Learn smart, Score High in KCET Maths 🚀 Matrices & Determinants Part 3....

Ajouté : 2026-05-30

[00:00:00]Hello dear audience, welcome back to my channel Maths with Gita.

[00:00:06]So in today's video, we continue with our next examples.

[00:00:11]That is KCET examples on two topics, matrices and determinants.

[00:00:16]Beginning with example number 21. Here, f of x is equal to they have given a determinant. In the first row we have x minus 3, 2x squared minus 18, 3x cubed minus 81. In the second row, x minus 5, 2x squared minus 50, 4x cubed minus 500.

[00:00:33]The third row they have given 1 2 3. And the question is here, what is f of 1 into f of 3 plus f of 3 into f of 5 plus f of 5 into f of 1. So if I calculate here, see in the first row we have x minus 3. 3 is there. So let us calculate first what is f of 3 here. So already they have given f of x to us, it's a determinant. So let us calculate f of 3.

[00:00:57]See when we put, that is x is equal to 3 here, you just compare. Here x, here what we have that is 3. So you put 3 here. 3 minus 3, we'll get in the first row first element that is zero. Here when we put 3, 3 squared 9, 9 twos are 18, 18 minus 18 yes it is again zero.

[00:01:17]Next here, 3 cubed 27, 27 threes are again 81. 80 minus 81 minus 81 is again zero. So in the first row of this determinant, all the elements are zero here. If we expand along first row, that is we'll get here that is determinant value zero. Or you can make use of a property of determinants also. If any one row contains all the elements zero, then the determinant value is how much here? Zero.

[00:01:43]>> [snorts] >> So f of 3 value is zero. Now similarly, in the second row we have x minus 5. So if I calculate f of 5 now, then what we get here is that is f of 5 U equal to determinant of we'll do the calculation directly students that is for the second row. So in the second row whatever X is there we put 5 U now. 5 minus 5 yes you will get here the element zero. Next 5 squared 25 25 twos are 50 yes again it is 50 minus 50 zero. Similarly you put here 5 U cubed 20 5 U 25 into 5 U that is 125 U into 4 it is again 500 500 minus 500 that is zero. So in the second row all the elements are zero here. So if I expand along second row yes this value is again zero. And this what we got here is f of 3 zero f of 5 U is zero. So here f of 1 we not calculated so we'll keep as it is but f of 3 is zero here anyhow zero into anything else what is answer zero. This is zero this is zero plus f of 5 U is again zero into f of 1 means it is again zero and hence students the entire answer is how much here that is zero and that is option two. This is example number 21.

[00:03:03]Beginning with that is example number 22. Yes here they have given that is matrix P and that is equal to matrix P is equal to 1 alpha 3 1 3 3 3 and 2 4 4. And they say that this matrix P is the adjoint of the matrix that is a 3 cross 3 and they have given here again that is determinant of A also it is 4. Then the question is here what is the value of alpha? So in this matrix P we have one unknown that is alpha that you should find now. First of all point to be noted P is the adjoint of the matrix A so you can write like this. Here P is nothing but adjoint of A and that is equal to what is matrix adjoint of A is given. 1 alpha 3 1 3 3 2 4 4 Now, again, what is given is determinant of A and that is equal to 4. Now, we know the relation between determinant of adjoint of A and determinant of A and the relation is determinant of adjoint of A is always equal to determinant of A raised to n minus 1. Now, let us find the determinant of the adjoint. That is you can expand here along first row plus minus plus. So, 1 into bracket leaving first row first column 3 into 4 12 minus 4 into 3. Yes, it is again 12.

[00:04:36]Minus alpha into bracket leaving first row second column 1 into 4 4 minus 2 3's a 6. Next, plus 3 into bracket leaving first row third column 1 into 4 4 minus 2 3's a 6. Is equal to the determinant of A is 4 here and the n stands for order. Here, the order is 3. 3 minus 1 means it is 2 4 squared. Later, you can write this as a 16. 12 minus 12 0 leave it. Multiply minus alpha. Minus alpha into this is 4 minus 6 is minus 2 plus 3 into bracket again 4 minus 6 is minus 2 that is equal to 4 squared that is 16.

[00:05:18]This is 2 alpha. This is minus 6 is equal to 16.

[00:05:24]So, 2 alpha is equal to 16 plus 6 which is 22. 2 alpha is equal to 22 and this alpha is equal to 22 divided by 2 and alpha value is 11. And that is option three.

