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.

Question

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.

crocolylec · Accepted Answer

(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 · Accepted Answer

Answer is given below (on video)

RizerMix · Accepted Answer

(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=[0−10201000−0.50.500001] (subtracting row 2 from each of the other rows)B6=[0−10001000−0.50.500001] (replacing column 4 by column 3)B7=[−100100−0.500011] (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·B3B=B2C=B1Therefore, the result can be written as a product of three matrices:B=ABCwhere A=B7·B6·B5·B4·B3, B=B2, and C=B1.

nick1337 · Accepted Answer

(a) Writing the result as a product of eight matrices:Let&#039;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+12b34−b24b12+12b32−b22b13+12b33−b232b11+12b31−2b21b24b22b232b2112b34−b2412b32−b2212b33−b2312b31−2b21b44−b24b42−b22b43−b232b41−2b21]6. Replace column 4 by column 3:B6=[b14+12b34−b24b12+12b32−b22b13+12b33−b23b13+12b33−b23b24b22b23b2312b34−b2412b32−b2212b33−b2312b33−b23b44−b24b42−b22b43−b23b43−b23]7. Delete column 1 (column dimension is reduced by 1):B7=[b12+12b32−b22b13+12b33−b23b13+12b33−b23b22b23b2312b32−b2212b33−b2312b33−b23b42−b22b43−b23b43−b23]Now, we can express the final matrix B7 as a product of eight matrices:B7=B·M1·M2·M3·M4·M5·M6·M7where each Mi represents the matrixcorresponding 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·Cwhere 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·M4B=M5·M6C=M7Thus, the final matrix B7 can be written as a product of ABC:B7=A·B·C

Don Sumner · Accepted Answer

Step 1:Let&#039;s denote the original matrix as B. We&#039;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=[001201000−11200001]6. Replace column 4 by column 3:C6=[001101000−112120001]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′=C7C6C5C4C3C2C1BStep 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&#039;s denote the combined matrices as A, B, and C.A=C7C6C5;B=C4C3C2;C=C1The resulting matrix can be expressed as a product of these three matrices applied to the original matrix:B′=ABCB

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

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