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Question: Let the matrix \( P=\left[ \begin{matrix} 3 & -1 & -2 \\\ 2 & 0 & \alpha \\\ 3 & -5 &...

Let the matrix P=[3−1−2 20α 3−50 ]P=\left[ \begin{matrix} 3 & -1 & -2 \\\ 2 & 0 & \alpha \\\ 3 & -5 & 0 \\\ \end{matrix} \right] , where α∈R \alpha \in R. Suppose Q=[qij] Q=\left[ {{q}_{ij}} \right]is matrix such
that PQ = kI, where k∈R k\in R, k≠0 k\ne 0and I is the identity matrix of order 3. If q23=−k8 {{q}_{23}}=\dfrac{-k}{8}​ and det⁡(Q)=k22 \det \left( Q \right)=\dfrac{{{k}^{2}}}{2}​, then
A) α=0,k=8\alpha =0,k=8
B) α−k+8=0\alpha -k+8=0
C) det⁡(Padj(Q))=29\det \left( Padj\left( Q \right) \right)={{2}^{9}}
D) det⁡(Qadj(P))=213\det \left( Qadj\left( P \right) \right)={{2}^{13}}

Explanation

Solution

This question can be solved by first finding the det of P matrix. Then you need to convert the given equation PQ = kI to another, by taking P to the right side. Here, you get the inverse of P. But the inverse of P is adj(P) divided by det(P). Then you get the Q matrix. Then by using the q23=−k8{{q}_{23}}=\dfrac{-k}{8}, you can find the unknown variables. Using the other equation, that is det⁡(Q)=k22\det \left( Q \right)=\dfrac{{{k}^{2}}}{2}, you can find the k variable. Then you can verify the options and then find the final answer.

Complete step by step solution:
First thing we need to do is find the det of the P matrix. Then we need to convert the given equation PQ = kI to another, by taking P to the right side. Here, you get the inverse of P. But the inverse of P is adj(P) divided by det(P).
det⁡(P)=∣3−1−2 20α 3−50 ∣\det \left( P \right)=\left| \begin{matrix} 3 & -1 & -2 \\\ 2 & 0 & \alpha \\\ 3 & -5 & 0 \\\ \end{matrix} \right|
⇒det⁡(P)=20+12α\Rightarrow \det \left( P \right)=20+12\alpha
We were given PQ = kI. Now we have to take P to the right side. Therefore, we get:
⇒PQ=kI\Rightarrow PQ=kI
⇒Q=kP−1I\Rightarrow Q=k{{P}^{-1}}I
⇒Q=kadj(P)∣P∣I\Rightarrow Q=k\dfrac{adj\left( P \right)}{\left| P \right|}I
⇒Q=k20+12α[5α10−α 3α0−3α−4 −10−120 ][100 010 001 ]\Rightarrow Q=\dfrac{k}{20+12\alpha }\left[ \begin{matrix} 5\alpha & 10 & -\alpha \\\ 3\alpha & 0 & -3\alpha -4 \\\ -10 & -12 & 0 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]
⇒Q=k20+12α[5α10−α 3α0−3α−4 −10−120 ]\Rightarrow Q=\dfrac{k}{20+12\alpha }\left[ \begin{matrix} 5\alpha & 10 & -\alpha \\\ 3\alpha & 0 & -3\alpha -4 \\\ -10 & -12 & 0 \\\ \end{matrix} \right]
Now , to find the unknown variables, we use the equation, q23=−k8{{q}_{23}}=\dfrac{-k}{8}
q23=k20+12α×(−3α−4){{q}_{23}}=\dfrac{k}{20+12\alpha}\times \left(-3\alpha -4\right). Equating both of them, we get
⇒q23=k20+12α×(−3α−4)=−k8\Rightarrow {{q}_{23}}=\dfrac{k}{20+12\alpha }\times \left( -3\alpha -4 \right)=\dfrac{-k}{8}
⇒−24α−32=−20−12α\Rightarrow -24\alpha -32=-20-12\alpha
⇒12α=−12\Rightarrow 12\alpha =-12
⇒α=−1\Rightarrow \alpha =-1
To find det(Q):
⇒det⁡(Q)=∣k20+12α∣∣[5α10−α 3α0−3α−4 −10−120 ]∣\Rightarrow \det \left( Q \right)=\left| \dfrac{k}{20+12\alpha } \right|\left| \left[ \begin{matrix} 5\alpha & 10 & -\alpha \\\ 3\alpha & 0 & -3\alpha -4 \\\ -10 & -12 & 0 \\\ \end{matrix} \right] \right|
⇒det⁡(Q)=k20+12α∣adj(P)∣\Rightarrow \det \left( Q \right)=\dfrac{k}{20+12\alpha }\left| adj\left( P \right) \right|
But det(adj(P)) can be given as :
⇒det⁡(adj(P))=(det⁡(P))n−1\Rightarrow \det \left( adj\left( P \right) \right)={{\left( \det \left( P \right) \right)}^{n-1}}
Here n is the no of rows or columns of the square matrix.
Therefore, we get:
⇒det⁡(adj(P))=(det⁡(P))2=(k)2=k2\Rightarrow \det \left( adj\left( P \right) \right)={{\left( \det \left( P \right) \right)}^{2}}={{\left( k \right)}^{2}}={{k}^{2}}
⇒det⁡(Q)=k20+12αk2\Rightarrow \det \left( Q \right)=\dfrac{k}{20+12\alpha }{{k}^{2}}
Now, to find k we use the equation, det⁡(Q)=k22\det \left( Q \right)=\dfrac{{{k}^{2}}}{2}. But det Q is given by, det⁡(Q)=1×k2×k20+12α\det \left( Q \right)=\dfrac{1\times {{k}^{2}}\times k}{20+12\alpha }. Equating both, we get,
⇒det⁡(Q)=1×k2×k20+12α=k22\Rightarrow \det \left( Q \right)=\dfrac{1\times {{k}^{2}}\times k}{20+12\alpha }=\dfrac{{{k}^{2}}}{2}
⇒k20+12α=12\Rightarrow \dfrac{k}{20+12\alpha }=\dfrac{1}{2}
⇒2k=20+12(−1)=8\Rightarrow 2k=20+12\left( -1 \right)=8
⇒k=4\Rightarrow k=4
det⁡(P)=20+12α=20−12=8\det \left( P \right)=20+12\alpha =20-12=8
det⁡(Q)=k22=162=8\det \left( Q \right)=\dfrac{k^{2}}{2}=\dfrac{16}{2}=8
Now we verify the options. We can clearly see A is wrong. For B,
4α−k+8=4(−1)−4+8=04\alpha -k+8=4\left( -1 \right)-4+8=0. Therefore, B is correct.
For C,
det⁡(Padj(Q))=det⁡(P)det⁡(Q)2=8×64=512=29\det \left( Padj\left( Q \right) \right)=\det \left( P \right)\det {{\left( Q \right)}^{2}}=8\times 64=512={{2}^{9}}
Therefore, C is correct.
For D,
det⁡(Qadj(P))=det⁡(Q)det⁡(P)2=8×64=512=29\det \left( Qadj\left( P \right) \right)=\det \left( Q \right)\det {{\left( P \right)}^{2}}=8\times 64=512={{2}^{9}}
Therefore, D is the wrong option.
Therefore, the final answers are option B and C.

Note: In order to answer these kinds of questions, you need to know the matrix multiplication, the determinant formula, the inverse matrix formula. You should also know how to calculate the adjacent of the matrix. Also, you should be careful while doing any substitutions and calculations.