Question

If A is $3 \times 3$ matrix then $|3A| = $ ……………….. $\left| A \right|$
A) $3$
B) $6$
C) $9$
D) $27$

Answer 1

Hint:If a constant is multiplied to a matrix then it should be multiplied with all the elements of matrix.After multiplying elements of matrix $A$ with $3$ then we have find determinant of that matrix and expand it.Compare the final answer with matrix $A$ we get required answer.

Complete step-by-step answer:
Let us consider a matrix A of order $3 \times 3$ , where ${a_{ij}}$ are elements of the matrix $A$
$A = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right)$
Applying determinant on both sides for A
$\det \left( {{A^{}}} \right) = \det \left( {\left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right)} \right)$
After expanding the determinant we get a simplified formula
$\det ({A^{}}){ = ^{}}\left( {{a_{11}}({a_{22}} \times {a_{33}} - {a_{23}} \times {a_{32}}) - {a_{12}}({a_{21}} \times {a_{33}} - {a_{23}} \times {a_{31}}) + {a_{13}}({a_{21}} \times {a_{32}} - {a_{31}} \times {a_{22}})} \right)$
Let us assume that every element of matrix a is multiplied with the constant value $K$ and gives rise to new matrix $A' = KA$
$A' = \left( {\begin{array}{*{20}{c}}
  {K{a_{11}}}&{K{a_{12}}}&{K{a_{13}}} \\
  {K{a_{21}}}&{K{a_{22}}}&{K{a_{23}}} \\
  {K{a_{31}}}&{K{a_{32}}}&{K{a_{33}}}
\end{array}} \right)$
Applying determinant on both sides for the matrix $A'$

$\det (A') = \det \left( {K{a_{11}}(K{a_{22}} \times K{a_{33}} - K{a_{23}} \times K{a_{32}}) - K{a_{12}}(K{a_{21}} \times K{a_{33}} - K{a_{23}} \times K{a_{31}}) + K{a_{13}}(K{a_{21}} \times K{a_{32}} - K{a_{31}} \times K{a_{22}})} \right)$
Taking $K$ the common in the above equation
$\det (A') = \left( {{K^3}{a_{11}}({a_{22}} \times {a_{33}} - {a_{23}} \times {a_{32}}) - {K^3}{a_{12}}({a_{21}} \times {a_{33}} - {a_{23}} \times {a_{31}}) + {K^3}{a_{13}}({a_{21}} \times {a_{32}} - {a_{31}} \times {a_{22}})} \right)$
$\det (A') = {K^3}\left( {{a_{11}}({a_{22}} \times {a_{33}} - {a_{23}} \times {a_{32}}) - {a_{12}}({a_{21}} \times {a_{33}} - {a_{23}} \times {a_{31}}) + {a_{13}}({a_{21}} \times {a_{32}} - {a_{31}} \times {a_{22}})} \right)$
$\dfrac{{\det (A')}}{{{K^3}}} = \left( {{a_{11}}({a_{22}} \times {a_{33}} - {a_{23}} \times {a_{32}}) - {a_{12}}({a_{21}} \times {a_{33}} - {a_{23}} \times {a_{31}}) + {a_{13}}({a_{21}} \times {a_{32}} - {a_{31}} \times {a_{22}})} \right)$
Equating the equations of $\det (A)$ and $\det (A')$
$\dfrac{{\det (A')}}{{{K^3}}} = \det (A)$
Cross multiplying ${K^3}$ with $\det (A)$
$\det ({A^/}) = {K^3} \times \det (A)$
Therefore we have obtained a formula for $3 \times 3$ matrix
By using the above formula let $A' = 3A$ because $A'$ is in the form of $KA$
$
  \det (3A) = {3^3} \times \det (A) \\
  \det (3A) = 27 \times \det (A) \\
 $
Therefore the answer for this question will be $27$

Additional Information: If two rows or columns of matrix are interchanged, then the sign of Matrix changes
2. Determinant of a matrix AB is equivalent to the determinant of the matrix A multiplied with determinant of matrix B
3. If each element of a row or a column of a square matrix is multiplied by a number $K$ , then the determinant of the matrix obtained is $K$ times the determinant of the given matrix

So, the correct answer is “Option D”.

Note:The formula obtained is only for a matrix of order $3 \times 3$ for other order matrix you should use the formula $\det (KA) = {K^n} \times \det (A)$ where n is defined as the order of the given matrix.