CBSE Class 12 > Class 12 Videos > Expanding determinant along 2nd 3rd row - Determinants
Expanding determinant along 2nd 3rd row - Determinants Video Lecture -
FAQs on Expanding determinant along 2nd 3rd row - Determinants
| 1. How do you expand a determinant along the second row? |  |
Ans. To expand a determinant along the second row, you can use the cofactor expansion method. Multiply each element in the second row by its corresponding cofactor, and then sum up the results. This will give you the expanded form of the determinant.
| 2. Can you explain the cofactor expansion method in determinants? | |
Ans. Yes, the cofactor expansion method is a technique used to expand determinants. It involves selecting a row or column and multiplying each element in that row or column by its corresponding cofactor. The cofactor is calculated by taking the determinant of the submatrix formed by excluding the row and column of the selected element. After multiplying and summing up the results, you obtain the expanded form of the determinant.
| 3. Is it possible to expand a determinant along the third row? | |
Ans. Yes, you can expand a determinant along any row or column. The process is similar to expanding along the second row. You multiply each element in the third row by its corresponding cofactor and then sum up the results to get the expanded form of the determinant.
| 4. Are there any other methods to expand determinants? | |
Ans. Yes, apart from the cofactor expansion method, there is another method called the Laplace expansion method. In this method, you select a row or column and calculate the sum of products of each element in that row or column with its corresponding minor (determinant of the submatrix formed by excluding the row and column of the selected element). This sum gives you the expanded form of the determinant.
| 5. What is the purpose of expanding a determinant along a row or column? | |
Ans. Expanding a determinant along a row or column helps in simplifying the determinant and making it easier to calculate. It allows you to express the determinant as a sum of simpler terms, which can be computed individually. This method is particularly useful when dealing with larger determinants, as it reduces the complexity of the calculations required.
Text Transcript from Video
hello friends this video on determinants
part 4 is brought you by exam fear calm
no more fear from exam before watching
this video please make sure that you
have watched part 1 - Part 3 let's
expand this using road - and see is it
this Sigma so this is my matrix this
time I will be expanding using this guy
because this is my go-to alright so for
this I'll take this guy and cut off
these guys I will get one side now I
select the second element cut off these
guys I will be the second side now I
take a third element in the same row
because I chose this row second row cut
off these elements I will get third set
so I have three sets correct so I will
add all the values in all the sets
so let me write set 1 set 1 what I get
is the formula is minus I 1 to the power
I plus J a IJ and this Madras the first
one is a to 1 that is minus 1 to the
power 2 plus 1 a to 1 and the métis we
get is this guy a 1 2 A 1 3 a32 and a33
similarly for step 2 we will get minus 1
to the power this guy is my step 1 step
2 and step 3 so here I plus J is 2 plus
2 and here it is a 2 is no it is 2 1 it
is 2 2 and now for this the elements are
a 1 1 a 1 3 a 1 1 a 1 3
and this guy is a 3 1 a 3 3 and the
determinant for step 3 I have a 2 3 so
2b minus 1 to the power 2 plus 3 a 2 3
and then we have the elements a 1 1 a 1
2
a 3-1 a 3-2 so a 1 1 a 1 2 a 3 1 a
correct so I have got 3 output in 3
steps that we solve it furthers this
becomes minus 1 into a 2 1 s minus a 2 1
into x l subtract remaining part that is
a 1 to a 3 3 minus a 1 3 a 3 correct so
this guy becomes minus 1 to the power 4
is 1 1 into a 2 2 is a 2 2 into x hands
subtract the remaining part a 1 1 a 3 3
minus a 1 3 a 3 1 here also same thing
minus 1 to the power 5 is minus minus 1
into a 2 2 is minus a 2 3 minus 1 into a
2 3 is 8 minus a 2 3 into multipliers a
1 1 a 3 to subtract remaining part a 1
to a 3 1 so I've got three values 3 sets
so determinant of a is nothing but you
can write it as this fashion also or
this fashion also is nothing but add all
outputs what you get if you add or
produce nothing bad minus a 2 1 into
this guy a 1 to a 3 3 minus a 1 3
and a three row correct
Plus this guy a to 2 into a 1 1 a 1 a 1
1 a 3 3 - this guy a 1 380 correct plus
again this more right this the last one
we have God - because it is minus here
so plus minus becomes minus so again
remove this part minus a 2 3 into a 1 1
a 3 2 minus a 1 2 8
now let me arrange this guy so the first
element will be a 1 1 so this guy has a
1 1 let me take this element this guy's
a 1 1 a 2 2 a 3 3
why a 1 1 into a 2 2 this guy class
again second element which has a 1 1
this guy so I'll take this guy element
so this is nothing but minus of this
this becomes minus a 1 1 a 2 3 a 3 2 so
a 1 limit is done so let me take a 1 2
now so a 1 2 let me take this guy this
guy I can take a 1 2 so this become
minus here is minus a12 a21 a 3-letter
second a 1 2 element this guy is second
a 1 2 element - - become plus so I'll
say plus a 1 2 into a 2 3 into a 3 1
I'm just arranging that that's all now
third is a 1 3 I we have taken so let's
take a 1 3 so this guy is a 1 3 actually
let's take this guy a 1 3
and this - man accomplice is come plus
a13 a21 a three - correct and the last
element left is this guy so we'll take
