Question

The number of linearly independent solutions of the system of equations

$\left[\begin{array}{ccc}1& 0& 2\\ 1& -1& 0\\ 2& -2& 0\end{array}\right]\left[\begin{array}{c}{X}_1\\ X_2\\ X_3\end{array}\right]=0$ is equal to

• A. 1
• B. 2
• C. 3
• D. 0

Answer 1

The correct option is A 1

$\left[\begin{array}{ccc}1& 0& 2\\ 1& -1& 0\\ 2& -2& 0\end{array}\right]\left[\begin{array}{c}{X}_1\\ X_2\\ X_3\end{array}\right]=0$ ... (1)

or AX = 0

Where Matrix A = $\left[\begin{array}{ccc}1& 0& 2\\ 1& -1& 0\\ 2& -2& 0\end{array}\right]$

$R_2\to R_2-R_1$

$R_3\to R_3-2R_1$

A $\sim$ $\left[\begin{array}{ccc}1& 0& 2\\ 0& -1& -2\\ 0& -2& -4\end{array}\right]$

$R_3\to R_3-2R_2$

A $\sim$ $\left[\begin{array}{ccc}1& 0& 2\\ 0& -1& -2\\ 0& 0& 0\end{array}\right]$

Rank of matrix A = 2. So r = 2
Number of columns in A is 3. So n = 3
So number of linearly independent solutions = n - r = 3-2 =1