Solveeit Logo
Login

Question

Question: Let \(M=\left[ \begin{matrix} 0 & 1 & a \\\ 1 & 2 & 3 \\\ 3 & b & 1 \\\ \end{matrix...

Let M=[01a 123 3b1 ]M=\left[ \begin{matrix} 0 & 1 & a \\\ 1 & 2 & 3 \\\ 3 & b & 1 \\\ \end{matrix} \right] and adj(M)=[−11−1 8−62 −53−1 ]adj\left( M \right)=\left[ \begin{matrix} -1 & 1 & -1 \\\ 8 & -6 & 2 \\\ -5 & 3 & -1 \\\ \end{matrix} \right] where a and b are real numbers.
Which of the following options is/are correct?
[a] det⁡(adj(M2))=81\det \left( adj\left( {{M}^{2}} \right) \right)=81
[b] a+b=3a+b=3
[c] If M[α β γ ]=[1 2 3 ]M\left[ \begin{matrix} \alpha \\\ \beta \\\ \gamma \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\\ 2 \\\ 3 \\\ \end{matrix} \right], then α+β+γ=3\alpha +\beta +\gamma =3
[d] adj(M−1)+adj(M)−1=−Madj\left( {{M}^{-1}} \right)+adj{{\left( M \right)}^{-1}}=-M

Explanation

Solution

Use the fact that Aadj(A)=det⁡(A)IAadj\left( A \right)=\det \left( A \right)I, where I is the identity matrix of same order as A. Hence determine the values of a, b and det(A). Hence verify which of the options are correct and which are incorrect.

Complete step-by-step answer:

