Question
Answers

Let two matrices A = $\left[ \begin{gathered}
  1\,\,\,\,2 \\
  3\,\,\,\,4 \\
\end{gathered} \right]$ and B = $\left[ \begin{gathered}
  {\text{a}}\,\,\,\,{\text{b}} \\
  {\text{c}}\,\,\,\,{\text{d}} \\
\end{gathered} \right]$ are two matrices such that they are commutative and c$ \ne $3b. Then, find the value of $\dfrac{{d - a}}{{3b - c}}$.

Answer Verified Verified
Hint: In order to solve this problem we will use the property commutative, since it is provided that the two matrices are commutative using this data and then equating the two obtained matrices after multiplication and solving to get the asked term you will reach the right answer.

Complete step-by-step answer:
We have A = $\left[ \begin{gathered}
  1\,\,\,\,2 \\
  3\,\,\,\,4 \\
\end{gathered} \right]$and B = $\left[ \begin{gathered}
  {\text{a}}\,\,\,\,{\text{b}} \\
  {\text{c}}\,\,\,\,{\text{d}} \\
\end{gathered} \right]$.
It's given that A and B are commutative.
It means AB = BA ……(1)
First we calculate AB.
So, AB = $\left[ \begin{gathered}
  1\,\,\,\,2 \\
  3\,\,\,\,4 \\
\end{gathered} \right]$$\left[ \begin{gathered}
  {\text{a}}\,\,\,\,{\text{b}} \\
  {\text{c}}\,\,\,\,{\text{d}} \\
\end{gathered} \right]$=$\left[ \begin{gathered}
  {\text{a + 2c}}\,\,\,\,\,\,\,\,{\text{b + 2d}} \\
  {\text{3a + 4c}}\,\,\,\,\,\,\,\,{\text{3b + 4d}} \\
\end{gathered} \right]$
Then we find BA.
So, BA =$\left[ \begin{gathered}
  {\text{a}}\,\,\,\,{\text{b}} \\
  {\text{c}}\,\,\,\,{\text{d}} \\
\end{gathered} \right]$$\left[ \begin{gathered}
  1\,\,\,\,2 \\
  3\,\,\,\,4 \\
\end{gathered} \right]$ = \[\left[ \begin{gathered}
  {\text{a + 3b}}\,\,\,\,{\text{2a + 4b}} \\
  {\text{c + 3d}}\,\,\,\,{\text{2c + 4d}} \\
\end{gathered} \right]\]
From (1) we can equate the value of AB and BA.
So, AB = BA
$\left[ \begin{gathered}
  {\text{a + 2c}}\,\,\,\,\,\,\,\,{\text{b + 2d}} \\
  {\text{3a + 4c}}\,\,\,\,\,\,\,\,{\text{3b + 4d}} \\
\end{gathered} \right]$=\[\left[ \begin{gathered}
  {\text{a + 3b}}\,\,\,\,{\text{2a + 4b}} \\
  {\text{c + 3d}}\,\,\,\,{\text{2c + 4d}} \\
\end{gathered} \right]\]
Now we can say, a+2c = a+3b
2c = 3b
So, c = $\dfrac{{{\text{3b}}}}{2}$ ……(2)
And also, b + 2d = 2a + 4b
2d – 2a = 3b
d - a = $\dfrac{{{\text{3b}}}}{{\text{2}}}$ ……(3)
Therefore, we can do $\dfrac{{{\text{d - a}}}}{{{\text{3b - c}}}}$= $\dfrac{{\dfrac{{\text{3}}}{{\text{2}}}{\text{b}}}}{{{\text{3b - }}\dfrac{{\text{3}}}{{\text{2}}}{\text{b}}}}$= $\dfrac{{\dfrac{{\text{3}}}{{\text{2}}}{\text{b}}}}{{\dfrac{{\text{3}}}{{\text{2}}}{\text{b}}}}$ = 1.
Hence, the value of $\dfrac{{{\text{d - a}}}}{{{\text{3b - c}}}}$= 1.

Note: Whenever you face such types of problems you have to use the properties of matrix. The properties used here is multiplication of matrices and addition of matrices. Then we have just solved the asked term by equating the matrix as it is given that the matrix is commutative. Doing this will give you the right answer.
Bookmark added to your notes.
View Notes
×