Matrix multiplication is pretty tough- so i will cover that in class. In the meantime , compute the following if 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} If the operation is not possible , write NOT POSSIBLE and be able to explain why a)A+B b)B+C c)2A

Matrix multiplication is pretty tough- so i will cover that in class. In the meantime , compute the following if 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} If the operation is not possible , write NOT POSSIBLE and be able to explain why a)A+B b)B+C c)2A

Question
Matrices
asked 2021-01-04
Matrix multiplication is pretty tough- so i will cover that in class. In the meantime , compute the following if
\(A=\begin{bmatrix}2&1&1 \\-1&-1&4 \end{bmatrix} , B=\begin{bmatrix}0 & 2 \\-4 & 1\\2&-3 \end{bmatrix} , C=\begin{bmatrix}6 & -1 \\3 & 0\\-2&5 \end{bmatrix} , D=\begin{bmatrix}2 & -3&4 \\-3& 1&-2 \end{bmatrix}\)
If the operation is not possible , write NOT POSSIBLE and be able to explain why
a)A+B
b)B+C
c)2A

Answers (1)

2021-01-05
Step 1
\(\text{Given : } A=\begin{bmatrix}2&1&1 \\-1&-1&4 \end{bmatrix} , B=\begin{bmatrix}0 & 2 \\-4 & 1\\2&-3 \end{bmatrix} , C=\begin{bmatrix}6 & -1 \\3 & 0\\-2&5 \end{bmatrix} , D=\begin{bmatrix}2 & -3&4 \\-3& 1&-2 \end{bmatrix}\)
Solution for question a:
To compute A+B:
Note that two matrices may be added if and only if they have the same dimension, that is, they must have the same number of rows and columns.
Here, Dimension of \(A=2 \times 3\)
And, Dimension of \(B=3 \times 2\)
Matrix A and B do not have the same dimension. Hence, matrix A and B cannot be added.
Therefore it is not possible to perform A+B.
Step 2
Solution for question b:
Here,
Dimension of matrix \(B=3 \times 2\)
Dimension of matrix \(C=3 \times 2\)
Both matrices B and C have the same dimension. Hence, matrix B and C can be added.
Further,
\(B+C=\begin{bmatrix}0 & 2 \\-4 & 1\\2&-3 \end{bmatrix}+\begin{bmatrix}6 & -1 \\3 & 0\\-2&5 \end{bmatrix}\)
\(=\begin{bmatrix}0+6 & 2+(-1) \\-4+3 & 1+0\\2+(-2)&(-3)+5 \end{bmatrix}\)
\(=\begin{bmatrix}6 &1 \\-1 &1\\0&2 \end{bmatrix}\)
Therefore,
\(B+C=\begin{bmatrix}6 &1 \\-1 &1\\0&2 \end{bmatrix}\)
Step 3
Solution for question c:
To compute 2A multiply each entry of the matrix A by 2.
\(2A=2\begin{bmatrix}2&1&1 \\-1&-1&4 \end{bmatrix}\)
\(=\begin{bmatrix}4&2&2 \\-2&-2&8 \end{bmatrix}\)
therefore,
\(2A=\begin{bmatrix}4&2&2 \\-2&-2&8 \end{bmatrix}\)
0

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