# If A=begin{bmatrix}1 & 1 3 & 4 end{bmatrix} , B=begin{bmatrix}2 1 end{bmatrix} ,C=begin{bmatrix}-7 & 1 0 & 4 end{bmatrix},D=begin{bmatrix}3 & 2 & 1 en

If
Find , if possible,
a) A+B , C-A and D-E b)AB, BA , CA , AC , DA , DB , BD , EB , BE and AE c) 7C , -3D and KE
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Yusuf Keller
Step 1
We can add and subtract matrices only which of them have same order.
For multiplication of two matrices it is necessary that no. of columns of first matrix must be equal to no of rows in second matrix.
Step 2
Here,

a) A+B , order of A is $2×2\ne$ order of $B\left(2×1\right)$
So, Not possible
C-A
$C-A=\left[\begin{array}{cc}-7& 1\\ 0& 4\end{array}\right]-\left[\begin{array}{cc}1& 1\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}-7-1& 1-1\\ 0-3& 4-4\end{array}\right]=\left[\begin{array}{cc}-8& 0\\ -3& 0\end{array}\right]$
$C-A=\left[\begin{array}{cc}-8& 0\\ -3& 0\end{array}\right]$
D-E
D-E have different orders
So,not possible
b) AB
$AB=\left[\begin{array}{cc}1& 1\\ 3& 4\end{array}\right]\left[\begin{array}{c}2\\ 1\end{array}\right]=\left[\begin{array}{c}2+1\\ 6+4\end{array}\right]=\left[\begin{array}{c}3\\ 10\end{array}\right]$
Hence
$AB=\left[\begin{array}{c}3\\ 10\end{array}\right]$

number of columns of B $\ne$ number of row in A
So,not possible
CA
$CA=\left[\begin{array}{cc}-7& 1\\ 0& 4\end{array}\right]\left[\begin{array}{cc}1& 1\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}-7+3& -7+4\\ 0+12& 0+16\end{array}\right]=\left[\begin{array}{cc}-4& -3\\ 12& 16\end{array}\right]$
Hence
$CA=\left[\begin{array}{cc}-4& -3\\ 12& 16\end{array}\right]$
AC
$AC=\left[\begin{array}{cc}1& 1\\ 3& 4\end{array}\right]\left[\begin{array}{cc}-7& 1\\ 0& 4\end{array}\right]=\left[\begin{array}{cc}-7& 1+4\\ -21& 3+16\end{array}\right]=\left[\begin{array}{cc}-7& 5\\ -21& 19\end{array}\right]$
Hence,
$AC=\left[\begin{array}{cc}-7& 5\\ -21& 19\end{array}\right]$
DA
$\left[D{\right]}_{1×3}\left[A{\right]}_{2×2}$
Number columns in [D] $\ne$ number rows in [A]
So,not possible DB . Not possible BD. Not possible
$EB.\left[E{\right]}_{2×3}\left[B{\right]}_{2×1}$
Not possible
BE
$\left[B{\right]}_{2×1}\left[E{\right]}_{2×3}$
Not possible
AE
$AE=\left[\begin{array}{cc}1& 1\\ 3& 4\end{array}\right]\left[\begin{array}{ccc}2& 3& 4\\ 1& 2& -1\end{array}\right]=$
$=\left[\begin{array}{ccc}2+1& 3+2& 4-1\\ 6+4& 9+8& 12-4\end{array}\right]=\left[\begin{array}{ccc}3& 5& 3\\ 10& 17& 8\end{array}\right]$
Hence,
$AE=\left[\begin{array}{ccc}3& 5& 3\\ 10& 17& 8\end{array}\right]$
(c)$7C=7\left[\begin{array}{cc}-7& 1\\ 0& 4\end{array}\right]=\left[\begin{array}{cc}-49& 7\\ 0& 28\end{array}\right]$
Jeffrey Jordon