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

2021 · Shift 1 · Q30
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. /2021 · Shift 1 · Q30

Matrices and Determinants question

2021 · Shift 1 · Q30

JEE AdvancedMathematicsMatrices and DeterminantsMultiple correct+4 / −2
For any 3 ×\times× 3 matrix M, let | M | denote the determinant of M. Let E=[12323481318]E = \left[ {\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 8 & {13} & {18} \\ \end{matrix} } \right]E=​128​2313​3418​​, P=[100001010]P = \left[ {\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{matrix} } \right]P=​100​001​010​​ and F=[13281813243]F = \left[ {\begin{matrix} 1 & 3 & 2 \\ 8 & {18} & {13} \\ 2 & 4 & 3 \\ \end{matrix} } \right]F=​182​3184​2133​​ If Q is a nonsingular matrix of order 3 ×\times× 3, then which of the following statements is(are) TRUE?
  1. A
    F = PEP and P2=[100010001]{P^2} = \left[ {\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} } \right]P2=​100​010​001​​
  2. B
    | EQ + PFQ −-− 1 | = | EQ | + | PFQ −-− 1 |
  3. C
    | (EF)3 | > | EF |2
  4. D
    Sum of the diagonal entries of P −-− 1EP + F is equal to the sum of diagonal entries of E + P −-− 1FP
View written solutionFree

Correct answer: A, B, D

  1. Given matrices
E=[12323481318],P=[100001010],F=[13281813243]E=\begin{bmatrix}1&2&3\\2&3&4\\8&13&18\end{bmatrix},\quad P=\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix},\quad F=\begin{bmatrix}1&3&2\\8&18&13\\2&4&3\end{bmatrix}E=​128​2313​3418​​,P=​100​001​010​​,F=​182​3184​2133​​

Also, QQQ is any nonsingular 3×33\times 33×3 matrix.

We check each option one by one.

  1. Option A: F=PEPF=PEPF=PEP and P2=IP^2=IP2=I

First compute PEPEPE:

Since left multiplication by PPP swaps row 2 and row 3,

PE=[12381318234]PE=\begin{bmatrix}1&2&3\\8&13&18\\2&3&4\end{bmatrix}PE=​182​2133​3184​​

Now multiply by PPP on the right. Right multiplication by PPP swaps column 2 and column 3:

PEP=[13281813243]=FPEP=\begin{bmatrix}1&3&2\\8&18&13\\2&4&3\end{bmatrix}=FPEP=​182​3184​2133​​=F

So, F=PEPF=PEPF=PEP is true.

Now compute P2P^2P2:

P2=[100001010][100001010]=[100010001]=IP^2= \begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix} \begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix} = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=IP2=​100​001​010​​​100​001​010​​=​100​010​001​​=I

Hence Option A is true.

  1. Option B: ∣EQ+PFQ−1∣=∣EQ∣+∣PFQ−1∣|EQ+PFQ^{-1}|=|EQ|+|PFQ^{-1}|∣EQ+PFQ−1∣=∣EQ∣+∣PFQ−1∣

The printed option is interpreted as

∣EQ+PFQ−1∣=∣EQ∣+∣PFQ−1∣|EQ+PFQ^{-1}|=|EQ|+|PFQ^{-1}|∣EQ+PFQ−1∣=∣EQ∣+∣PFQ−1∣

Now use F=PEPF=PEPF=PEP and P−1=PP^{-1}=PP−1=P.

Then

PFQ−1=P(PEP)Q−1=P2EPQ−1=EPQ−1PFQ^{-1}=P(PEP)Q^{-1}=P^2EPQ^{-1}=EPQ^{-1}PFQ−1=P(PEP)Q−1=P2EPQ−1=EPQ−1

So

EQ+PFQ−1=EQ+EPQ−1=E(Q+PQ−1)EQ+PFQ^{-1}=EQ+EPQ^{-1}=E(Q+PQ^{-1})EQ+PFQ−1=EQ+EPQ−1=E(Q+PQ−1)

Hence

∣EQ+PFQ−1∣=∣E∣ ∣Q+PQ−1∣|EQ+PFQ^{-1}|=|E|\,|Q+PQ^{-1}|∣EQ+PFQ−1∣=∣E∣∣Q+PQ−1∣

Let us first find ∣E∣|E|∣E∣:

∣E∣=∣12323481318∣|E|= \begin{vmatrix} 1&2&3\\ 2&3&4\\ 8&13&18 \end{vmatrix}∣E∣=​128​2313​3418​​

Expanding along first row,

∣E∣=1(54−52)−2(36−32)+3(26−24)=2−8+6=0|E|=1(54-52)-2(36-32)+3(26-24)=2-8+6=0∣E∣=1(54−52)−2(36−32)+3(26−24)=2−8+6=0

So ∣E∣=0|E|=0∣E∣=0.

