Question

[tex] Find a,b,c,d, if \left[\begin{array}{ccc}a+b&4\\3&c+d\end{array}\right] + \left[\begin{array}{ccc}6&a\\2d&-1\end{array}\right]= \left[\begin{array}{ccc}3b&3a\\3d&3c\end{array}\right] [\tex]

Answer 1

Answer:

sorry i don't that question

Answer 2

Answer:

a=2,b=4,c=1,d=3

Step-by-step explanation:

In this Question,

We have been given that

$\begin{bmatrix} a+b&4 \\ 3& c+d\end{bmatrix}+\begin{bmatrix} 6&a \\ 2d& -1\end{bmatrix}=\begin{bmatrix} 3b&3a \\ 3d& 3c\end{bmatrix}$

We have to find the values of a,b,c,d

$\begin{bmatrix} a+b&4 \\ 3& c+d\end{bmatrix}+\begin{bmatrix} 6&a \\ 2d& -1\end{bmatrix}=\begin{bmatrix} 3b&3a \\ 3d& 3c\end{bmatrix}$

Equating L.H.S with R.H.S we get,

a+b+6=3b ....(i)

4+a=3a     ......(ii)

3+2d=3d   .......(iii)

c+d-1=3c   .......(iv)

Solving Equation (ii) i.e. 2a=4 which implies a =2

Solving Equation (iii) i.e. 3+2d=3d which implies d=3

Now putting the value of d in quation (iv) we get c+3-1=3c

                                                                                 2c=2 which imples c=1

Again putting the value of a in Equation (1) we get 2+b+6=3b

                                                                                    8=2b which imples b =4

Hence Value of a is 2,Value of b is 4,Value of c is 1,Value of d is 3