Question

\[\left| {\begin{array}{*{20}{c}}
  {{{\left( {1 + a} \right)}^2}}&{{{\left( {1 + 2a} \right)}^2}}&{{{\left( {1 + 3a} \right)}^2}} \\
  {{{\left( {2 + a} \right)}^2}}&{{{\left( {2 + 2a} \right)}^2}}&{{{\left( {2 + 3a} \right)}^2}} \\
  {{{\left( {3 + a} \right)}^2}}&{{{\left( {3 + 2a} \right)}^2}}&{{{\left( {3 + 3a} \right)}^2}}
\end{array}} \right| = 648a?\]
$a = ?$
1)$7$
2) $8$
3)$9$
4) $10$

Answer 1

Hint: We have to find the value of ‘a’ firstly we have to solve the determinant from the left hand side. Since it is very complex to expand along the row and column of the matrix. So we have to apply elementary operations on rows and columns of the matrix. This will reduce the matrix into the simplest form. Now it will be easy to find the determinant. In last we equate this with right hand side and find the value of ‘a’

Complete step-by-step answer:
We have given determinant with variable ‘a’ and we have to find this variable.
Firstly we will take left hand side.
L.H.S. \[\left| {\begin{array}{*{20}{c}}
  {{{\left( {1 + a} \right)}^2}}&{{{\left( {1 + 2a} \right)}^2}}&{{{\left( {1 + 3a} \right)}^2}} \\
  {{{\left( {2 + a} \right)}^2}}&{{{\left( {2 + 2a} \right)}^2}}&{{{\left( {2 + 3a} \right)}^2}} \\
  {{{\left( {3 + a} \right)}^2}}&{{{\left( {3 + 2a} \right)}^2}}&{{{\left( {3 + 3a} \right)}^2}}
\end{array}} \right|\]
We apply formula ${\left( {1 + a} \right)^2} = {x^2} + {y^2} + 2xy$ each element
L.H.S. \[\left| {\begin{array}{*{20}{c}}
  {1 + {a^2} + 2a}&{1 + 4{a^2} + 4a}&{1 + 9{a^2} + 6a} \\
  {4 + {a^2} + 4a}&{4 + 4{a^2} + 8a}&{4 + 9{a^2} + 12a} \\
  {9 + {a^2} + 6a}&{9 + 4{a^2} + 12a}&{9 + 9{a^2} + 18a}
\end{array}} \right|\]
Now we apply row operations on Row $3$and row $2$.
Operation on ${R_2}$ is ${R_2} \to {R_2} - {R_1}$
And operation on ${R_3}$is ${R_3} \to {R_3} - {R_1}$
L.H.S. \[\left| {\begin{array}{*{20}{c}}
  {1 + {a^2} + 2a}&{1 + 4{a^2} + 4a}&{1 + 9{a^2} + 6a} \\
  {3 + 2a}&{3 + 4a}&{3 + 6a} \\
  {8 + 4a}&{8 + 8a}&{8 + 12a}
\end{array}} \right|\]
Now we apply operation on ${R_3}$as ${R_3} \to {R_3} - 2{R_2}$
L.H.S. \[\left| {\begin{array}{*{20}{c}}
  {1 + {a^2} + 2a}&{1 + 4{a^2} + 4a}&{1 + 9{a^2} + 6a} \\
  {3 + 2a}&{3 + 4a}&{3 + 6a} \\
  2&2&2
\end{array}} \right|\]

Applying operation of ${C_2}$ and ${C_3}$ as
${C_2} \to {C_2} - {C_1}$ and ${C_3} \to {C_3} - {C_1}$
L.H.S. \[\left| {\begin{array}{*{20}{c}}
  {1 + {a^2} + 2a}&{3{a^2} + 2a}&{8{a^2} + 4a} \\
  {3 + 2a}&{2a}&{4a} \\
  2&0&0
\end{array}} \right|\]
Now expanding L.H.S. along ${R_3}$
L.H.S. $ = 2\left[ {4a(3{a^2} + 2a) - 2a(8{a^2} + 4a)} \right]$
     \[ = 2\left[ {12{a^3} + 8{a^2} - 16{a^3} - 8{a^2})} \right]\]
     \[ = 2\left[ { - 4{a^2}} \right]\]
     \[ = - 8{a^3}\]
L.H.S. \[ = - 8{a^3}\]
We have given R.H.S. \[ = - 648a\]
Therefore \[ - 8{a^3} = 648a\]
$\Rightarrow$ \[8{a^2} = 648\]
$\Rightarrow$ \[{a^2} = \dfrac{{648}}{8}\]
$\Rightarrow$ \[{a^2} = 81\]
$\Rightarrow$ \[{a^2} = \sqrt {81} \]
$\Rightarrow$ \[{a^2} = \pm 9\]
So option $\left( 3 \right)$ is correct

Note: Matrix is a set of numbers arranged in row and column. So as to form a rectangular array. The numbers are called elements or entries of the matrix. It has various applications in various branches of mathematics.
Operations on matrix: There are three kinds of elementary matrix operations. Interchange two rows or two columns. Multiply each element in column/row by non zero element and add the result to another row.