Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Matrices and Determinants question

2019 · Shift 2 · Q29
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Mathematics
  4. /Matrices and Determinants
  5. /2019 · Shift 2 · Q29

Matrices and Determinants question

2019 · Shift 2 · Q29

JEE AdvancedMathematicsMatrices and DeterminantsNumerical+3 / −1
Suppose det ∣∑k=0nk∑k=0nnCkk2∑k=0nnCk.k∑k=0nnCk3k∣=0\left| {\begin{matrix} {\sum\limits_{k = 0}^n k } & {\sum\limits_{k = 0}^n {{}^n{C_k}{k^2}} } \\ {\sum\limits_{k = 0}^n {{}^n{C_k}.k} } & {\sum\limits_{k = 0}^n {{}^n{C_k}{3^k}} } \\ \end{matrix} } \right| = 0​k=0∑n​kk=0∑n​nCk​.k​k=0∑n​nCk​k2k=0∑n​nCk​3k​​=0 holds for some positive integer n. Then ∑k=0nnCkk+1\sum\limits_{k = 0}^n {{{{}^n{C_k}} \over {k + 1}}}k=0∑n​k+1nCk​​ equals ..............
Numerical answer
View written solutionFree

Correct answer: 6.20

  1. Write each matrix entry in closed form

The determinant is

∣∑k=0nk∑k=0nnCkk2∑k=0nnCkk∑k=0nnCk3k∣=0.\begin{vmatrix} \sum_{k=0}^n k & \sum_{k=0}^n {^nC_k}k^2 \\ \sum_{k=0}^n {^nC_k}k & \sum_{k=0}^n {^nC_k}3^k \end{vmatrix}=0.​∑k=0n​k∑k=0n​nCk​k​∑k=0n​nCk​k2∑k=0n​nCk​3k​​=0.

Let us simplify all four sums.

  1. Evaluate the sums

(i) ∑k=0nk\sum_{k=0}^n k∑k=0n​k

∑k=0nk=n(n+1)2.\sum_{k=0}^n k=\frac{n(n+1)}{2}.k=0∑n​k=2n(n+1)​.

(ii) ∑k=0nnCkk\sum_{k=0}^n {^nC_k}k∑k=0n​nCk​k

Using the standard identity,

∑k=0nnCkk=n2n−1.\sum_{k=0}^n {^nC_k}k=n2^{n-1}.k=0∑n​nCk​k=n2n−1.

(iii) ∑k=0nnCkk2\sum_{k=0}^n {^nC_k}k^2∑k=0n​nCk​k2

Use

∑k=0nnCkk(k−1)=n(n−1)2n−2,\sum_{k=0}^n {^nC_k}k(k-1)=n(n-1)2^{n-2},k=0∑n​nCk​k(k−1)=n(n−1)2n−2,

and

∑k=0nnCkk=n2n−1.\sum_{k=0}^n {^nC_k}k=n2^{n-1}.k=0∑n​nCk​k=n2n−1.

So,

∑k=0nnCkk2=∑nCk[k(k−1)+k]=n(n−1)2n−2+n2n−1.\sum_{k=0}^n {^nC_k}k^2 =\sum {^nC_k}[k(k-1)+k] =n(n-1)2^{n-2}+n2^{n-1}.k=0∑n​nCk​k2=∑nCk​[k(k−1)+k]=n(n−1)2n−2+n2n−1.

Factorizing,

∑k=0nnCkk2=n(n+1)2n−2.\sum_{k=0}^n {^nC_k}k^2 =n(n+1)2^{n-2}.k=0∑n​nCk​k2=n(n+1)2n−2.

(iv) ∑k=0nnCk3k\sum_{k=0}^n {^nC_k}3^k∑k=0n​nCk​3k

By binomial theorem,

∑k=0nnCk3k=(1+3)n=4n.\sum_{k=0}^n {^nC_k}3^k=(1+3)^n=4^n.k=0∑n​nCk​3k=(1+3)n=4n.
  1. Substitute into the determinant

The determinant becomes

∣n(n+1)2n(n+1)2n−2n2n−14n∣=0.\begin{vmatrix} \dfrac{n(n+1)}{2} & n(n+1)2^{n-2} \\ n2^{n-1} & 4^n \end{vmatrix}=0.​2n(n+1)​n2n−1​n(n+1)2n−24n​​=0.

So,

n(n+1)2⋅4n−n(n+1)2n−2⋅n2n−1=0.\frac{n(n+1)}{2}\cdot 4^n - n(n+1)2^{n-2}\cdot n2^{n-1}=0.2n(n+1)​⋅4n−n(n+1)2n−2⋅n2n−1=0.