this plus - become minus a 1 3 a 2 2 and
a so this is mine hence so if you
compare this guy with the one which we
obtained in the last line by expanding
Row 1 you see that they are exactly same
they are exactly same
thus you can say that if you expand
using Roman or Row 2 you get the same
answer for determinate similarly we can
do for Row 3 so for Row 3 what we can do
we select this Row 3 and then we select
this elements first women be selected we
got these guys to move second element be
selected we got these guys you move and
third element we select n we got this so
I got 2 cross 2 determinant I can use
the same formula minus 1 to the power I
plus J a I J and the determinant I get 3
determinant add them all and you see
that you get same answer here also I not
solve this for you can solve this
you'll get same answer so you can see
that for all rows 1 2 or 3 you get same
determinant similarly we get thank you
visit exam viacom to watch free
education videos try free online tests
get the best quality study materials
study from the best tutors and mentors
and much more thanks again
part 4 is brought you by exam fear calm
no more fear from exam before watching
this video please make sure that you
have watched part 1 - Part 3 let's
expand this using road - and see is it
this Sigma so this is my matrix this
time I will be expanding using this guy
because this is my go-to alright so for
this I'll take this guy and cut off
these guys I will get one side now I
select the second element cut off these
guys I will be the second side now I
take a third element in the same row
because I chose this row second row cut
off these elements I will get third set
so I have three sets correct so I will
add all the values in all the sets
so let me write set 1 set 1 what I get
is the formula is minus I 1 to the power
I plus J a IJ and this Madras the first
one is a to 1 that is minus 1 to the
power 2 plus 1 a to 1 and the métis we
get is this guy a 1 2 A 1 3 a32 and a33
similarly for step 2 we will get minus 1
to the power this guy is my step 1 step
2 and step 3 so here I plus J is 2 plus
2 and here it is a 2 is no it is 2 1 it
is 2 2 and now for this the elements are
a 1 1 a 1 3 a 1 1 a 1 3
and this guy is a 3 1 a 3 3 and the
determinant for step 3 I have a 2 3 so
2b minus 1 to the power 2 plus 3 a 2 3
and then we have the elements a 1 1 a 1
2
a 3-1 a 3-2 so a 1 1 a 1 2 a 3 1 a
correct so I have got 3 output in 3
steps that we solve it furthers this
becomes minus 1 into a 2 1 s minus a 2 1
into x l subtract remaining part that is
a 1 to a 3 3 minus a 1 3 a 3 correct so
this guy becomes minus 1 to the power 4
is 1 1 into a 2 2 is a 2 2 into x hands
subtract the remaining part a 1 1 a 3 3
minus a 1 3 a 3 1 here also same thing
minus 1 to the power 5 is minus minus 1
into a 2 2 is minus a 2 3 minus 1 into a
2 3 is 8 minus a 2 3 into multipliers a
1 1 a 3 to subtract remaining part a 1
to a 3 1 so I've got three values 3 sets
so determinant of a is nothing but you
can write it as this fashion also or
this fashion also is nothing but add all
outputs what you get if you add or
produce nothing bad minus a 2 1 into
this guy a 1 to a 3 3 minus a 1 3
and a three row correct
Plus this guy a to 2 into a 1 1 a 1 a 1
1 a 3 3 - this guy a 1 380 correct plus
again this more right this the last one
we have God - because it is minus here
so plus minus becomes minus so again
remove this part minus a 2 3 into a 1 1
a 3 2 minus a 1 2 8
now let me arrange this guy so the first
element will be a 1 1 so this guy has a
1 1 let me take this element this guy's
a 1 1 a 2 2 a 3 3
why a 1 1 into a 2 2 this guy class
again second element which has a 1 1
this guy so I'll take this guy element
so this is nothing but minus of this
this becomes minus a 1 1 a 2 3 a 3 2 so
a 1 limit is done so let me take a 1 2
now so a 1 2 let me take this guy this
guy I can take a 1 2 so this become
minus here is minus a12 a21 a 3-letter
second a 1 2 element this guy is second
a 1 2 element - - become plus so I'll
say plus a 1 2 into a 2 3 into a 3 1
I'm just arranging that that's all now
third is a 1 3 I we have taken so let's
take a 1 3 so this guy is a 1 3 actually
let's take this guy a 1 3
and this - man accomplice is come plus
a13 a21 a three - correct and the last
element left is this guy so we'll take
this plus - become minus a 1 3 a 2 2 and
a so this is mine hence so if you
compare this guy with the one which we
obtained in the last line by expanding
Row 1 you see that they are exactly same
they are exactly same
thus you can say that if you expand
using Roman or Row 2 you get the same
answer for determinate similarly we can
do for Row 3 so for Row 3 what we can do
we select this Row 3 and then we select
this elements first women be selected we
got these guys to move second element be
selected we got these guys you move and
third element we select n we got this so
I got 2 cross 2 determinant I can use
the same formula minus 1 to the power I
plus J a I J and the determinant I get 3
determinant add them all and you see
that you get same answer here also I not
solve this for you can solve this
you'll get same answer so you can see
that for all rows 1 2 or 3 you get same
determinant similarly we get thank you
visit exam viacom to watch free
education videos try free online tests
get the best quality study materials
study from the best tutors and mentors
and much more thanks again
About this Video
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Apr 19, 2026 Last updated
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