Let A = [ ll 2 & 4 3 & 2 ], B = [ ll 1 & 3 -2 & 5 ], C = [ cc -2 & 5 3 & 4 ]. Find each of the following: 1. A + B 2. A - B 3. 3A - C 4. AB 5. BA

1. A + B = [ ll 2 & 4 3 & 2 ]+ [ cc 1 & 3 -2 & 5 ] = [ cc 2+1 & 4+3 3-2 & 2+5 ] = [ ll 3 & 7 1 & 7 ] 2. A – B = [ ll 2 & 4 3 & 2 ]- [ cc 1 & 3 -2 & 5 ] = [ cc 2-1 & 4-3 3-(-2) & 2-5 ] = [ cc 1 & 1 5 & -3 ] 3. We have, 3A - C = 3 [ cc 2 & 4 3 & 2 ]- [ cc -2 & 5 3 & 4 ] = [ ll 3 2 & 3 4 3 3 & 3 2 ]- [ cc -2 & 5 3 & 4 ] = [ cc 6 & 12 9 & 6 ]- [ cc -2 & 5 3 & 4 ] = [ cc 6+2 & 12-5 9-3 & 6-4 ] = [ ll 8 & 7 6 & 2 ] 4. We have, AB = [ cc 2 & 4 3 & 2 ] [ cc 1 & 3 -2 & 5 ] = [ ll 2(1)+4(-2) & 2(3)+4(5) 3(1)+2(-2) & 3(3)+2(5) ] = [ ll 2-8 & 6+20 3-4 & 9+10 ] = [ cc -6 & 26 -1 & 19 ] 5. We have BA = [ cc 1 & 3 -2 & 5 ] [ ll 2 & 4 3 & 2 ] = [ cc 1(2)+3(3) & 1(4)+3(2) -2(2)+5(3) & -2(4)+5(2) ] = [ rr 2+9 & 4+6 -4+15 & -8+10 ] = [ ll 11 & 10 11 & 2 ]

Let A = [ ll 2 & 4 3 & 2 ], B = [ ll 1 & 3 -2 & 5 ], C = [ cc -2 …

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Question

Let A = $\left[\begin{array}{ll} {2} & {4} \\ {3} & {2} \end{array}\right]$, B = $\left[\begin{array}{ll} {1} & {3} \\ {-2} & {5} \end{array}\right]$, C = $\left[\begin{array}{cc} {-2} & {5} \\ {3} & {4} \end{array}\right]$. Find each of the following:

A + B
A - B
3A - C
AB
BA

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Answer

A + B = $\left[\begin{array}{ll} {2} & {4} \\ {3} & {2} \end{array}\right]+\left[\begin{array}{cc} {1} & {3} \\ {-2} & {5} \end{array}\right]$
= $\left[\begin{array}{cc} {2+1} & {4+3} \\ {3-2} & {2+5} \end{array}\right]$
= $\left[\begin{array}{ll} {3} & {7} \\ {1} & {7} \end{array}\right]$
A – B = $\left[\begin{array}{ll} {2} & {4} \\ {3} & {2} \end{array}\right]-\left[\begin{array}{cc} {1} & {3} \\ {-2} & {5} \end{array}\right]$
= $\left[\begin{array}{cc} {2-1} & {4-3} \\ {3-(-2)} & {2-5} \end{array}\right]$
= $\left[\begin{array}{cc} {1} & {1} \\ {5} & {-3} \end{array}\right]$
We have,
3A - C = $3\left[\begin{array}{cc} {2} & {4} \\ {3} & {2} \end{array}\right]-\left[\begin{array}{cc} {-2} & {5} \\ {3} & {4} \end{array}\right]$
= $\left[\begin{array}{ll} {3 \times 2} & {3 \times 4} \\ {3 \times 3} & {3 \times 2} \end{array}\right]-\left[\begin{array}{cc} {-2} & {5} \\ {3} & {4} \end{array}\right]$
= $\left[\begin{array}{cc} {6} & {12} \\ {9} & {6} \end{array}\right]-\left[\begin{array}{cc} {-2} & {5} \\ {3} & {4} \end{array}\right]$
= $\left[\begin{array}{cc} {6+2} & {12-5} \\ {9-3} & {6-4} \end{array}\right]$
= $\left[\begin{array}{ll} {8} & {7} \\ {6} & {2} \end{array}\right]$
We have,
AB = $\left[\begin{array}{cc} {2} & {4} \\ {3} & {2} \end{array}\right]\left[\begin{array}{cc} {1} & {3} \\ {-2} & {5} \end{array}\right]$
= $\left[\begin{array}{ll} {2(1)+4(-2)} & {2(3)+4(5)} \\ {3(1)+2(-2)} & {3(3)+2(5)} \end{array}\right]$
= $\left[\begin{array}{ll} {2-8} & {6+20} \\ {3-4} & {9+10} \end{array}\right]$
= $\left[\begin{array}{cc} {-6} & {26} \\ {-1} & {19} \end{array}\right]$
We have
BA = $\left[\begin{array}{cc} {1} & {3} \\ {-2} & {5} \end{array}\right]\left[\begin{array}{ll} {2} & {4} \\ {3} & {2} \end{array}\right]$
= $\left[\begin{array}{cc} {1(2)+3(3)} & {1(4)+3(2)} \\ {-2(2)+5(3)} & {-2(4)+5(2)} \end{array}\right]$
= $\left[\begin{array}{rr} {2+9} & {4+6} \\ {-4+15} & {-8+10} \end{array}\right]$
= $\left[\begin{array}{ll} {11} & {10} \\ {11} & {2} \end{array}\right]$