GO Classes DPP | GATE CS | Linear Algebra | System of Linear Equations

Answer 1

Let

$A=\begin{bmatrix}
1 & 2 \\
3 & 5 \\
2 & 3
\end{bmatrix}$

and we want to know for which vectors $b$ the system $Ax=b$ is consistent.

Write the columns of $A$ as

$c_1=\begin{bmatrix}1\\3\\2\end{bmatrix}$, $\quad c_2=\begin{bmatrix}2\\5\\3\end{bmatrix}$.

Then $Ax=b$ has a solution exactly when $b$ is a linear combination of $c_1$ and $c_2$, so $b$ must lie in the column space of $A$.

A quick way is to find a relation among the three entries of any vector in the column space.

Let $y=\begin{bmatrix}a\\b\\c\end{bmatrix}$ satisfy $y^T A=0$. Then

$a+3b+2c=0$

and

$2a+5b+3c=0$.

Solving these gives $a=-b$ and $c=-b$, so one such vector is

$y=\begin{bmatrix}-1\\1\\-1\end{bmatrix}$.

Therefore every vector $b$ in the column space must satisfy

$y^T b=0$,

that is,

$-b_1+b_2-b_3=0$,

or equivalently,

$b_2=b_1+b_3$.

Now check each option:

• A. $\begin{bmatrix}1\\2\\1\end{bmatrix}$: here $2=1+1$, so it works.
   
• B. $\begin{bmatrix}2\\4\\3\end{bmatrix}$: here $4\ne 2+3$, so it does not work.
   
• C. $\begin{bmatrix}3\\8\\5\end{bmatrix}$: here $8=3+5$, so it works.
   
• D. $\begin{bmatrix}1\\3\\2\end{bmatrix}$: here $3=1+2$, so it works.

So the system has a solution for A, C, and D.

Answer 2

For (Ax = b), a solution exists only when (b) is a linear combination of the columns of (A).
Vectors A, C, and D are linear combinations of the columns of the given matrix, so the answer is A, C, and D.

Comments

It's correct this solution is better understanding

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Answer 3

A,C,D