[00:05:41]>> [clears throat] >> This is example number 22.

[00:05:48]Next, example number 23. Here first they have given A is equal to Okay, here A is equal to they have defined determinant of X 1 1 X means A stands for here determinant of this. And then B is a determinant of that is for a matrix 3 cross 3. Then the question is here, what is derivative of B with respect to X? So, here the A and B are represented in the form of determinants, but the question is here students find the derivative. So, you can take this even example in the concept derivatives also.

[00:06:25]So, let us find first what is A is given here? It is a determinant of X 1 1 X.

[00:06:31]So, we know determinant how it is solved for 2 cross 2 product of diagonal elements minus product of non-diagonal elements. X into X is X squared minus 1 into 1 is 1. Now, let us go with that is a matrix so the the element B you can say or simply you can read it as a B. So, what is B is equal to given? That is a determinant of X 1 1 1 X 1 1 1 that is X. Now, expanding along that is the first row plus minus plus.

[00:07:05]So, first X into bracket leaving first row first column X into X X squared minus 1. Next minus 1 into bracket leaving first row second column 1 into X X minus 1. Next plus 1 into bracket leaving first row third column 1 into 1 1 minus X.

[00:07:26]So, now this is X cubed minus X multiply X inside X cubed minus X. This is minus X minus minus plus 1. Now, 1 into 1 minus X is again how much? 1 minus X. Now, if I do this calculation, what do we get here?

[00:07:44]x³ Now, -x -x -x we have three times and that is -3x and this is how much? + 2.

[00:07:57]This is the value of that is B. Now, let us differentiate with respect to x. So, B differentiation with respect to x is dB by dx. Now, x³ differentiation is 3x² 3x differentiation is 3. 2 is a constant and hence differentiation is 0. 3 is common here, so the remaining is x² - 1.

[00:08:19]Now, dB by dx is equal to Okay, 3. Now, what is x² - 1 already we got here? x² - 1 is nothing but A and thus we got final answer dB by dx is 3A and that is option 1.

[00:08:37]This is example number 23.

[00:08:41]Now, move to the next example, that is question number 24. Here they have given f of x is equal to determinant of They have They have given one function to us which is in the form of determinant.

[00:08:52]First row we have cos x x 1.

[00:08:54]>> [snorts] >> Second row 2 sin x x 2x. Third row sin x x and x. The question is here, what is the limit of f of x divided by x² as x tends to 0.

[00:09:04]>> [snorts] >> So, here the function is given students in the form of determinant and somehow collected this example in this concept.

[00:09:10]Even you can take this example even in the limits also. So, here anyhow, first we have to solve this determinant and we apply the limit as x tends to 0. So, let us consider that is They have asked the students here, what is limit of f of x divided by x² as x tends to 0. Now, this is equal to limit. So, 1 by x² I'll write here separately. 1 by x² as x tends to 0.

[00:09:39]Uh f of x we substitute now. That is cos x x 1 next 2 sin x x 2 x and sin x x and x Now here you can expand this by using first that is using first row plus minus plus. So when we expand students it will be a little bit lengthy. So here now we make use of one of the property of the determinant. And the property is what we do. See in determinant students whenever you have if I take one example suppose in the first row I'm having 2 4 6.

[00:10:19]Second row third row whatever it may be.

[00:10:22]Here when I see the elements of the first row 2 4 6 in this students the element 2 is common. So whenever we have determinant now the common element can be taken either from any row or either from any column. That is what the property we have in determinants. In matrix whenever we are taking common we take that common from every element of the matrix. But in determinant students it is not the same. In determinants what we do is we take the element common element common either element common either from any row any row or any column. You should remember this. In determinants I mean here if I take this.

[00:11:11]Here now we can take two common from the first row. Then what is remaining? 1 2 3. Second row 1 2 3 1 1 1. When you simplify this or this determinant you will get students same answer. You can try this. You are getting actually answer zero. Okay? For both. So you do this calculation or this calculation you will get answer same. And that's why what we can say the scalar can be taken out from the determinant either from any row or from any column. So, the same duty we do here now. So, what we do is Uh we'll take this again uh limit of limit of 1 by x squared. Now, first row I'll not disturb as it is we write cos x x and 1.

[00:12:01]From the second row what we do now? Here x is there, here 2x means you can take x common. Here no x means what we do is I'll take x common students from the second row as well as from the third row also. So, total 2x are out. So, it is x into x x squared you will get. Then what you will get here? Two remains, sin x.