We have
M=[01a 123 3b1 ]M=\left[ \begin{matrix} 0 & 1 & a \\\ 1 & 2 & 3 \\\ 3 & b & 1 \\\ \end{matrix} \right] and adj(M)=[−11−1 8−62 −53−1 ]adj\left( M \right)=\left[ \begin{matrix} -1 & 1 & -1 \\\ 8 & -6 & 2 \\\ -5 & 3 & -1 \\\ \end{matrix} \right]
We know that Aadj(A)=det⁡(A)IAadj\left( A \right)=\det \left( A \right)I, where I is the identity matrix of same order as A.
Hence, we have
[01a 123 3b1 ][−11−1 8−62 −53−1 ]=det⁡(M)[100 010 001 ]\left[ \begin{matrix} 0 & 1 & a \\\ 1 & 2 & 3 \\\ 3 & b & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} -1 & 1 & -1 \\\ 8 & -6 & 2 \\\ -5 & 3 & -1 \\\ \end{matrix} \right]=\det \left( M \right)\left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]
Hence, we have
[8−5a3(a−2)2−a 0−20 8(b−1)6(1−b)2(−2+y) ]=[det⁡(M)00 0det⁡(M)0 00det⁡(M) ]\left[ \begin{matrix} 8-5a & 3\left( a-2 \right) & 2-a \\\ 0 & -2 & 0 \\\ 8\left( b-1 \right) & 6\left( 1-b \right) & 2\left( -2+y \right) \\\ \end{matrix} \right]=\left[ \begin{matrix} \det \left( M \right) & 0 & 0 \\\ 0 & \det \left( M \right) & 0 \\\ 0 & 0 & \det \left( M \right) \\\ \end{matrix} \right]
Hence, we have
3(a−2)=0⇒a=2 det⁡(M)=−2 8(b−1)=0⇒b=1 \begin{aligned} & 3\left( a-2 \right)=0\Rightarrow a=2 \\\ & \det \left( M \right)=-2 \\\ & 8\left( b-1 \right)=0\Rightarrow b=1 \\\ \end{aligned}
Checking option [a] We know that det⁡(adj(A))=det⁡(A)n−1\det \left( adj\left( A \right) \right)=\det {{\left( A \right)}^{n-1}} where n is the order of A.
Hence, we have
det⁡(adj(M2))=det⁡(M2)3−1=det⁡(M2)2\det \left( adj\left( {{M}^{2}} \right) \right)=\det {{\left( {{M}^{2}} \right)}^{3-1}}=\det {{\left( {{M}^{2}} \right)}^{2}}
We know that det⁡(M2)=det⁡(M)2\det \left( {{M}^{2}} \right)=\det {{\left( M \right)}^{2}}
Hence, we have
det⁡(adj(M2))=det⁡(M)4=(−2)4=16\det \left( adj\left( {{M}^{2}} \right) \right)=\det {{\left( M \right)}^{4}}={{\left( -2 \right)}^{4}}=16
Checking option [b]: We have a =2 and b= 1. Hence, we have a+b = 2+1 = 3
Checking option [c]:
We have M[α β γ ]=[1 2 3 ]    (i)M\left[ \begin{matrix} \alpha \\\ \beta \\\ \gamma \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\\ 2 \\\ 3 \\\ \end{matrix} \right]\ \ \ \ \left( i \right)
Since det⁡(M)≠0\det \left( M \right)\ne 0, we have M−1{{M}^{-1}} exists.
Pre-multiplying equation (i) by M−1{{M}^{-1}}, we get
[α β γ ]=M−1[1 2 3 ]\left[ \begin{matrix} \alpha \\\ \beta \\\ \gamma \\\ \end{matrix} \right]={{M}^{-1}}\left[ \begin{matrix} 1 \\\ 2 \\\ 3 \\\ \end{matrix} \right]
We know that A−1=1det⁡(A)adj(A){{A}^{-1}}=\dfrac{1}{\det \left( A \right)}adj\left( A \right)
Hence, we have
M−1=1−2[−11−1 8−62 −53−1 ]=[12−1212 −43−1 52−3212 ]{{M}^{-1}}=\dfrac{1}{-2}\left[ \begin{matrix} -1 & 1 & -1 \\\ 8 & -6 & 2 \\\ -5 & 3 & -1 \\\ \end{matrix} \right]=\left[ \begin{matrix} \dfrac{1}{2} & \dfrac{-1}{2} & \dfrac{1}{2} \\\ -4 & 3 & -1 \\\ \dfrac{5}{2} & \dfrac{-3}{2} & \dfrac{1}{2} \\\ \end{matrix} \right]
Hence, we have
[α β γ ]=[12−1212 −43−1 52−3212 ][1 2 3 ]=[1 −1 1 ]\left[ \begin{matrix} \alpha \\\ \beta \\\ \gamma \\\ \end{matrix} \right]=\left[ \begin{matrix} \dfrac{1}{2} & \dfrac{-1}{2} & \dfrac{1}{2} \\\ -4 & 3 & -1 \\\ \dfrac{5}{2} & \dfrac{-3}{2} & \dfrac{1}{2} \\\ \end{matrix} \right]\left[ \begin{matrix} 1 \\\ 2 \\\ 3 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\\ -1 \\\ 1 \\\ \end{matrix} \right]
Hence, we have
α=1,β=−1,γ=1\alpha =1,\beta =-1,\gamma =1
Hence, we have
α+β+γ=1\alpha +\beta +\gamma =1
Checking option [d]:
We know that Aadj(A)=det⁡(A)IAadj\left( A \right)=\det \left( A \right)I
Hence, we have
M−1adj(M−1)=det⁡(M−1)I{{M}^{-1}}adj\left( {{M}^{-1}} \right)=\det \left( {{M}^{-1}} \right)I
Pre-multiplying both sides by M, we get
adj(M−1)=1det⁡(M)M=−M2adj\left( {{M}^{-1}} \right)=\dfrac{1}{\det \left( M \right)}M=-\dfrac{M}{2}
We know that AA−1=IA{{A}^{-1}}=I
Hence, we have
adj(M)adj(M)−1=Iadj\left( M \right)adj{{\left( M \right)}^{-1}}=I
Pre-multiplying both sides by M, we get
Madj(M)adj(M)−1=M det⁡(M)Iadj(M)−1=M ⇒adj(M)−1=Mdet⁡(M)=−M2 \begin{aligned} & Madj\left( M \right)adj{{\left( M \right)}^{-1}}=M \\\ & \det \left( M \right)Iadj{{\left( M \right)}^{-1}}=M \\\ & \Rightarrow adj{{\left( M \right)}^{-1}}=\dfrac{M}{\det \left( M \right)}=-\dfrac{M}{2} \\\ \end{aligned}
Hence, we have
adj(M−1)+adj(M)−1=−M2−M2=−Madj\left( {{M}^{-1}} \right)+adj{{\left( M \right)}^{-1}}=-\dfrac{M}{2}-\dfrac{M}{2}=-M

So, the correct answers are “Option b and d”.

Note: [1] The properties of adjoint of a matrix are very important in matrix algebra and students are advised to remember them. In particular the following two properties are very important
[a] Aadj(A)=adj(A)A=det⁡(A)IAadj\left( A \right)=adj\left( A \right)A=\det \left( A \right)I
[b] det⁡(adj(A))=det⁡(A)n−1\det \left( adj\left( A \right) \right)=\det {{\left( A \right)}^{n-1}}
The calculations in this problem would have been very long and tedious if we had not used these two properties.