A=begin{bmatrix}2& 1&1 -1 & -1&4 end{bmatrix} B=begin{bmatrix}0& 2 -4 & 12 & -3 end{bmatrix} C=begin{bmatrix}6& -1 3 & 0-2 & 5 end{bmatrix} D=begin{bmatrix}2& -3&4 -3 & 1&-2 end{bmatrix} a)A-3D b)B+frac{1}{2} c) C+ frac{1}{2}B (a),(b),(c) need to be solved

banganX 2021-02-05 Answered
A=[211114]B=[024123]C=[613025]D=[234312]
a)A3D
b)B+12
c) C+12B
(a),(b),(c) need to be solved
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Expert Answer

un4t5o4v
Answered 2021-02-06 Author has 105 answers
Step 1
The addition or subtraction of two matrices A and B are possible only if the number of rows and columns of the matrices A and Bare equal.
Suppose that the matrices A and B have same number of rows and columns. Then, A+B is calculated by adding corresponding elements of the matrices A and B, and A-B is calculated by subtracting corresponding elements of the matrices B from A.
Step 2
a)We have, A=[211114] and D=[234312]. So , 3D=[3×23×(3)3×43×(3)3×13×(2)]=[6912936]
Now ,
A3D=[211114][6912936]
=[261(9)1121(9)134(6)]
=[410118410]
Step 3
b. We have,
B=[024123] Since the matrix B contains 3 rows and 2 columns while 12 is a scalar which is a matrix of one row and one column, so the operation B+12 is not possible. Because the addition or subtraction of two matrices A and B are possible only if the number of rows and columns of the matrices A and Bare equal.
Step 4
c) We have,
B=[024123]C=[613025] So , 12B=[12×012×212×(4)12×112×212×(3)]=[01212132]
Now,
[613025]+[01212132]
=[6+01+13+(2)0+122+15+(32)]
=[6032121532]
=[6011212×532]
=[6011211032]
=[60112172]
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Answered 2021-09-14
Jeffrey Jordon
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