[00:12:22]We don't have x means we'll divide by x here. Here x I've taken common, one remains. Here x I've taken common means two remains. Similarly here, we don't have x hence we'll write sin x divided by x. Here x is common, one you will get. Here x is common, one you will get.

[00:12:40]Now, what happens is this x squared and this x squared you can cancel. Now, we apply limit as x tends to zero. Now, wherever x is there, that is we put now limit as x tends to zero. So, here x is there, we put cos x means zero now. Cos zero Yes, cos table we know for zero angle we have value one. Here x is there, so you put zero. Here constant we have, that it is independent of the limit, so as it is one you write. Now, two I'll keep as it is. All of you know sin x divided by x as x tends to zero, yes it is one of the standard theorem in the limits which is called as sandwich theorem. Limit of sin x by x as x tends to zero, yes it's yes its value is one. So, here two is there.

[00:13:27]Two into one means it is two. Now, this is constant. Keep as it is because constants are independent of the limit.

[00:13:34]Again, here sin x by x is there as x tends to zero, this value is one. One one. Yes, we simplify this now. You will get what actually the limit of f of x divided by x squared.

[00:13:45]Expanding along first row, plus minus plus. So, one into bracket leaving first row first column, one into one one minus one into two two. This is zero, leave it. Plus one into bracket leaving first row third column, two minus one. So, this is minus one. This is plus one. And hence, what is answer here? Zero. Yes, and that is option two.

[00:14:10]This is example number 24.

[00:14:15]Beginning with example number 25. Here, A and B are the two matrices such that AB is equal to B and BA is equal to A.

[00:14:23]Then, the question is here, what is the value of A squared plus B squared? Okay.

[00:14:28]A squared plus B squared, this is equal to A squared can be written as A into A plus B squared can be written as B into B.

[00:14:38]Further, this A I'll keep as it is. Now, this this matrix A value we write from this given data. They have given A is nothing but BA. So, we'll substitute A means BA.

[00:14:51]Plus B as it is. Now, here we have B and the B value is AB. So, let us substitute that is AB. Now, we know matrix multiplication satisfy associative law. A into BA can be written as AB into A. Similarly here, BA into B.

[00:15:12]Now, again from the given data, AB means B. This A I'll keep as it is. BA means A. This B we keep as it is. Again, further, BA is A plus AB is a B. So, the answer is A plus B, and that is option three.

[00:15:30]This is example number 25.

[00:15:34]Next, we go with example number 26.

[00:15:37]Here, they have given one matrix to A.

[00:15:39]That is, in the first row, we have 2 - K, 2. In the second row, 1, 3 - K.

[00:15:45]This matrix A is a singular matrix, then find the value of 5K - K squared. So, here students, they have mentioned that A is a singular matrix, and we know a matrix is said to be singular matrix whenever students its determinant value is zero. So, already A is a singular matrix. This implies determinant of A is equal to zero. Now, let us find the determinant. It's a 2 cross 2 matrix.

[00:16:14]Product of diagonal elements. So, 2 - K into 3 - K minus product of non-diagonal elements.

[00:16:23]1 into 2, yes, we'll get 2 is equal to zero. Now, we multiply these two brackets. 2 * 3 = 6. 2 into - K is - 2K.

[00:16:34]- K into 3 is - 3K. Minus minus plus K into K, K squared.

[00:16:40]>> [snorts] >> And this is - 2 is equal to zero. Now, 6 - 2 is 4. - 2K - 3K is - 5K plus K squared is equal to zero. Now, what we do is 4 I'll keep here only. Bring this - 5K, it becomes plus 5K. Plus K squared you take right side, it becomes - K squared. And they have asked this value only. What is the value of 5K - K squared? And we got 5K - K squared value is 4, and that is option three.

[00:17:18]This is example number 26.

[00:17:22]Beginning with example number 27, here they have given the area of triangle with the vertices. Okay, triangle vertices are given -3 0, 3 0, and 0 K.

[00:17:32]And the area is also given students, that is nine square units. Then the question is here, what is the value of K? Even such examples we have that is in the theory, that is board theory, whatever is there, PU board syllabus. In that also we have determinants chapter, finding the value of K. So, already they have given that area of triangle area of triangle is equal to nine square units. So, area of the triangle is given by 1/2 into determinant of -3. Take the vertices, X coordinate, Y coordinate, -3 0, 3 0, and then 0 K.

[00:18:12]Here in the third column, we take 1 1 1.

