How to Quickly Find the Rank of a 4 x 6 Matrix Using Column Operations

[A.Magnus]

Is there any shortcut to find the rank of this $4 \times 6$ matrix quickly?

$$A =
\begin{pmatrix}
-3 &2 &-1 &-2 &7 &-1\\
9 &2 &27 &18 &7 &-9\\
3 &2 &1 &0 &7 &-1\\
6 &2 &8 &4 &-7 &-4\\
\end{pmatrix}$$

The above is a sample question for semester final test. If it were a homework, of course I don't mind using the lengthy row operations to reduce $A$ to row echelon form, or using graphic or online calculator. But these resources are unavailable in an exam setting, and time is always an issue. Any help or hints would be very much appreciated; thank you in advance for all your gracious help. ~MA

 

[Opalg]

Is there any shortcut to find the rank of this $4 \times 6$ matrix quickly?

$$A =
\begin{pmatrix}
-3 &2 &-1 &-2 &7 &-1\\
9 &2 &27 &18 &7 &-9\\
3 &2 &1 &0 &7 &-1\\
6 &2 &8 &4 &-7 &-4\\
\end{pmatrix}$$

The above is a sample question for semester final test. If it were a homework, of course I don't mind using the lengthy row operations to reduce $A$ to row echelon form, or using graphic or online calculator. But these resources are unavailable in an exam setting, and time is always an issue. Any help or hints would be very much appreciated; thank you in advance for all your gracious help. ~MA

I think that the most painless way to do this is to use the fact that the rank is the dimension of the space spanned by the columns. Denoting this space by $C$, notice that $C$ contains columns 2 and 5 of $A$, and therefore it also contains their scalar multiples $\begin{bmatrix}1\\1\\1\\1\end{bmatrix}$ and $\begin{bmatrix}1\\1\\1\\-1\end{bmatrix}$. Subtract one of these from the other and then take a scalar multiple, to see that $C$ contains the vector $e_4 = \begin{bmatrix}0\\0\\0\\1\end{bmatrix}$ from the standard basis of $\mathbb{R}^4.$

Now add twice column 6 of $A$ to column 2, add a multiple of $e_4$, and then take a scalar multiple, to see that $C$ contains $e_2 = \begin{bmatrix}0\\1\\0\\0\end{bmatrix}.$ Continue in that way, to see that $C$ contains all the vectors in the standard basis, and is therefore the whole of $\mathbb{R}^4.$

Conclusion: $A$ has rank $4$.

 

[A.Magnus]

I think that the most painless way to do this is to use the fact that the rank is the dimension of the space spanned by the columns. Denoting this space by $C$, notice that $C$ contains columns 2 and 5 of $A$, and therefore it also contains their scalar multiples $\begin{bmatrix}1\\1\\1\\1\end{bmatrix}$ and $\begin{bmatrix}1\\1\\1\\-1\end{bmatrix}$. Subtract one of these from the other and then take a scalar multiple, to see that $C$ contains the vector $e_4 = \begin{bmatrix}0\\0\\0\\1\end{bmatrix}$ from the standard basis of $\mathbb{R}^4.$

Now add twice column 6 of $A$ to column 2, add a multiple of $e_4$, and then take a scalar multiple, to see that $C$ contains $e_2 = \begin{bmatrix}0\\1\\0\\0\end{bmatrix}.$ Continue in that way, to see that $C$ contains all the vectors in the standard basis, and is therefore the whole of $\mathbb{R}^4.$

Conclusion: $A$ has rank $4$.

Genius! Thanks for your gracious helping hand! ~MA