Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Matrices
+1 vote

# Find the values of a,b,c and d,if $3\begin{bmatrix}a & b\\c & d\end{bmatrix}=\begin{bmatrix}a & 6\\-1 & 2d\end{bmatrix}+\begin{bmatrix}4 & a+b\\c+d & 3\end{bmatrix}$

Toolbox:
• The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
• If the order of 2 matrices are equal, their corresponding elements are equal, i.e, if $A_{ij}=B_{ij}$, then any element $a_{ij}$ in matrix A is equal to corresponding element $b_{ij}$ in matrix B.
Step1
Given
$\begin{bmatrix}3a & 3b\\3c & 3d\end{bmatrix}=\begin{bmatrix}a & 6\\-1 & 2d\end{bmatrix}+\begin{bmatrix}4 & a+b\\c+d & 3\end{bmatrix}$
The given matrices are equal hence their corresponding elements should be equal.
$\begin{bmatrix}3a & 3b\\3c & 3d\end{bmatrix}=\begin{bmatrix}a+4 & 6+a+b\\-1+c+d & 2d+3\end{bmatrix}$
$\Rightarrow$ 3a=a+4------(1)
$\quad$3b=6+a+b-----(2)
$\quad$3c=-1+c+d-----(3)
$\quad$3d=2d+3-----(4)
Step2:
From equation (1) we have,
3a=a+4
3a-a=4
2a=4
$a=\frac{4}{2}$
a=2.
Step3:
From equation (2) we have,
3b=6+a+b
3b=6+2+b [Substitute the value of a in equation(2)]
3b=8+b
3b-b=8.
2b=8.
b=8/2
b=4.
Step4:
From equation (4) we have,
3d=2d+3
3d-2d=3
d=3
Step5:
From equation (3) we have,
3c=-1+c+d.
Substitute the value of d in equation (3)
3c=-1+c+3
3c-c=-1+3
2c=2
c=1