[00:18:15]So, area is nine square units means the this determinant value may be plus or minus nine. Because whether determinant value is plus nine or minus nine, the area is always taken the absolute value of this plus or minus nine, hence the area is nine means the determinant value may be plus or may be minus, so you can take plus or minus nine. Now, 1/2 into Okay, here students, we expand along the second column, because in second column already we have two zeros. So, minus plus minus. This is zero, this is zero, leave it. Directly we'll get minus K into bracket, minus K into bracket. K is present in third row, second column, so you leave third row, second column. -3 into 1, -3 minus 3 into 1, yes it is three. Is equal to plus or minus nine.

[00:19:08]Now, this is -3 -3 is -6. -6 into -K is minus minus + Yes, you will get 6k. Now, this two is in the denominator. Bring right side, it becomes that is it comes to the numerator. 9 * 2 you will get that is 18. So, k value is plus or minus 18 divided by 6. And thus, we got two k values here. Either + 3 or either that is - 3. So, here in the options, they have given three. Hence, you can take this as a second option. I mean, uh three is the right answer here. So, the second option is correct. This is example number 27.

[00:19:50]Now, beginning with example number 28.

[00:19:52]Here, they have given delta is equal to determinant of In the first column, they have given 1 1 1. In the second column, a b c. And in the third column, a squared b squared c squared. And delta 1 is equal to one more determinant is given in which the first row contains 1 1 1. Second row contains b c c a a b.

[00:20:10]And third row contains that is a b c.

[00:20:13]Then, the question is here, delta 1 is I mean, they have asked here, what is the value of delta 1? When we see these options, students, all are delta 1 is equal to given. So, what we do now?

[00:20:25]We'll take this delta 1. What is delta 1 given? Determinant of 1 1 1 b c c a a b. And in the third row, they have given a b c.

[00:20:38]Now, what we do is delta 1 is equal to Here, they have given students, when you see the options, delta 1 is expressed that is in the form of delta. Means, you should try to get this value now in this form. Hence, what you should do is dividing and multiplying by We'll do a b c. Okay? From a b c, we divide. And even a b c we multiply. So, that there is no any change in the equation. So, you can write 1 1 1 b c c a a b. And a Now, what we do is 1 / ABC we'll keep here only this. And now, these whatever the scalars are there, I told we can take scalar common, students, from either any row or any column. Similarly, any scalar is multiplied to any row or any column. So, A I'll multiply to first column. So, automatically I'll get here A squared. That is the reason. B we multiply to second column, and C we multiply to third column. So, that you will get like this here. A ABC, and here A squared.

[00:21:45]B Here, when I multiply B, yes, it is again ABC, and you will get here B squared. Here, C Here, ABC, and you will get here C squared. That is multiplying C to the third column.

[00:22:00]Now, 1 / ABC. Now, if I observe here, in the second Okay, in the second row, every element is ABC, so you can take from the second row common, that is ABC.

[00:22:12]Because I said in the determinant, the common can be taken either from any row or from any column. So, here from the row second row, ABC is common, so let us take out. So, the remaining is here ABC.

[00:22:24]Here remains is 1 1 1, and this [snorts] as it is A squared, B squared, C squared. Now, here, this ABC from the numerator and from the denominator get cancelled. Now, here, no, we apply a property. That is, in determinants, we have a property.

[00:22:42]Uh the property says is, if we interchange, if we interchange any two rows if we interchange any two rows or any two columns any two columns, determinant value changes by sign.

[00:23:04]Determinant value changes by sign.

[00:23:10]Changes by sign means if it is plus, it becomes minus. If it is minus, it becomes plus. So, okay. Now, we have here nothing means a plus sign. Okay.

[00:23:21]Now, what I do is the first row and the second row I'll interchange. So, what you will get first here? Minus uh determinant of uh second row we'll write in the first row now.

[00:23:33]And first row elements we write in the second row. Third row we keep as it is.

[00:23:38]Now, after this now, we apply one more property that what we have is determinant of A is same as determinant of its transpose. So, I'll write now transpose of this. First row I'll write it as a first column. Second row I'll write it as a second column. And third row I'll write it as a third column. And if you compare now, this is minus of this is nothing but the given that is delta. Same to same we got here. And hence you can write the answer this as a delta. And what final answer we got is delta dash is equal to or delta one is equal to minus [snorts] delta. So, delta one is equal to minus delta and that is the third option.

[00:24:21]This is example number 28.

