Let B be a 4x4 matrix to which we apply the following operations: 1. double column 1, 2. halve row 3, 3. add row 3 to row 1, 4. interchange columns 1

Josalynn

Josalynn

Answered question

2021-02-08

Let B be a 4x4 matrix to which we apply the following operations:
1. double column 1,
2. halve row 3,
3. add row 3 to row 1,
4. interchange columns 1 and 4,
5. subtract row 2 from each of the other rows,
6. replace column 4 by column 3,
7. delete column 1 (column dimension is reduced by 1).
(a) Write the result as a product of eight matrices.
(b) Write it again as a product of ABC (same B) of three matrices.

Answer & Explanation

crocolylec

crocolylec

Skilled2021-02-09Added 100 answers


(a) To write the result as a product of 8 matrices.
Perform the following operations to get the final answer.
1.(BP)
2.Q(BP)
4.R(Q(BP))
5.R(Q(BP))S
6.T(R(Q(BP))S)
7.T(R(Q(BP))S)U
8.T(R(Q(BP))S)UV

(b) To write the result as a product of 3 matrices.
Multiply the matrices TRQ and name it as A. Multiply the matrices PSUV and name it as C. The representation ABC is the required product.
Jeffrey Jordon

Jeffrey Jordon

Expert2022-08-23Added 2605 answers

Answer is given below (on video)

RizerMix

RizerMix

Expert2023-05-22Added 656 answers

(a) To obtain the result as a product of eight matrices, we can apply each operation one by one:
B1=[2000010000100001] (doubling column 1)
B2=[20000100000.500001] (halving row 3)
B3=[20000100200.500001] (adding row 3 to row 1)
B4=[00020100000.500001] (interchanging columns 1 and 4)
B5=[0102010000.50.500001] (subtracting row 2 from each of the other rows)
B6=[0100010000.50.500001] (replacing column 4 by column 3)
B7=[1001000.500011] (deleting column 1)
Therefore, the result can be written as a product of eight matrices:
B=B7·B6·B5·B4·B3·B2·B1
(b) To express the result as a product of three matrices (ABC), we can combine some consecutive operations:
A=B7·B6·B5·B4·B3
B=B2
C=B1
Therefore, the result can be written as a product of three matrices:
B=ABC
where A=B7·B6·B5·B4·B3, B=B2, and C=B1.
nick1337

nick1337

Expert2023-05-22Added 777 answers

(a) Writing the result as a product of eight matrices:
Let's start with the original matrix B, which is a 4x4 matrix:
B=[b11b12b13b14b21b22b23b24b31b32b33b34b41b42b43b44]
1. Double column 1:
B1=[2b11b12b13b142b21b22b23b242b31b32b33b342b41b42b43b44]
2. Halve row 3:
B2=[2b11b12b13b142b21b22b23b2412b3112b3212b3312b342b41b42b43b44]
3. Add row 3 to row 1:
B3=[2b11+12b31b12+12b32b13+12b33b14+12b342b21b22b23b2412b3112b3212b3312b342b41b42b43b44]
4. Interchange columns 1 and 4:
B4=[b14+12b34b12+12b32b13+12b332b11+12b31b24b22b232b2112b3412b3212b3312b31b44b42b432b41]
5. Subtract row 2 from each of the other rows:
B5=[b14+12b34b24b12+12b32b22b13+12b33b232b11+12b312b21b24b22b232b2112b34b2412b32b2212b33b2312b312b21b44b24b42b22b43b232b412b21]
6. Replace column 4 by column 3:
B6=[b14+12b34b24b12+12b32b22b13+12b33b23b13+12b33b23b24b22b23b2312b34b2412b32b2212b33b2312b33b23b44b24b42b22b43b23b43b23]
7. Delete column 1 (column dimension is reduced by 1):
B7=[b12+12b32b22b13+12b33b23b13+12b33b23b22b23b2312b32b2212b33b2312b33b23b42b22b43b23b43b23]
Now, we can express the final matrix B7 as a product of eight matrices:
B7=B·M1·M2·M3·M4·M5·M6·M7
where each Mi represents the matrix
corresponding to the operation applied in step i.
(b) Writing the result as a product of ABC of three matrices:
Since we have the sequence of operations applied to matrix B, we can express the final matrix B7 as a product of three matrices: A, B, and C.
B7=A·B·C
where A represents the product of the first four matrices (M1, M2, M3, M4), B represents the product of the next two matrices (M5, M6), and C represents the last matrix (M7).
A=M1·M2·M3·M4
B=M5·M6
C=M7
Thus, the final matrix B7 can be written as a product of ABC:
B7=A·B·C
Don Sumner

Don Sumner

Skilled2023-05-22Added 184 answers

Step 1:
Let's denote the original matrix as B. We'll apply the given operations step by step and represent each operation as a separate matrix.
1. Double column 1:
C1=[2000010000100001]
2. Halve row 3:
C2=[10000100001200001]
3. Add row 3 to row 1:
C3=[10100100001200001]
4. Interchange columns 1 and 4:
C4=[00120100001200001]
5. Subtract row 2 from each of the other rows:
C5=[00120100011200001]
6. Replace column 4 by column 3:
C6=[001101000112120001]
7. Delete column 1 (column dimension is reduced by 1):
C7=[0110112001001]
Finally, the resulting matrix after applying all the operations can be obtained by multiplying all the matrices together:
B=C7C6C5C4C3C2C1B
Step 2:
(b) Writing the result as a product of ABC (same B) of three matrices:
We can combine multiple consecutive matrices together to obtain a simplified representation. Let's denote the combined matrices as A, B, and C.
A=C7C6C5;B=C4C3C2;C=C1
The resulting matrix can be expressed as a product of these three matrices applied to the original matrix:
B=ABCB

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