Question

If \[A = \left[ {\begin{array}{*{20}{c}}
  0&2 \\
  3&{ - 4}
\end{array}} \right]\] and \[kA = \left[ {\begin{array}{*{20}{c}}
  0&{3a} \\
  {2b}&{24}
\end{array}} \right]\], then the values of k, a, b are respectively
A. -6, -12, -18
B. -6, 4, 9
C. -6, -4, -9
D. 6,12,18

Answer 1

Hint: In the given problem we find substitute the matrix \[A = \left[ {\begin{array}{*{20}{c}}
  0&2 \\
  3&{ - 4}
\end{array}} \right]\] in \[kA = \left[ {\begin{array}{*{20}{c}}
  0&{3a} \\
  {2b}&{24}
\end{array}} \right]\]. We then multiply the scalar k with all the elements of matrix A. The corresponding terms in both the matrices are then equated to find the unknowns.

Complete step by step solution:
We are given that,
\[kA = \left[ {\begin{array}{*{20}{c}}
  0&{3a} \\
  {2b}&{24}
\end{array}} \right]\] and \[A = \left[ {\begin{array}{*{20}{c}}
  0&2 \\
  3&{ - 4}
\end{array}} \right]\]
Substituting matrix A in kA we get,
\[k\left[ {\begin{array}{*{20}{c}}
  0&2 \\
  3&{ - 4}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&{3a} \\
  {2b}&{24}
\end{array}} \right]\]
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  0&{2k} \\
  {3k}&{ - 4k}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&{3a} \\
  {2b}&{24}
\end{array}} \right]\]
Equating the corresponding terms we get,
\[2k = 3a,\,\,\,\,3k = 2b,\,\,\,\, - 4k = 24\]
\[ \Rightarrow a = \dfrac{{2k}}{3},\,\,\,b = \dfrac{{3k}}{2},\,\,\,k = - 6\]
\[ \Rightarrow k = - 6,\,\,\,a = - 4,\,\,\,b = - 9\]

Option C. is the correct answer.

Note: A matrix can be multiplied with scalar numbers. If A = \[{\left[ {{a_{ij}}} \right]_{m \times n}}\] (a matrix of size \[m \times n\]) and k is a scalar, each of the elements of A is multiplied by k, resulting in kA=\[{\left[ {k{a_{ij}}} \right]_{m \times n}}\] to create the resultant matrix. One must make sure the signs are written correctly when writing the value of a, b and k.