Question

If \[A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&1\\0&2&{ - 3}\\2&1&0\end{array}} \right]\] , \[B = \left( {adj A} \right)\], and \[C = 5A\]. Then find the value of \[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}}\].
A. 5
B. \[25\]
C. \[ - 1\]
D. 1
E. \[125\]

Answer 1

Hint: First, calculate the determinant of the given \[3 \times 3\] matrix \[A\]. Simplify the required expression by using the given equations. Then apply the properties of the determinant of the matrix and solve it to get the required answer.

Formula used:
The determinant of a \[3 \times 3\] matrix \[A = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right]\] is: \[\left| A \right| = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{32}}{a_{23}}} \right) - {a_{12}}\left( {{a_{21}}{a_{33}} - {a_{31}}{a_{23}}} \right) + {a_{13}}\left( {{a_{21}}{a_{32}} - {a_{31}}{a_{22}}} \right)\]
The determinant of a scalar \[m\] times a matrix \[A\] of order \[n\] is: \[\left| {mA} \right| = {m^n}\left| A \right|\]
\[adj\left( {adj A} \right) = {\left| A \right|^{n - 2}}A\]
\[\left| {adj\left( {adj A} \right)} \right| = \left| {{{\left| A \right|}^{n - 2}}A} \right| = {\left( {{{\left| A \right|}^{n - 2}}} \right)^n}\left| A \right|\]

Complete step by step solution:
The given matrices are \[A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&1\\0&2&{ - 3}\\2&1&0\end{array}} \right]\], \[B = \left( {adj A} \right)\], and \[C = 5A\].

Let’s calculate the determinant of the given matrix \[A\].
Apply the formula of the determinant of a \[3 \times 3\] matrix \[\left| A \right| = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{32}}{a_{23}}} \right) - {a_{12}}\left( {{a_{21}}{a_{33}} - {a_{31}}{a_{23}}} \right) + {a_{13}}\left( {{a_{21}}{a_{32}} - {a_{31}}{a_{23}}} \right)\].
We get,
\[\left| A \right| = 1\left( {2 \times 0 - 1 \times \left( { - 3} \right)} \right) - \left( { - 1} \right)\left( {0 \times 0 - 2 \times \left( { - 3} \right)} \right) + 1\left( {0 \times 1 - 2 \times 2} \right)\]
\[ \Rightarrow \left| A \right| = 1\left( 3 \right) + 1\left( 6 \right) + 1\left( { - 4} \right)\]
\[ \Rightarrow \left| A \right| = 3 + 6 - 4\]
\[ \Rightarrow \left| A \right| = 5\]

Now solve the required expression \[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}}\] by using the given equations \[B = \left( {adj A} \right)\], and \[C = 5A\].
\[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{\left| {adj \left( {adj A} \right)} \right|}}{{\left| {5A} \right|}}\]
Use the properties for the determinant.
\[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{\left| {{{\left| A \right|}^{n - 2}}A} \right|}}{{{5^n}\left| A \right|}}\]
\[ \Rightarrow \dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{{{\left( {{{\left| A \right|}^{n - 2}}} \right)}^n}\left| A \right|}}{{{5^n}\left| A \right|}}\]
Here, the order of the square matrix is 3.
Then,
\[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{{{\left( {{{\left| A \right|}^{3 - 2}}} \right)}^3}\left| A \right|}}{{{5^3}\left| A \right|}}\]
\[ \Rightarrow \dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{{{\left( {{{\left| A \right|}^1}} \right)}^3}\left| A \right|}}{{{5^3}\left| A \right|}}\]
\[ \Rightarrow \dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{{{\left| A \right|}^3}\left| A \right|}}{{{5^3}\left| A \right|}}\]
Now substitute the value of the determinant in the above equation.
\[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = \dfrac{{{5^3} \times 5}}{{{5^3} \times 5}}\]
Cancel out the common terms from the numerator and the denominator.
\[\dfrac{{\left| {adj B} \right|}}{{\left| C \right|}} = 1\]
Hence the correct option is D.

Note: Students should keep in mind that the adjoint matrix of any matrix is the transpose of its cofactor matrix. While solving the questions related to the properties of the adjoint matrix, students can make note of the following important adjoint properties:
\[A\left( {adj A} \right) = \left( {adj A} \right)A = \left| A \right|I\]
\[\left| {adj A} \right| = {\left| A \right|^{n - 1}}\], where \[n\] is the order of the square matrix
\[adj\left( {adj A} \right) = {\left| A \right|^{n - 2}}A\]
\[adj\left( {AB} \right) = adj\left( B \right)adj\left( A \right)\]