Opérations sur les matrices : Addition et soustraction - Exercice 4

12 min
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COMPETENCE‾  :  Calculer{\color{red}\underline{COMPETENCE}\;:\;Calculer}
On considère les matrices suivantes : A=(−2341)A=\left(\begin{array}{cc} {-2} & {3} \\ {4} & {1} \end{array}\right) et B=(562−5)B=\left(\begin{array}{cc} {5} & {6} \\ {2} & {-5} \end{array}\right)
Question 1

Calculer : A+BA+B

Correction
A+B=(−2341)+(562−5)A+B=\left(\begin{array}{cc} {\red{-2}} & {\blue{3}} \\ {\green{4}} & {\pink{1}} \end{array}\right)+\left(\begin{array}{cc} {\red{5}} & {\blue{6}} \\ {\green{2}} & {\pink{-5}} \end{array}\right)
A+B=(−2+53+64+21+(−5))A+B=\left(\begin{array}{cc} {\red{-2}+\red{5}} & {\blue{3}+\blue{6}} \\ {\green{4}+\green{2}} & {\pink{1}+\left(\pink{-5}\right)} \end{array}\right)
A+B=(396−4)A+B=\left(\begin{array}{cc} {3} & {9} \\ {6} & {-4} \end{array}\right)
Question 2

Calculer : A−BA-B

Correction
A−B=(−2341)−(562−5)A-B=\left(\begin{array}{cc} {\red{-2}} & {\blue{3}} \\ {\green{4}} & {\pink{1}} \end{array}\right)-\left(\begin{array}{cc} {\red{5}} & {\blue{6}} \\ {\green{2}} & {\pink{-5}} \end{array}\right)
A−B=(−2−53−64−21−(−5))A-B=\left(\begin{array}{cc} {\red{-2}-\red{5}} & {\blue{3}-\blue{6}} \\ {\green{4}-\green{2}} & {\pink{1}-\left(\pink{-5}\right)} \end{array}\right)
A−B=(−7−326)A-B=\left(\begin{array}{cc} {-7} & {-3} \\ {2} & {6} \end{array}\right)
Question 3

Calculer : 3A3A

Correction
3A=3×(−2341)\purple{3}A=\purple{3}\times\left(\begin{array}{cc} {-2} & {3} \\ {4} & {1} \end{array}\right)
3A=(−2×33×34×31×3)\purple{3}A=\left(\begin{array}{cc} {-2\times\purple{3}} & {3\times\purple{3}} \\ {4\times\purple{3}} & {1\times\purple{3}} \end{array}\right)
3A=(−69123)\purple{3}A=\left(\begin{array}{cc} {-6} & {9} \\ {12} & {3} \end{array}\right)
Question 4

Calculer : 4B4B

Correction
4B=4×(562−5){\color{blue}{4}}B={\color{blue}{4}}\times\left(\begin{array}{cc} {5} & {6} \\ {2} & {-5} \end{array}\right)
4B=(5×46×42×4−5×4){\color{blue}{4}}B=\left(\begin{array}{cc} {5\times\color{blue}{4}} & {6\times\color{blue}{4}} \\ {2\times\color{blue}{4}} & {-5\times\color{blue}{4}} \end{array}\right)
4B=(20248−20){\color{blue}{4}}B=\left(\begin{array}{cc} {20} & {24} \\ {8} & {-20} \end{array}\right)
Question 5

Calculer : 3A−4B3A-4B

Correction
D'après les questions précédentes, nous savons que : 3A=(−69123)3A=\left(\begin{array}{cc} {-6} & {9} \\ {12} & {3} \end{array}\right) et 4B=(20248−20)4B=\left(\begin{array}{cc} {20} & {24} \\ {8} & {-20} \end{array}\right)
On a :
3A−4B=(−69123)−(20248−20)3A-4B=\left(\begin{array}{cc} {\red{-6}} & {\blue{9}} \\ {\green{12}} & {\pink{3}} \end{array}\right)-\left(\begin{array}{cc} {\red{20}} & {\blue{24}} \\ {\green{8}} & {\pink{-20}} \end{array}\right)
3A−4B=(−6−209−2412−83−(−20))3A-4B=\left(\begin{array}{cc} {\red{-6}-\red{20}} & {\blue{9}-\blue{24}} \\ {\green{12}-\green{8}} & {\pink{3}-\left(\pink{-20}\right)} \end{array}\right)
3A−4B=(−26−15423)3A-4B=\left(\begin{array}{cc} {-26} & {-15} \\ {4} & {23} \end{array}\right)

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