Spaces:
Sleeping
Sleeping
SrijitMukherjee
commited on
Commit
•
3b6b630
1
Parent(s):
12964a1
Upload your_file.csv
Browse files- your_file.csv +38 -0
your_file.csv
ADDED
@@ -0,0 +1,38 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
exam name,year,problem number,problem,solution,correct option
|
2 |
+
IIT JAM MS,2023,Problem 2,"Let $M$ = $M_1M_2$ , where $M_1$ and $M_2$ are two 3 × 3 distinct matrices. Consider the following two statements:
|
3 |
+
(I) The rows of $M$ are linear combinations of rows of $M_2$.
|
4 |
+
(II) The columns of $M$ are linear combinations of columns of $M_1$.
|
5 |
+
|
6 |
+
Then,
|
7 |
+
|
8 |
+
(A) only (I) is TRUE
|
9 |
+
|
10 |
+
(B) only (II) is TRUE
|
11 |
+
|
12 |
+
(C) both (I) and (II) are TRUE
|
13 |
+
|
14 |
+
(D) neither (I) nor (II) is TRUE","Let $M_1$= $\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$ and $M_2$ = $\begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{bmatrix}$
|
15 |
+
|
16 |
+
We can represent the matrix $M_1$ using it’s column vectors and $M_2$ using it’s row vectors as shown:
|
17 |
+
|
18 |
+
Let $M_1$ = $\begin{bmatrix} \vec{a_{1}}& \vec{a_{2}} &\vec{a_{3}}\\ \end{bmatrix}$ and $M_2$ = $\begin{bmatrix} \vec{b_{1}}\\ \vec{b_{2}}\\ \vec{b_{3}}\\ \end{bmatrix}$, where $\vec{a_i}$ and $\vec{b_i}$ is in $R^3$
|
19 |
+
|
20 |
+
The matrix multiplication $M_1M_2$ yields a 3x3 matrix $M$.
|
21 |
+
|
22 |
+
This matrix can be written in two different ways.
|
23 |
+
|
24 |
+
### Column View of $M$
|
25 |
+
|
26 |
+
$M$ = $\begin{bmatrix} \sum_{i=1}^{3} b_{i1}\vec{a_{i}} & \sum_{i=1}^{3} b_{i2}\vec{a_{i}} & \sum_{i=1}^{3} b_{i3}\vec{a_{i}}\\ \end{bmatrix}$. This shows that the each column of $M$ is a linear combinations of the columns of $M_1$.
|
27 |
+
|
28 |
+
### Row View of $M$
|
29 |
+
|
30 |
+
$M$ = $\begin{bmatrix} \sum_{i=1}^{3} a_{1i}\vec{b_{i}} \\ \sum_{i=1}^{3} b_{2i}\vec{b_{i}} \\ \sum_{i=1}^{3} b_{3i}\vec{b_{i}}\\ \end{bmatrix}$.
|
31 |
+
|
32 |
+
This shows that the each row of $M$ is a linear combination of the rows of $M_2$.
|
33 |
+
|
34 |
+
You can algebraically verify this using the below matrix multiplication formula.
|
35 |
+
|
36 |
+
![Untitled](https://prod-files-secure.s3.us-west-2.amazonaws.com/aafb3d2f-7f27-4753-a17f-8ceef5abd2ad/a8943351-cd19-44c7-a996-c6438c21eb3f/Untitled.png)
|
37 |
+
|
38 |
+
However, you can extend the proof similarly to show that this is true for any $n \times n$ matrix.",C
|