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# Find the values of a,b,c and d from the question:$\begin{bmatrix}a-b & 2a+c\\2a-b & 3c+d\end{bmatrix}$ = $\begin{bmatrix}-1 & 5\\0 & 13\end{bmatrix}$

$\begin{array}{1 1} a=1 b=2 c=3 d=4 \\ a=3 b=4 c=1 d=2 \\ a=0 b=2 c=3 d=1\\ a=2 b=3 c=0 d=1 \end{array}$

Toolbox:
• 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.
Given $\begin{bmatrix}a-b & 2a+c\\2a-b & 3c+d\end{bmatrix}=\begin{bmatrix}-1 & 5 \\0& 13\end{bmatrix}.$ Since these matrices are equal, we can an obtain the value of a,b,c and d by comparing the matrices corresponding elements.
By comparing the given two matrices of equal order, we can see that:
$a-b=-1$ (i)
$2a-b=0$ (ii)
$2a+c = 5$ (iii)
$3c+d = 13$ (iv)
Subtracting (i) and (ii), we get $a-b - (2a-b) = -1-0 \rightarrow -a = -1 \rightarrow a = 1$.
Substituting $a=1$ in (i), we get $1-b=-1 \rightarrow b= 2$.
Substituting $a=1$ in (iii), we get $2 + c = 5 \rightarrow c = 5-2 = 3$.
Substituting $c=3$ in (iv), we get $3x3+d = 13 \rightarrow d = 13-9 = 4$.
Solving for $a,b,c,d$ we get $(a,b,c,d) = (1,2,3,4)$.