Josalynn

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.

crocolylec

(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

Answer is given below (on video)

RizerMix

(a) To obtain the result as a product of eight matrices, we can apply each operation one by one:
${B}_{1}=\left[\begin{array}{cccc}2& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$ (doubling column 1)
${B}_{2}=\left[\begin{array}{cccc}2& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0.5& 0\\ 0& 0& 0& 1\end{array}\right]$ (halving row 3)
${B}_{3}=\left[\begin{array}{cccc}2& 0& 0& 0\\ 0& 1& 0& 0\\ 2& 0& 0.5& 0\\ 0& 0& 0& 1\end{array}\right]$ (adding row 3 to row 1)
${B}_{4}=\left[\begin{array}{cccc}0& 0& 0& 2\\ 0& 1& 0& 0\\ 0& 0& 0.5& 0\\ 0& 0& 0& 1\end{array}\right]$ (interchanging columns 1 and 4)
${B}_{5}=\left[\begin{array}{cccc}0& -1& 0& 2\\ 0& 1& 0& 0\\ 0& -0.5& 0.5& 0\\ 0& 0& 0& 1\end{array}\right]$ (subtracting row 2 from each of the other rows)
${B}_{6}=\left[\begin{array}{cccc}0& -1& 0& 0\\ 0& 1& 0& 0\\ 0& -0.5& 0.5& 0\\ 0& 0& 0& 1\end{array}\right]$ (replacing column 4 by column 3)
${B}_{7}=\left[\begin{array}{ccc}-1& 0& 0\\ 1& 0& 0\\ -0.5& 0& 0\\ 0& 1& 1\end{array}\right]$ (deleting column 1)
Therefore, the result can be written as a product of eight matrices:
$B={B}_{7}·{B}_{6}·{B}_{5}·{B}_{4}·{B}_{3}·{B}_{2}·{B}_{1}$
(b) To express the result as a product of three matrices (ABC), we can combine some consecutive operations:
$A={B}_{7}·{B}_{6}·{B}_{5}·{B}_{4}·{B}_{3}$
$B={B}_{2}$
$C={B}_{1}$
Therefore, the result can be written as a product of three matrices:
$B=ABC$
where $A={B}_{7}·{B}_{6}·{B}_{5}·{B}_{4}·{B}_{3}$, $B={B}_{2}$, and $C={B}_{1}$.

