GO Classes DA 2025 | Weekly Quiz 6 | Change of Basis & Linear Transformation | Question: 9

Question

Suppose that $a \neq 0$ and $a \neq b$. Which equation below is the equation relating $a, b$ and $c$ so that the vectors

$$
\left[\begin{array}{l}
1 \\
1 \\
a
\end{array}\right], \quad\left[\begin{array}{l}
1 \\
2 \\
b
\end{array}\right], \quad\left[\begin{array}{c}
-1 \\
2 \\
c
\end{array}\right]
$$
form a basis for $\mathbb{R}^{3}$ ?

• A. $3 a-4 b+c \neq 0$
• B. $4 a-3 b+c \neq 0$
• C. $3 a-4 b-c \neq 0$
• D. $4 a-3 b-c \neq 0$

Answer 1

To make the basis of $R^3$, vectors have to be Linearly Independent

LI Matrix of 3x3 will have 3 pivot variables in row-echelon form

Let's make the above matrix to row-echelon form

---

$\begin{bmatrix} 1 & 1 & -1 \\ 1 & 2 & 2 \\ a & b & c \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & -1 \\ 1 & 2 & 2 \\ a-k.1 & b-k.1 & c+k.1 \end{bmatrix}$

$R^3 \rightarrow R^3 - k.R^1$

$a-k = 0 \Rightarrow a = k$

Continuing the row operation

$\begin{bmatrix} 1 & 1 & -1 \\ 1 & 2 & 2 \\ a-k.1 & b-k.1 & c+k.1 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & -1 \\ 0 & 1 & 3 \\ a-k.1 & b-k.1 & c+k.1 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & -1 \\ 0 & 1 & 3 \\ a-k.1 & b-k.1 - m.1 & c+k.1-m.3 \end{bmatrix}$

$R^2 \rightarrow R^2 - R^1 $
$R^3 \rightarrow R^3 - m.R^1$

So,
$b-k-m = 0 \Rightarrow b-a-m=0 \Rightarrow m=b-a$

To make the matrix have 3 pivots, 3rd row 3rd column element has to be a non-zero element

So, $c+k-3m \neq0$

$c+a-3(b-a) \neq0$

$c+a-3b+3a \neq0$

$c-3b+4a \neq0$
Option B is the correct option