VIPSolutionsgive answer in 2 step with explanation at the end of each step and final answer at the end: 1 . 4 Suppose the matrix B = ( [ 2 , 3 , 1 , 0 ] , [ 1 , 2 , 0 , 1 ] , [ 0 , 0 , 0 , 1 ] , [ 1 , 0 , 1 , 0 ] ) is partitioned such that B = ( [ B 1 1 , B 1 2 ] , [ B 2 1 , B 2 2 ] ) i ) Write down the sub - matrices B 1 1 , B 1 2 , B 2 1 and B 2 2 ii ) Determine the transpose of each submatrix. ( 4 ) iii ) Express B T in terms of the submatrices.2 31 0 14 Suppose the matrix B= ( ie eres such that B = (Bs 2) 1010 ) Write down the sub-matrices By, Bz, B;and By, @ ii) Determine the transpose of each submatrix. @ ii) Express BT in terms of the submatrices. 3)
Question:
give answer in 2 step with explanation at the end of each step and final answer at the end:
1
.
4
Suppose the matrix
B
=
(
[
2
,
3
,
1
,
0
]
,
[
1
,
2
,
0
,
1
]
,
[
0
,
0
,
0
,
1
]
,
[
1
,
0
,
1
,
0
]
)
is partitioned such that
B
=
(
[
B
1
1
,
B
1
2
]
,
[
B
2
1
,
B
2
2
]
)
i
)
Write down the sub
-
matrices
B
1
1
,
B
1
2
,
B
2
1
and
B
2
2
ii
)
Determine the transpose of each submatrix.
(
4
)
iii
)
Express
B
T
in terms of the submatrices.
2 31 0
14 Suppose the matrix B= ( ie eres such that B = (Bs 2)
1010
) Write down the sub-matrices By, Bz, B;and By, @
ii) Determine the transpose of each submatrix. @
ii) Express BT in terms of the submatrices. 3)
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Asked by: Test User 100237
Created at: 2025-05-18 06:09:32
Asked by: Test User 100237
Created at: 2025-05-18 06:09:32
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Let's solve the problem step by step as requested:
---
### **Step 1: Write down the sub-matrices \( B_{11}, B_{12}, B_{21} \), and \( B_{22} \)**
Given matrix:
\[
B = \begin{pmatrix}
2 & 3 & 1 & 0 \\
1 & 2 & 0 & 1 \\
0 & 0 & 0 & 1 \\
1 & 0 & 1 & 0
\end{pmatrix}
\]
Partitioning \( B \) into four 2x2 sub-matrices:
\[
B = \begin{pmatrix}
B_{11} & B_{12} \\
B_{21} & B_{22}
\end{pmatrix}
\]
where each \( B_{ij} \) is a 2x2 matrix.
So:
- \( B_{11} \) is the top left 2x2:
\(\begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}\)
- \( B_{12} \) is the top right 2x2:
\(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
- \( B_{21} \) is the bottom left 2x2:
\(\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}\)
- \( B_{22} \) is the bottom right 2x2:
\(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\)
**Explanation:**
We split the 4x4 matrix into four 2x2 blocks, breaking at row 2 and column 2.
---
### **Step 2: Determine the transpose of each submatrix**
- \( B_{11}^T = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \)
- \( B_{12}^T = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \) (same as original since it's diagonal)
- \( B_{21}^T = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \)
- \( B_{22}^T = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \) (same as original)
**Explanation:**
To find the transpose, switch the rows and columns for each submatrix.
---
### **Step 3: Express \( B^T \) in terms of the submatrices**
The transpose of a partitioned matrix is:
\[
B^T = \begin{pmatrix}
B_{11}^T & B_{21}^T \\
B_{12}^T & B_{22}^T
\end{pmatrix}
\]
**Explanation:**
When transposing a block matrix, the off-diagonal blocks swap positions and each block is transposed.
---
### **Final Answers:**
#### **i) Submatrices**
\[
B_{11} = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}, \quad
B_{12} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\]
\[
B_{21} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \quad
B_{22} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
\]
#### **ii) Transposes**
\[
B_{11}^T = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}, \quad
B_{12}^T = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\]
\[
B_{21}^T = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad
B_{22}^T = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
\]
#### **iii) Expression for \( B^T \)**
\[
B^T = \begin{pmatrix}
B_{11}^T & B_{21}^T \\
B_{12}^T & B_{22}^T
\end{pmatrix}
\]
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