nick1337

(a) Writing the result as a product of eight matrices:
Let's start with the original matrix B, which is a 4x4 matrix:
$B=\left[\begin{array}{cccc}{b}_{11}& {b}_{12}& {b}_{13}& {b}_{14}\\ {b}_{21}& {b}_{22}& {b}_{23}& {b}_{24}\\ {b}_{31}& {b}_{32}& {b}_{33}& {b}_{34}\\ {b}_{41}& {b}_{42}& {b}_{43}& {b}_{44}\end{array}\right]$
1. Double column 1:
${B}_{1}=\left[\begin{array}{cccc}2{b}_{11}& {b}_{12}& {b}_{13}& {b}_{14}\\ 2{b}_{21}& {b}_{22}& {b}_{23}& {b}_{24}\\ 2{b}_{31}& {b}_{32}& {b}_{33}& {b}_{34}\\ 2{b}_{41}& {b}_{42}& {b}_{43}& {b}_{44}\end{array}\right]$
2. Halve row 3:
${B}_{2}=\left[\begin{array}{cccc}2{b}_{11}& {b}_{12}& {b}_{13}& {b}_{14}\\ 2{b}_{21}& {b}_{22}& {b}_{23}& {b}_{24}\\ \frac{1}{2}{b}_{31}& \frac{1}{2}{b}_{32}& \frac{1}{2}{b}_{33}& \frac{1}{2}{b}_{34}\\ 2{b}_{41}& {b}_{42}& {b}_{43}& {b}_{44}\end{array}\right]$
3. Add row 3 to row 1:
${B}_{3}=\left[\begin{array}{cccc}2{b}_{11}+\frac{1}{2}{b}_{31}& {b}_{12}+\frac{1}{2}{b}_{32}& {b}_{13}+\frac{1}{2}{b}_{33}& {b}_{14}+\frac{1}{2}{b}_{34}\\ 2{b}_{21}& {b}_{22}& {b}_{23}& {b}_{24}\\ \frac{1}{2}{b}_{31}& \frac{1}{2}{b}_{32}& \frac{1}{2}{b}_{33}& \frac{1}{2}{b}_{34}\\ 2{b}_{41}& {b}_{42}& {b}_{43}& {b}_{44}\end{array}\right]$
4. Interchange columns 1 and 4:
${B}_{4}=\left[\begin{array}{cccc}{b}_{14}+\frac{1}{2}{b}_{34}& {b}_{12}+\frac{1}{2}{b}_{32}& {b}_{13}+\frac{1}{2}{b}_{33}& 2{b}_{11}+\frac{1}{2}{b}_{31}\\ {b}_{24}& {b}_{22}& {b}_{23}& 2{b}_{21}\\ \frac{1}{2}{b}_{34}& \frac{1}{2}{b}_{32}& \frac{1}{2}{b}_{33}& \frac{1}{2}{b}_{31}\\ {b}_{44}& {b}_{42}& {b}_{43}& 2{b}_{41}\end{array}\right]$
5. Subtract row 2 from each of the other rows:
${B}_{5}=\left[\begin{array}{cccc}{b}_{14}+\frac{1}{2}{b}_{34}-{b}_{24}& {b}_{12}+\frac{1}{2}{b}_{32}-{b}_{22}& {b}_{13}+\frac{1}{2}{b}_{33}-{b}_{23}& 2{b}_{11}+\frac{1}{2}{b}_{31}-2{b}_{21}\\ {b}_{24}& {b}_{22}& {b}_{23}& 2{b}_{21}\\ \frac{1}{2}{b}_{34}-{b}_{24}& \frac{1}{2}{b}_{32}-{b}_{22}& \frac{1}{2}{b}_{33}-{b}_{23}& \frac{1}{2}{b}_{31}-2{b}_{21}\\ {b}_{44}-{b}_{24}& {b}_{42}-{b}_{22}& {b}_{43}-{b}_{23}& 2{b}_{41}-2{b}_{21}\end{array}\right]$
6. Replace column 4 by column 3:
${B}_{6}=\left[\begin{array}{cccc}{b}_{14}+\frac{1}{2}{b}_{34}-{b}_{24}& {b}_{12}+\frac{1}{2}{b}_{32}-{b}_{22}& {b}_{13}+\frac{1}{2}{b}_{33}-{b}_{23}& {b}_{13}+\frac{1}{2}{b}_{33}-{b}_{23}\\ {b}_{24}& {b}_{22}& {b}_{23}& {b}_{23}\\ \frac{1}{2}{b}_{34}-{b}_{24}& \frac{1}{2}{b}_{32}-{b}_{22}& \frac{1}{2}{b}_{33}-{b}_{23}& \frac{1}{2}{b}_{33}-{b}_{23}\\ {b}_{44}-{b}_{24}& {b}_{42}-{b}_{22}& {b}_{43}-{b}_{23}& {b}_{43}-{b}_{23}\end{array}\right]$
7. Delete column 1 (column dimension is reduced by 1):
${B}_{7}=\left[\begin{array}{ccc}{b}_{12}+\frac{1}{2}{b}_{32}-{b}_{22}& {b}_{13}+\frac{1}{2}{b}_{33}-{b}_{23}& {b}_{13}+\frac{1}{2}{b}_{33}-{b}_{23}\\ {b}_{22}& {b}_{23}& {b}_{23}\\ \frac{1}{2}{b}_{32}-{b}_{22}& \frac{1}{2}{b}_{33}-{b}_{23}& \frac{1}{2}{b}_{33}-{b}_{23}\\ {b}_{42}-{b}_{22}& {b}_{43}-{b}_{23}& {b}_{43}-{b}_{23}\end{array}\right]$
Now, we can express the final matrix ${B}_{7}$ as a product of eight matrices:
${B}_{7}=B·{M}_{1}·{M}_{2}·{M}_{3}·{M}_{4}·{M}_{5}·{M}_{6}·{M}_{7}$
where each ${M}_{i}$ 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 ${B}_{7}$ as a product of three matrices: A, B, and C.
${B}_{7}=A·B·C$
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).
$A={M}_{1}·{M}_{2}·{M}_{3}·{M}_{4}$
$B={M}_{5}·{M}_{6}$
$C={M}_{7}$
Thus, the final matrix ${B}_{7}$ can be written as a product of ABC:
${B}_{7}=A·B·C$

Don Sumner

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:
${C}_{1}=\left[\begin{array}{cccc}2& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$
2. Halve row 3:
${C}_{2}=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& \frac{1}{2}& 0\\ 0& 0& 0& 1\end{array}\right]$
3. Add row 3 to row 1:
${C}_{3}=\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 0\\ 0& 0& \frac{1}{2}& 0\\ 0& 0& 0& 1\end{array}\right]$
4. Interchange columns 1 and 4:
${C}_{4}=\left[\begin{array}{cccc}0& 0& 1& 2\\ 0& 1& 0& 0\\ 0& 0& \frac{1}{2}& 0\\ 0& 0& 0& 1\end{array}\right]$
5. Subtract row 2 from each of the other rows:
${C}_{5}=\left[\begin{array}{cccc}0& 0& 1& 2\\ 0& 1& 0& 0\\ 0& -1& \frac{1}{2}& 0\\ 0& 0& 0& 1\end{array}\right]$
6. Replace column 4 by column 3:
${C}_{6}=\left[\begin{array}{cccc}0& 0& 1& 1\\ 0& 1& 0& 0\\ 0& -1& \frac{1}{2}& \frac{1}{2}\\ 0& 0& 0& 1\end{array}\right]$
7. Delete column 1 (column dimension is reduced by 1):
${C}_{7}=\left[\begin{array}{ccc}0& 1& 1\\ 0& 1& \frac{1}{2}\\ 0& 0& 1\\ 0& 0& 1\end{array}\right]$
Finally, the resulting matrix after applying all the operations can be obtained by multiplying all the matrices together:
${B}^{\prime }={C}_{7}{C}_{6}{C}_{5}{C}_{4}{C}_{3}{C}_{2}{C}_{1}B$
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 $C$.
$A={C}_{7}{C}_{6}{C}_{5}\phantom{\rule{1em}{0ex}};\phantom{\rule{1em}{0ex}}B={C}_{4}{C}_{3}{C}_{2}\phantom{\rule{1em}{0ex}};\phantom{\rule{1em}{0ex}}C={C}_{1}$
The resulting matrix can be expressed as a product of these three matrices applied to the original matrix:
${B}^{\prime }=ABCB$

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