Answer

Verified

449.1k+ 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

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Give 10 examples for herbs , shrubs , climbers , creepers

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE