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

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,

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.