[00:24:25]Let us move to the next example, question number 29. Here they have given one matrix equation to us. X into matrix that is column matrix three two plus Y into a column matrix one minus one is equal to a column matrix 15 five. Then the question is here, what are the values of X and Y? So, when I multiply this X inside students, you will get three X. This size is a two X. Okay, this is first matrix. Plus matrix equation is given here, okay? So, Y you multiply inside, you will get Y minus Y.

[00:24:55]Is equal to here the matrix that is 15 5. Now left hand side two matrices are there with algebraic addition. So we do addition of these two matrix now. What matrix addition says corresponding elements you have to add. Here we have a element 3X plus here we have the corresponding element Y. And this is 2X plus into minus or you can write plus what is the element here corresponding that is minus Y is equal to 15 5. Now between these two matrices students we have a algebra equal to. So when two matrices are equal again we have a relation what is whenever two matrices are same the corresponding elements are equal. Means 3X plus Y is equal to 15.

[00:25:41]2X plus into minus you can write minus Y is equal to 5. Now what we do is these two equations we added. 3X plus 2X is 5X. Here plus Y minus Y get cancelled is equal to 15 plus 5 is 20. And thus X is equal to 20 divided by 5. So X value is 4. Yes we got X value. Now we put this X value in any one of these two equation.

[00:26:11]So you can take 3X plus Y is equal to 15 or you can use second one also. So 3X Y Y value I need. So 15 minus 3X I'll take. Put X is equal to 4. So this is 3 4s are 12 and this Y value is 3.

[00:26:29]Thus X is 4 Y is 3 and that is option 4.

[00:26:35]This is example number 29.

[00:26:39]Now question number 30. The matrix A is equal to 1 tan alpha by 2 minus tan alpha by 2 1 and matrix A into B. So here multiplication is given A * B = I.

[00:26:54]Then, the question is here, what is the matrix B? Okay, first of all, A * B = I means here inverse of both the matrices exist.

[00:27:05]A * B = I in the sense A inverse is B and as well as B inverse is what? A. So, here they have asked what is B. So, B is nothing but it is A inverse.

[00:27:17]Whenever students A * B or B * A is equal to I, know, we say that A and B are inverse of each other. So, here B is same as what we can say A inverse. So, let us find now A inverse, then automatically we'll get B. A inverse is nothing but according to the formula 1 / determinant of A into adjoint of A.

[00:27:41]So, now 1 / determinant of A. The determinant of A is nothing but here is the matrix A already. So, let us find the determinant now. Product of diagonal elements minus product of non-diagonal elements. 1 * 1 1 minus So, here already one more minus sign is there. So, this minus minus becomes plus. tan alpha by 2 into tan alpha by 2 is tan squared alpha by 2. Into adjoint of A. Adjoint of A for 2 cross 2 can be written directly.

[00:28:12]There is a a shortcut. Interchange diagonal elements. So, diagonal elements are already 1 1. So, again you will get 1 1 only. And change the sign of non-diagonal elements. So, this becomes minus and this becomes plus. So, you will get minus tan alpha by 2 and this is plus tan alpha by 2. Now, 1 / Okay, 1 + tan squared alpha by 2. This looks like a trigonometric identity 1 + tan squared theta. 1 + [snorts] tan squared theta is sec square theta. So, we'll write this as a sec square alpha by 2. And when I observe this, okay, that is this matrix, this matrix is actually students, see, it's a transpose of this matrix. If you observe your first column, that we have here as a first row. And hence, what we can say, this is actually A dash. And further, you know 1 / sec theta is cos theta. So, here sec square is there, hence you can write cos square alpha by 2 into A dash. This is the matrix B.

[00:29:25]So, and that is option A dash means it's a transpose. So, you can take this transpose. Sometimes it is denoted like this also, AT. But it's same as A dash. So, that is option two is the right answer. Here, one thing you remember, A into B is equal to I means both are inverse of each other. So, B can be written as A inverse or A can written as B inverse. Or in another way, if I say I want [snorts] B here, so take this matrix A right hand side. Whenever we take students any matrix left or right, it changes to its inverse. So, take matrix A to the right hand side, it becomes A inverse. So, you will get B is equal to I into A inverse. And again, I into A inverse is how much? A inverse.

[00:30:08]Apply the formula for finding A inverse, so you will get this answer. This is example number 13.

[00:30:15]So, with this, we conclude today's video.

[00:30:19]Till then, take care, stay happy, and spread happiness. So, don't forget to subscribe my channel Maths with Gita. Thank you.