A=[211−1−14]B=[02−412−3]C=[6−130−25]D=[2−34−31−2] a)A−3D b)B+12 c) C+12B (a),(b),(c) need to be solved

Question

A=[211−1−14]B=[02−412−3]C=[6−130−25]D=[2−34−31−2]  a)A−3D  b)B+12  c) C+12B  (a),(b),(c) need to be solved

un4t5o4v · Accepted Answer

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=[211−1−14] and D=[2−34−31−2]. So , 3D=[3×23×(−3)3×43×(−3)3×13×(−2)]=[6−912−93−6]  Now ,  A−3D=[211−1−14]−[6−912−93−6]  =[2−61−(−9)1−12−1−(−9)−1−34−(−6)]  =[−410−118−410]  Step 3  b. We have,  B=[02−412−3] 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=[02−412−3]C=[6−130−25] So , 12B=[12×012×212×(−4)12×112×212×(−3)]=[01−2121−32]  Now,  [6−130−25]+[01−2121−32]  =[6+0−1+13+(−2)0+12−2+15+(−32)]  =[603−212−15−32]  =[60112−12×5−32]  =[60112−110−32]  =[60112−172]

Accepted Answer

[10​12​]X=[342103][31​40​23​]

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

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

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