Answer
Verified
422.4k+ views
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.
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE