Question

Consider the matrices

$\mathbf{A}\mathbf{=}\left[\begin{array}{cccccc}1& 0& 2& 0& 4& 0\\ 0& 1& 3& 0& 5& 0\\ 0& 0& 0& 1& 6& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right]\mathbf{}\mathbf{and}\mathbf{}\mathbf{B}\mathbf{=}\left[\begin{array}{cccccc}1& 0& 2& 0& 0& 4\\ 0& 1& 3& 0& 0& 5\\ 0& 0& 0& 1& 0& 6\\ 0& 0& 0& 0& 1& 7\end{array}\right]$

Show that the kernels of the matrices A and B are different.

Short answer

$$\begin{gathered} ker(A)=span\left\{\left[\begin{array}{c}-2\\ -3\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ -5\\ 0\\ -6\\ 1\\ 0\end{array}\right]\right\}\mathrm{and}\ker \left(\mathrm{B}\right)=\mathrm{span}\left\{\left[\begin{array}{c}-2\\ -3\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ -5\\ 0\\ -6\\ 1\\ 0\end{array}\right]\right\} \\ \Rightarrow \ker (\mathrm{A})\ne \ker (\mathrm{B}) \end{gathered}$$

Step-by-step solution

 Finding the ker(A)

Consider $A\overrightarrow{x}=0$

$\Rightarrow \left[\begin{array}{cccccc}1& 0& 2& 0& 4& 0\\ 0& 1& 3& 0& 5& 0\\ 0& 0& 0& 1& 6& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\\ x_6\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right]$

$$\begin{gathered} \Rightarrow x_1+2x_3+4x_5=0 \\ {x}_2+3x_3+5x_5=0 \\ {x}_4+6x_5=0 \\ {x}_6=0 \end{gathered}$$

Since the pivot elements are in first, second, fourth, and sixth columns, therefore$x_3\mathrm{and}{x}_5$ are free variables.

Let$x_3=k_{1}and{x}_5=k_2$

$$\begin{gathered} \Rightarrow x_1=-2k_1-4k_2 \\ {x}_2=-3k_1-5k_2 \\ {x}_4=-6k_2 \\ {x}_6=0 \end{gathered}$$

$\Rightarrow \left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\\ x_6\end{array}\right]=k_1\left[\begin{array}{c}-2\\ -3\\ 1\\ 0\\ 0\\ 0\end{array}\right]+k_2\left[\begin{array}{c}-4\\ -5\\ 0\\ -6\\ 1\\ 0\end{array}\right]$

$\Rightarrow ker(A)=span\left\{\left[\begin{array}{c}-2\\ -3\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ -5\\ 0\\ -6\\ 1\\ 0\end{array}\right]\right\}$

 Finding the ker(B)

Let $Bx⃗=0$

$\Rightarrow \left[\begin{array}{cccccc}1& 0& 2& 0& 0& 4\\ 0& 1& 3& 0& 0& 5\\ 0& & & 1& 0& 6\\ 0& 0& 0& 0& 1& 7\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\\ x_6\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right]$

$$\begin{gathered} \Rightarrow x_1+2x_3+4x_6=0 \\ {x}_2+3x_3+5x_6=0 \\ {x}_4+6x_6=0 \\ {x}_5+7x_6=0 \end{gathered}$$

Since the pivot elements are in first, second, fourth, and fifth columns, therefore $x_3\mathrm{and}{x}_6$are free variables.

Let $x_3=k_1\mathrm{and}{x}_6=k_2$

$$\begin{gathered} \Rightarrow x_1=-2k_1-4k_2 \\ {x}_2=-3k_1-5k_2 \\ {x}_4=-6k_2 \\ {x}_5=-7k_2 \end{gathered}$$

role="math" localid="1659371280384" $$\begin{gathered} \Rightarrow \left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\\ x_6\end{array}\right]=k_1\left[\begin{array}{c}-2\\ -3\\ 1\\ 0\\ 0\\ 0\end{array}\right]+k_2\left[\begin{array}{c}-4\\ -5\\ 0\\ -6\\ -7\\ 1\end{array}\right] \\ \Rightarrow ker(B)=span\left\{\left[\begin{array}{c}-2\\ -3\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ -5\\ 0\\ -6\\ -7\\ 1\end{array}\right]\right\} \end{gathered}$$

 Final Answer

Since, for the matrix A and B,

$ker(A)=span\left\{\left[\begin{array}{c}-2\\ -3\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ -5\\ 0\\ -6\\ 1\\ 0\end{array}\right]\right\}$ and$ker(B)=span\left\{\left[\begin{array}{c}-2\\ -3\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ -5\\ 0\\ -6\\ -7\\ 1\end{array}\right]\right\}$

$\Rightarrow ker(A)\ne ker(B)$