Linear Algebra Mock Question

Question

Answer 1

Option A,

let eqn 1 be: -m1x+y=b1

let eqn 2 be: -m2x+y=b2

eqn1 - eqn2 => (m2-m1)x = b1-b2

now, if there is only 1 unique solution, that means,

in c1V1=0, there is no ci that is not equal to zero, therefore c1=0 => m2-m1 = 0 => m2==m1
and b1-b2 =0, => b1==b2.

I know my answer is not very clear, but let me know what you didnt understand

 

Comments

Wrong Solution

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could you kindly share the correct solution

Answer 2

c

Comments

how ?

Answer 3

Method 1:

The system has exactly one solution that means

$rank(A) = rank(A:b) = n$

if rank(A) = 0 then there is either zero solution or no solution

$-m_1 + m_2 = 0 \Rightarrow m_1 = m_2$

so for unique solution $m_1 \neq m_2$ 

 

method 2:

Exactly one solution means number of linearly independent vector equal to number of vectors(given square matrix)

$\begin{bmatrix} -m_1 & 1 \\ -m_2 & 1\end{bmatrix}$

 

if $m_1 = m_2 $ 

then there exists some k such that vector $\begin{bmatrix} m_1 \\ m_2 \end{bmatrix}$ is a linear combination of vector $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ i.e. $\begin{bmatrix} m_1 \\ m_2 \end{bmatrix} = k \begin{bmatrix} 1 \\ 1 \end{bmatrix}$

Comments

@Himanshu P Dev