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Matrices and Determinants question

2016 · Shift 1 · Q33
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  5. /2016 · Shift 1 · Q33

Matrices and Determinants question

2016 · Shift 1 · Q33

JEE AdvancedMathematicsMatrices and DeterminantsMultiple correct+4 / −2
Let P=[3−1−220α3−50]P = \left[ {\begin{matrix} 3 & { - 1} & { - 2} \\ 2 & 0 & \alpha \\ 3 & { - 5} & 0 \\ \end{matrix} } \right]P=​323​−10−5​−2α0​​, where α∈\alpha\inα∈ R. Suppose Q=[qij]Q = [{q_{ij}}]Q=[qij​] is a matrix such that PQ = kl, where k ∈\in∈ R, k eee 0 and I is the identity matrix of order 3. If q23=−k8{q_{23}} = - {k \over 8}q23​=−8k​ and det⁡(Q)=k22\det (Q) = {{{k^2}} \over 2}det(Q)=2k2​, then
  1. A
    α\alphaα = 0, k = 8
  2. B
    4α−k+8=04\alpha - k + 8 = 04α−k+8=0
  3. C
    det⁡(Padj(Q))=29\det (Padj(Q)) = {2^9}det(Padj(Q))=29
  4. D
    det⁡(Qadj(P))=213\det (Qadj(P)) = {2^{13}}det(Qadj(P))=213
View written solutionFree

Correct answer: B, C

  1. Interpret the condition PQ=kIPQ = kIPQ=kI

Given

P=[3−1−220α3−50],P=\begin{bmatrix} 3&-1&-2\\ 2&0&\alpha\\ 3&-5&0 \end{bmatrix},P=​323​−10−5​−2α0​​,

and Q=[qij]Q=[q_{ij}]Q=[qij​] satisfies PQ=kI,PQ=kI,PQ=kI, with k≠0k\neq 0k=0.

This implies Q=kP−1,Q=kP^{-1},Q=kP−1, so PPP must be invertible.

Also,

\implies Q=\frac{k\,\operatorname{adj}(P)}{\det P}.$$ Hence each entry of $Q$ is $$q_{ij}=\frac{k\,C_{ji}}{\det P},$$ where $C_{ji}$ is the cofactor of $P$. --- 2. **Use the condition $q_{23}=-\dfrac{k}{8}$** We need $q_{23}$, i.e. row 2 column 3 of $Q$. From $$Q=\frac{k\,\operatorname{adj}(P)}{\det P},$$ we get $$q_{23}=\frac{k\,C_{32}}{\det P}.$$ Now compute $C_{32}$ of $P$: Delete row 3 and column 2 from $P$:

\begin{vmatrix} 3 & -2\ 2 & \alpha \end{vmatrix}=3\alpha-(-4)=3\alpha+4.

Since $$C_{32}=(-1)^{3+2}(3\alpha+4)=-(3\alpha+4),$$ we have $$q_{23}=\frac{k\,[-(3\alpha+4)]}{\det P}.

Given q23=−k8,q_{23}=-\frac{k}{8},q23​=−8k​, and k≠0k\neq 0k=0, so

  ⟹  det⁡P=8(3α+4).(1)\implies \det P=8(3\alpha+4). \qquad (1)⟹detP=8(3α+4).(1)
  1. Compute det⁡P\det PdetP directly

Expand along the first row:

det⁡P=3∣0α−50∣−(−1)∣2α30∣+(−2)∣203−5∣.\det P= 3\begin{vmatrix}0&\alpha\\-5&0\end{vmatrix} -(-1)\begin{vmatrix}2&\alpha\\3&0\end{vmatrix} +(-2)\begin{vmatrix}2&0\\3&-5\end{vmatrix}.detP=3​0−5​α0​​−(−1)​23​α0​​+(−2)​23​0−5​​.

Now,

∣0α−50∣=5α,\begin{vmatrix}0&\alpha\\-5&0\end{vmatrix}=5\alpha,​0−5​α0​​=5α, ∣2α30∣=−3α,\begin{vmatrix}2&\alpha\\3&0\end{vmatrix}=-3\alpha,​23​α0​​=−3α, ∣203−5∣=−10.\begin{vmatrix}2&0\\3&-5\end{vmatrix}=-10.​23​0−5​​=−10.

Thus

det⁡P=3(5α)+(−3α)+(−2)(−10)=15α−3α+20=12α+20.\det P=3(5\alpha)+(-3\alpha)+(-2)(-10)=15\alpha-3\alpha+20=12\alpha+20.detP=3(5α)+(−3α)+(−2)(−10)=15α−3α+20=12α+20.

So

From (1) and (2):

Hence

  ⟹  α=−1.\implies \alpha=-1.⟹α=−1.

Then

  1. Use det⁡(Q)=k22\det(Q)=\dfrac{k^2}{2}det(Q)=2k2​

Since Q=kP−1,Q=kP^{-1},Q=kP−1, for a 3×33\times 33×3 matrix,

Given det⁡(Q)=k22,\det(Q)=\frac{k^2}{2},det(Q)=2k2​, so

Because k≠0k\neq 0k=0,

  ⟹  2k=det⁡P.\implies 2k=\det P.⟹2k=detP.

Since det⁡P=8\det P=8detP=8,

Thus, α=−1,k=4.\boxed{\alpha=-1,\quad k=4.}α=−1,k=4.​

  1. Check each option

Option A: α=0, k=8\alpha=0,\ k=8α=0, k=8

But we found α=−1,k=4.\alpha=-1,\quad k=4.α=−1,k=4. So A is false.

Option B: 4α−k+8=04\alpha-k+8=04α−k+8=0

Substitute values: 4(−1)−4+8=−4−4+8=0.4(-1)-4+8=-4-4+8=0.4(−1)−4+8=−4−4+8=0. So B is true.

Option C: det⁡(P adj⁡(Q))=29\det(P\,\operatorname{adj}(Q))=2^9det(Padj(Q))=29

Use det⁡(P adj⁡(Q))=det⁡(P)det⁡(adj⁡(Q)).\det(P\,\operatorname{adj}(Q))=\det(P)\det(\operatorname{adj}(Q)).det(Padj(Q))=det(P)det(adj(Q)). For a 3×33\times 33×3 matrix QQQ, det⁡(adj⁡(Q))=(det⁡Q)3−1=(det⁡Q)2.\det(\operatorname{adj}(Q))=(\det Q)^{3-1}=(\det Q)^2.det(adj(Q))=(detQ)3−1=(detQ)2. Now

Hence

Therefore

So C is true.

Option D: det⁡(Q adj⁡(P))=213\det(Q\,\operatorname{adj}(P))=2^{13}det(Qadj(P))=213

Similarly, det⁡(Q adj⁡(P))=det⁡(Q)det⁡(adj⁡(P)).\det(Q\,\operatorname{adj}(P))=\det(Q)\det(\operatorname{adj}(P)).det(Qadj(P))=det(Q)det(adj(P)). Since PPP is 3×33\times 33×3,

Also,

So

Thus D is false.

  1. Final answer

The correct options are B, C.\boxed{\text{B, C}}.B, C​.

This matches the stored correct answer.

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