Question

If given that: \[A = \left[ {\begin{array}{*{20}{c}}
  1&2&{ - 1} \\
  { - 1}&1&2 \\
  2&{ - 1}&1
\end{array}} \right]\]. Then what will be the value of det(adj(adjA))?
A. \[{14^2}\]
B. \[{14^4}\]
C. Cannot be determined
D. None of these

Answer 1

Hint:The given problem revolves around the concepts of matrices and determinants. As a result, using the condition inverse matrix that is \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adjA\], multiplying it with the respective matrix ‘A’ so as to get the desired solution. Hence using the certain conditions/rules, substituting the value of its determinant, the required solution is obtained.

Complete step by step answer:
Since, we have given that $A$ is the matrix existing the values that,
\[A = \left[ {\begin{array}{*{20}{c}}
  1&2&{ - 1} \\
  { - 1}&1&2 \\
  2&{ - 1}&1
\end{array}} \right]\]

Hence, the determinant of A is
\[A = \left[ {\begin{array}{*{20}{c}}
  1&2&{ - 1} \\
  { - 1}&1&2 \\
  2&{ - 1}&1
\end{array}} \right] = \left| {\begin{array}{*{20}{c}}
  1&2&{ - 1} \\
  { - 1}&1&2 \\
  2&{ - 1}&1
\end{array}} \right|\]
By the definition of the determinant that is, we get
\[\left| A \right| = 1\left[ {1 - \left( { - 2} \right)} \right] - 2\left( { - 1 - 4} \right) + \left( { - 1} \right)\left( {1 - 2} \right)\]
As a result, solving mathematically, we get
\[\left| A \right| = 1\left( {1 + 2} \right) - 2\left( { - 5} \right) - 1\left( { - 1} \right)\]
\[\Rightarrow \left| A \right| = 3 + 10 + 1\]
Hence, we get
\[\left| A \right| = 14\] … (i)

Now, since to find adj(adjA)?
Considering the adj(A), we know that,inverse of determinant A is the inverse of its respective determinant multiplied by adjacent of respective matrix that is ‘A’ (where, adj(A) is represent as \[adj\left( A \right) = \left[ {\begin{array}{*{20}{c}}
  1&{ - 1}&2 \\
  2&1&{ - 1} \\
  { - 1}&2&1
\end{array}} \right]\]
\[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adjA\]

Mathematically, the above equation can also be written as,
\[{A^{ - 1}}\left| A \right| = adjA\]
Multiplying and dividing the equation by ‘A’, we get
\[A{A^{ - 1}}\left| A \right| = AadjA\]
We know that, any given matrix multiplied by its inverse is always the identity matrix that is \[\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]\] respectively i.e. \[A{A^{ - 1}} = I\]!

Hence, the equation becomes
\[I\left| A \right| = AadjA\]
Taking the determinant on both the sides, we get
\[\left| {\left| A \right|I} \right| = \left| {AadjA} \right|\]
Since, any matrix/determinant multiplied by an Identity matrix/determinant is always the respective/same matrix of determinant i.e. \[AI = A\],
\[\left| {\left| A \right|} \right| = \left| {AadjA} \right|\]
Separating the terms, we get
\[{\left| A \right|^n} = \left| A \right|\left| {adjA} \right|\]
Where, n is the order of the matrix
Here, \[n = 3\]

Dividing the equation by \[\left| A \right|\], we get
\[\dfrac{{{{\left| A \right|}^n}}}{{\left| A \right|}} = \left| {adjA} \right|\]
\[\Rightarrow {\left| A \right|^{n - 1}} = \left| {adjA} \right|\] … (ii)
Similarly,
For adj(adjA),
Multiplying equation (ii) by its adj, we get
\[{\left| {adjA} \right|^{n - 1}} = \left| {adj\left( {adjA} \right)} \right|\]
\[\Rightarrow \left| {adj\left( {adjA} \right)} \right| = {\left| {adjA} \right|^{n - 1}}\]
From (ii),
\[\left| {adj\left( {adjA} \right)} \right| = {\left| A \right|^{\left( {n - 1} \right)\left( {n - 1} \right)}}\]

Hence, solving the equation mathematically, we get
\[\left| {adj\left( {adjA} \right)} \right| = {\left| A \right|^{{{\left( {n - 1} \right)}^2}}}\]
Expanding the bracket that is \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\], we get
\[\left| {adj\left( {adjA} \right)} \right| = {\left| A \right|^{{n^2} - 2n + 1}}\]
As a result,
From (i), substituting \[\left| A \right| = 14\], we get
\[\left| {adj\left( {adjA} \right)} \right| = {14^{{n^2} - 2n + 1}}\]
Since, order is \[3 \to n = 3\]
\[\left| {adj\left( {adjA} \right)} \right| = {14^{{3^2} - 2 \times 3 + 1}} = {14^{9 - 6 + 1}}\]
\[\therefore \left| {adj\left( {adjA} \right)} \right| = {14^4}\]

Hence, option B is correct.

Note:Remember that \[\left| {\left| A \right|} \right|\] is always equal to \[{\left| A \right|^n}\]. One must be able to know the certain rules or the conditions used in matrices/determinants such as \[AI = A\], \[A{A^{ - 1}} = I\], etc. As a result, knowing the certain expansion rules like \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\], \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\], etc. would be better, so as to be sure of our final answer.