6 - 3 A 6 - 3 A B - 2 6 B - 2 6 2 C3 2 C3 Okay. So a a transpose is equal to so a transpose is equal to a 1x 7 of 6 - 3 a b - 2 6 2 c 3 a transpose 1 by 7 6 b 2 - 3 - 2 c a 6 3 Okay. So multiply 1x 1 1x 7 1x 7 1 by 49 first row first column 6 into 6 36 - 3 into - 3 + 9 a into a a² + a square first row second column 6 b + 6 6 b + 6 + 6 a next first row Third column 12 6 2's are 12 - 3 C + 3 A second row first column 6 B + 6 A 6 B + 6 A + 6 A Okay, next second row second column B into B² + 4 + 36 Six.

[01:05:49]Next. Second row, third column. B into 2 upper 2 B sorry second row third column. 2 B - 2 C + 6 into 3 18. Next third row first column 2 into 6 12 uh - 3 C + 3 A. Third row second column 2 B - 2 C + 3 into 6 18. Third row third column 2 2's are 4 + C into C C² + 3 3's are 9. So addition subtraction solve.

[01:06:35]So 1x 49 into 36 + 9 45. So, a square + 45.

[01:06:51]So, 6 a + 6 b + 6 like terms 3 a - 3 c + 12 like terms. So, 6 a + 6 b + 6 36 + 4. So b ² + 40 2 b - 2 c + 18 like terms. So 3 a - 3 c + 12 2 b - 2 c + 18 like terms 9 + 4 13 so c² + 13 a transpose so apose i3 matrix 1x 49 of Okay.

[01:07:56]I3 1 0 0 0 1 0 0 0. Okay.

[01:08:03]A transposeal I division.

[01:08:14]So next step by 49 left 49 into 49 49* 0. So 49 0 0 0 49 0 0 0 49 the mat square A square B square C square A square + 45 So A square + 45 49 a square 49 - 35 allow the four.

[01:09:14]Next b² + 40 49. So b ² + 40 = 49. B² value 49 - 49.

[01:09:24]Next C² C² + 13 is equal to 49. So C square.

[01:09:32]So 13 - 3 6 4 - 1. So a square bus next set of equations 6 a + 6 b + 6 a + 6 b + 6 = 0. Next equations require 3 a - 3 z + 12 is equal to 0. So 3 a + sorry 3 a - 3 c + 12 is equal to 0. Next equation 2 b - 2 c + 18 is equal to 0.

[01:10:27]So equation first 6 a + 6 b + 6= 0 a + b + 1 = 0 6 0 by 6. So in the equation a + b + 1 = a + 1 the first equation second equation 3 a - 3 c + 12 = 0.

[01:11:13]Second equation 3 a - 3 c + 12 is equal to 0.

[01:11:22]3 of a minus c + 4 = 0.

[01:11:28]by a c= a minus c + 4. So four a minus c minus 4 a. Okay. So third equation third equation 2 b - 2 c + 18 is equal to 0. 2 b - c + 9 = 0. B - cus first equation second equation.

[01:12:05]First equation minus second equation.

[01:12:08]First equation a + b equal to minus. So a + b equal to minus1. Second equation a minus c minus of a minus c -4us of -4 a + a + c. So -1 minus of minus plus 4 plus a minus a cancel b + c is equal to 4 - b uh sorry second equation minus sorry third equation minus fourth equation third equation b minus c= -9 9. So b - c = - 9. Fourth equation minus of b + c - 3 sorry - b - c - b - c - 3 - 9 - 12 + b - cancel. So - 2 C - 12 - cancel C 12 by 2. So 12 C value fourth B + C= 3 B + C = 3 B 3 C B = 3 - 6 So -3 Next a B= A + B = So Bus A= -1 - B -1us of -1 + 3. So A so A B C A B C.

[01:14:44]Hence find a inverse a b 2 - 3 6 2 - 3 6 matute values. So 6 - 3 - 2 6 2 3 So okay next inverse information.

[01:15:46]So, So a transpose is equal to 1 by 7 - 3 2 6 - 3 2 - 3 - 2 6 - 3 - 2 6 2 6 3 2 6 A inverse Okay. So next step a inverse is equal to 1 by 7 of 6 - 3 2 - 3 - 2 6 2 6.1 Thank