Factor out n(n+1)2n−2n(n+1)2^{n-2}n(n+1)2n−2:

n(n+1)2n−2(4n2n−1−n)=0.n(n+1)2^{n-2}\left(\frac{4^n}{2^{n-1}}-n\right)=0.n(n+1)2n−2(2n−14n​−n)=0.

Now,

4n2n−1=22n2n−1=2n+1.\frac{4^n}{2^{n-1}}=\frac{2^{2n}}{2^{n-1}}=2^{n+1}.2n−14n​=2n−122n​=2n+1.

Hence,

n(n+1)2n−2(2n+1−n)=0.n(n+1)2^{n-2}(2^{n+1}-n)=0.n(n+1)2n−2(2n+1−n)=0.

Since nnn is a positive integer, we need

2n+1=n.2^{n+1}=n.2n+1=n.

This is impossible for positive integer nnn.

So let us compute the determinant directly more carefully:

n(n+1)2⋅4n−n(n+1)2n−2⋅n2n−1=0.\frac{n(n+1)}{2}\cdot 4^n - n(n+1)2^{n-2}\cdot n2^{n-1}=0.2n(n+1)​⋅4n−n(n+1)2n−2⋅n2n−1=0.

Using 4n=22n4^n=2^{2n}4n=22n,

n(n+1)222n−n2(n+1)22n−3=0.\frac{n(n+1)}{2}2^{2n}-n^2(n+1)2^{2n-3}=0.2n(n+1)​22n−n2(n+1)22n−3=0.

Factor n(n+1)22n−3n(n+1)2^{2n-3}n(n+1)22n−3:

n(n+1)22n−3(4−n)=0.n(n+1)2^{2n-3}(4-n)=0.n(n+1)22n−3(4−n)=0.

Therefore,

n=4.n=4.n=4.
  1. Now evaluate ∑k=0nnCkk+1\sum_{k=0}^n \dfrac{{^nC_k}}{k+1}∑k=0n​k+1nCk​​

We need

∑k=044Ckk+1.\sum_{k=0}^4 \frac{{^4C_k}}{k+1}.k=0∑4​k+14Ck​​.

Use the identity

∑k=0nnCkk+1=2n+1−1n+1.\sum_{k=0}^n \frac{{^nC_k}}{k+1}=\frac{2^{n+1}-1}{n+1}.k=0∑n​k+1nCk​​=n+12n+1−1​.

For n=4n=4n=4,

∑k=044Ckk+1=25−15=315.\sum_{k=0}^4 \frac{{^4C_k}}{k+1}=\frac{2^5-1}{5}=\frac{31}{5}.k=0∑4​k+14Ck​​=525−1​=531​.

Thus the required value is

315.\boxed{\frac{31}{5}}.531​​.

Since this is an integer-type entry but the result is decimal in stored answer format,

315=6.2.\frac{31}{5}=6.2.531​=6.2.
  1. Comparison with stored answer

Stored correct answer: 6.206.206.20

Our derived answer: 315=6.2\dfrac{31}{5}=6.2531​=6.2

These are equal.

PreviousNext

More from Matrices and Determinants

  • Let S be the set of all column matrices ​b1​b2​b3​​​ such that b1​,b2​,b3​∈R and the system of equations (in real variables) ​−x+2y+5z=b1​2x−4y+3z=b2​x−2y+2z=b3​​…2018 · Multiple correct
  • Let P be a matrix of order 3 × 3 such that all the entries in P are from the set {− 1, 0, 1}. Then, the maximum possible value of the determinant of P is ............ .2018 · Numerical
  • Which of the following is(are) NOT the square of a 3 × 3 matrix with real entries?2017 · Multiple correct
  • For a real number α, if the system ​1αα2​α1α​α2α1​​​xyz​​=​1−11​​…2017 · Numerical
  • How many 3 × 3 matrices M with entries from {0, 1, 2} are there, for which the sum of the diagonal entries of MTM is 5?2017 · MCQ
  • Let P=​323​−10−5​−2α0​​, where α∈ R. Suppose Q=[qij​] is a matrix such that PQ = kl, where k ∈ R, k e…2016 · Multiple correct
  • The total number of distinct x ∈ R for which ​x2x3x​x24x29x2​1+x31+8x31+27x3​​=10 is ​…2016 · Numerical
  • Let z=2−1+3​i​, where i=−1​, and r, s ∈{1, 2, 3}. Let P=[(−z)rz2s​z2szr​] and I be the identity…2016 · Numerical