Therefore,

∣EQ+PFQ−1∣=0|EQ+PFQ^{-1}|=0∣EQ+PFQ−1∣=0

Also,

∣EQ∣=∣E∣∣Q∣=0|EQ|=|E||Q|=0∣EQ∣=∣E∣∣Q∣=0

and

∣PFQ−1∣=∣P∣∣F∣∣Q−1∣|PFQ^{-1}|=|P||F||Q^{-1}|∣PFQ−1∣=∣P∣∣F∣∣Q−1∣

Now

∣F∣=∣PEP∣=∣P∣∣E∣∣P∣=∣P∣2∣E∣=0|F|=|PEP|=|P||E||P|=|P|^2|E|=0∣F∣=∣PEP∣=∣P∣∣E∣∣P∣=∣P∣2∣E∣=0

So

∣PFQ−1∣=0|PFQ^{-1}|=0∣PFQ−1∣=0

Thus

∣EQ∣+∣PFQ−1∣=0+0=0|EQ|+|PFQ^{-1}|=0+0=0∣EQ∣+∣PFQ−1∣=0+0=0

Hence both sides are equal. Therefore Option B is true.

  1. Option C: ∣(EF)3∣>∣EF∣2|(EF)^3|>|EF|^2∣(EF)3∣>∣EF∣2

Using determinant properties,

∣(EF)3∣=∣EF∣3|(EF)^3|=|EF|^3∣(EF)3∣=∣EF∣3

So the inequality becomes

∣EF∣3>∣EF∣2|EF|^3>|EF|^2∣EF∣3>∣EF∣2

Now

∣EF∣=∣E∣∣F∣=0⋅0=0|EF|=|E||F|=0\cdot 0=0∣EF∣=∣E∣∣F∣=0⋅0=0

Hence

∣(EF)3∣=0,∣EF∣2=0|(EF)^3|=0, \qquad |EF|^2=0∣(EF)3∣=0,∣EF∣2=0

So the statement becomes

0>00>00>0

which is false.

Therefore Option C is false.

  1. Option D: Sum of diagonal entries of P−1EP+FP^{-1}EP+FP−1EP+F equals that of E+P−1FPE+P^{-1}FPE+P−1FP

The sum of diagonal entries is the trace. So we compare

tr⁡(P−1EP+F)andtr⁡(E+P−1FP)\operatorname{tr}(P^{-1}EP+F) \quad \text{and} \quad \operatorname{tr}(E+P^{-1}FP)tr(P−1EP+F)andtr(E+P−1FP)

Now P−1=PP^{-1}=PP−1=P because P2=IP^2=IP2=I.

Also, trace is invariant under similarity:

tr⁡(P−1EP)=tr⁡(E),tr⁡(P−1FP)=tr⁡(F)\operatorname{tr}(P^{-1}EP)=\operatorname{tr}(E), \qquad \operatorname{tr}(P^{-1}FP)=\operatorname{tr}(F)tr(P−1EP)=tr(E),tr(P−1FP)=tr(F)

Therefore,

tr⁡(P−1EP+F)=tr⁡(P−1EP)+tr⁡(F)=tr⁡(E)+tr⁡(F)\operatorname{tr}(P^{-1}EP+F)=\operatorname{tr}(P^{-1}EP)+\operatorname{tr}(F)=\operatorname{tr}(E)+\operatorname{tr}(F)tr(P−1EP+F)=tr(P−1EP)+tr(F)=tr(E)+tr(F)

and

tr⁡(E+P−1FP)=tr⁡(E)+tr⁡(P−1FP)=tr⁡(E)+tr⁡(F)\operatorname{tr}(E+P^{-1}FP)=\operatorname{tr}(E)+\operatorname{tr}(P^{-1}FP)=\operatorname{tr}(E)+\operatorname{tr}(F)tr(E+P−1FP)=tr(E)+tr(P−1FP)=tr(E)+tr(F)

Hence they are equal. So Option D is true.

  1. Final conclusion

The true statements are:

  • A
  • B
  • D

Option C is false.

This matches the stored correct answer.

PreviousNext

More from Matrices and Determinants

  • For any 3 × 3 matrix M, let |M| denote the determinant of M. Let I be the 3 × 3 identity matrix. Let E and F be two 3 × 3 matrices such that (I − EF) is invertible. If G = (I − EF) − 1, then which of the…2021 · Multiple correct
  • Let M be a 3 × 3 invertible matrix with real entries and let I denote the 3 × 3 identity matrix. If M − 1 = adj(adj M), then which of the following statements is/are ALWAYS TRUE?2020 · Multiple correct
  • The trace of a square matrix is defined to be the sum of its diagonal entries. If A is a 2 × 2 matrix such that the trace of A is 3 and the trace of A3 is − 18, then the value of the determinant of A is .............2020 · Numerical
  • Let M=[sin4θ1+cos2θ​−1−sin2θcos4θ​]=αI+βM−1, where α=α…2019 · MCQ
  • Let M=​013​12b​a31​​ and adj M=​−18−5​1−63​−12−1​​…2019 · Multiple correct
  • Let x ∈ R and let P=​100​120​123​​, Q=​20x​x4x​x06​​…2019 · Multiple correct
  • P1​=I=​100​010​001​​,P2​=​100​001​010​​,P3​=​010​100​001​​,P4​=​001​100​010​​,P5​=​010​001​100​​,P6​=​001​010​100​​…2019 · Multiple correct
  • Suppose det ​k=0∑n​kk=0∑n​nCk​.k​k=0∑n​nCk​k2k=0∑n​nCk​3k​​=0 holds…2019 · Numerical