# Find the matrices: a)A + B b) A - B c) -4A d)3A + 2B. A=begin{bmatrix}3&1 &1-1&2&5 end{bmatrix} , B=begin{bmatrix}2&-3 &6-3&1&-4 end{bmatrix}

Find the matrices:
a)A + B
b) A - B
c) -4A
d)3A + 2B.
$A=\left[\begin{array}{ccc}3& 1& 1\\ -1& 2& 5\end{array}\right],B=\left[\begin{array}{ccc}2& -3& 6\\ -3& 1& -4\end{array}\right]$
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tabuordy
Step 1
Consider the given matrices
$A=\left[\begin{array}{ccc}3& 1& 1\\ -1& 2& 5\end{array}\right],B=\left[\begin{array}{ccc}2& -3& 6\\ -3& 1& -4\end{array}\right]$
Step 2
(a) part
Now,
$A+B=\left[\begin{array}{ccc}3& 1& 1\\ -1& 2& 5\end{array}\right]+\left[\begin{array}{ccc}2& -3& 6\\ -3& 1& -4\end{array}\right]$
$=\left[\begin{array}{ccc}3+2& 1+\left(-3\right)& 1+6\\ -1+\left(-3\right)& 2+1& 5+\left(-4\right)\end{array}\right]$
$=\left[\begin{array}{ccc}5& 1-3& 7\\ -1-3& 3& 1\end{array}\right]$
$=\left[\begin{array}{ccc}5& -2& 7\\ -4& 3& 1\end{array}\right]$
Hence, the required sum of matrices A and B is $A+B=\left[\begin{array}{ccc}5& -2& 7\\ -4& 3& 1\end{array}\right]$
Step 3
(b) part
To find the difference of matirces A-B
$A-B=\left[\begin{array}{ccc}3& 1& 1\\ -1& 2& 5\end{array}\right]-\left[\begin{array}{ccc}2& -3& 6\\ -3& 1& -4\end{array}\right]$
$=\left[\begin{array}{ccc}3-2& 1-\left(-3\right)& 1-6\\ -1-\left(-3\right)& 2-1& 5-\left(-4\right)\end{array}\right]$
$=\left[\begin{array}{ccc}1& 1+3& -5\\ -1+3& 1& 5+4\end{array}\right]$
$=\left[\begin{array}{ccc}1& 4& -5\\ 2& 1& 9\end{array}\right]$
Hence, the required sum of matrices A and B is $A-B=\left[\begin{array}{ccc}1& 4& -5\\ 2& 1& 9\end{array}\right]$
Step 4
(c) part
Since, $A=\left[\begin{array}{ccc}3& 1& 1\\ -1& 2& 5\end{array}\right]$ So , $-4A=-4×\left[\begin{array}{ccc}3& 1& 1\\ -1& 2& 5\end{array}\right]$
$=\left[\begin{array}{ccc}-4×3& -4×1& -4×1\\ -4×\left(-1\right)& -4×2& -4×5\end{array}\right]$
$\left[\begin{array}{ccc}-12& -4& -4\\ 4& -8& -20\end{array}\right]$ hence, the required matrix −4A is $-4A=\left[\begin{array}{ccc}-12& -4& -4\\ 4& -8& -20\end{array}\right]$
Step 5
(d) part
First find 3A,
Since, $A=\left[\begin{array}{ccc}3& 1& 1\\ -1& 2& 5\end{array}\right]$ Then , $3A=3×\left[\begin{array}{ccc}3& 1& 1\\ -1& 2& 5\end{array}\right]$
$=\left[\begin{array}{ccc}3×3& 3×1& 3×1\\ 3×\left(-1\right)& 3×2& 3×5\end{array}\right]$
$\left[\begin{array}{ccc}9& 3& 3\\ -3& 6& 15\end{array}\right]$
Now find matrix 2B,
Since, $B=\left[\begin{array}{ccc}2& -3& 6\\ -3& 1& -4\end{array}\right]$
Then, $2B=2×\left[\begin{array}{ccc}2& -3& 6\\ -3& 1& -4\end{array}\right]$
Jeffrey Jordon