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Matrices and Determinants question

2021 · 24 Feb · Shift 1 · Q36
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Matrices and Determinants question

2021 · 24 Feb · Shift 1 · Q36

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
Let P = [3−1−220α3−50]\left[ {\begin{matrix} 3 & { - 1} & { - 2} \\ 2 & 0 & \alpha \\ 3 & { - 5} & 0 \\ \end{matrix} } \right]​323​−10−5​−2α0​​, where α∈\alpha \inα∈ R. Suppose Q = [ qij] is a matrix satisfying PQ = kl3 for some non-zero k ∈\in∈ R. If q23 =−k8- {k \over 8}−8k​ and |Q| =k22{{{k^2}} \over 2}2k2​, then a2 + k2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 17

  1. Interpret the condition

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=kI3,PQ=kI_3,PQ=kI3​,

where k≠0k\neq 0k=0.

Since PQ=kIPQ=kIPQ=kI, matrix PPP must be invertible, and

Q=kP−1.Q=kP^{-1}.Q=kP−1.

Also,

∣PQ∣=∣kI3∣=k3.|PQ|=|kI_3|=k^3.∣PQ∣=∣kI3​∣=k3.

But ∣PQ∣=∣P∣ ∣Q∣|PQ|=|P|\,|Q|∣PQ∣=∣P∣∣Q∣, so

∣P∣ ∣Q∣=k3.|P|\,|Q|=k^3.∣P∣∣Q∣=k3.

Given

∣Q∣=k22,|Q|=\frac{k^2}{2},∣Q∣=2k2​,

therefore

∣P∣⋅k22=k3.|P|\cdot \frac{k^2}{2}=k^3.∣P∣⋅2k2​=k3.

Since k≠0k\neq 0k=0, divide by k2k^2k2:

∣P∣2=k⇒k=∣P∣2.\frac{|P|}{2}=k \quad\Rightarrow\quad k=\frac{|P|}{2}.2∣P∣​=k⇒k=2∣P∣​.
  1. Compute ∣P∣|P|∣P∣

Expand along the first row:

∣P∣=3∣0α−50∣−(−1)∣2α30∣+(−2)∣203−5∣.|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}.∣P∣=3​0−5​α0​​−(−1)​23​α0​​+(−2)​23​0−5​​.

Now,

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

So,

∣P∣=3(5α)+1(−3α)+(−2)(−10)=15α−3α+20=12α+20.|P|=3(5\alpha)+1(-3\alpha)+(-2)(-10)=15\alpha-3\alpha+20=12\alpha+20.∣P∣=3(5α)+1(−3α)+(−2)(−10)=15α−3α+20=12α+20.

Hence,

k=12α+202=6α+10.k=\frac{12\alpha+20}{2}=6\alpha+10.k=212α+20​=6α+10.
  1. Use the condition on q23q_{23}q23​

Since

Q=kP−1=kadj⁡(P)∣P∣,Q=kP^{-1}=k\frac{\operatorname{adj}(P)}{|P|},Q=kP−1=k∣P∣adj(P)​,

and because k=∣P∣/2k=|P|/2k=∣P∣/2 from above, we get

Q=12adj⁡(P).Q=\frac{1}{2}\operatorname{adj}(P).Q=21​adj(P).

So each entry of QQQ is half the corresponding cofactor-transpose entry.

We need q23q_{23}q23​.

By definition,

q_{23}=\frac12 (\operatorname{adj}(P))_{23}= rac12 C_{32},

because (adj⁡(P))ij=Cji(\operatorname{adj}(P))_{ij}=C_{ji}(adj(P))ij​=Cji​.

Now compute cofactor C32C_{32}C32​:

C32=(−1)3+2M32=−M32.C_{32}=(-1)^{3+2}M_{32}=-M_{32}.C32​=(−1)3+2M32​=−M32​.

Delete row 3 and column 2 from PPP:

M32=∣3−22α∣=3α+4.M_{32}=\begin{vmatrix}3&-2\\2&\alpha\end{vmatrix}=3\alpha+4.M32​=​32​−2α​​=3α+4.

Thus,

C32=−(3α+4).C_{32}=-(3\alpha+4).C32​=−(3α+4).

Therefore,

q23=12(−(3α+4))=−3α+42.q_{23}=\frac12\bigl(-(3\alpha+4)\bigr)=-\frac{3\alpha+4}{2}.q23​=21​(−(3α+4))=−23α+4​.

Given in the question,

q23=−k8.q_{23}=-\frac{k}{8}.q23​=−8k​.

So,

−3α+42=−k8⇒4(3α+4)=k.-\frac{3\alpha+4}{2}=-\frac{k}{8} \quad\Rightarrow\quad 4(3\alpha+4)=k.−23α+4​=−8k​⇒4(3α+4)=k.

Hence,

k=12α+16.k=12\alpha+16.k=12α+16.

But from Step 2,

k=6α+10.k=6\alpha+10.k=6α+10.

Equate:

12α+16=6α+10⇒6α=−6⇒α=−1.12\alpha+16=6\alpha+10 \quad\Rightarrow\quad 6\alpha=-6 \quad\Rightarrow\quad \alpha=-1.12α+16=6α+10⇒6α=−6⇒α=−1.

Then

k=6(−1)+10=4.k=6(-1)+10=4.k=6(−1)+10=4.
  1. Compute α2+k2\alpha^2+k^2α2+k2
α2+k2=(−1)2+42=1+16=17.\alpha^2+k^2=(-1)^2+4^2=1+16=17.α2+k2=(−1)2+42=1+16=17.
  1. Compare with stored answer

Derived answer is

17.17.17.

This matches the stored correct answer.

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