• عربي

Need Help?

Subscribe to Linear Algebra

###### \${selected_topic_name}
• Notes

Determine whether $b$ is in $col(A),$ whether $w$ is in row(A) and whether $X$ is in null(A):

$A=\left(\begin{array}{ccc}{1} & {0} & {-1} \\ {1} & {1} & {1}\end{array}\right) \quad b=\left(\begin{array}{l}{3} \\ {2}\end{array}\right) \quad w=\left(\begin{array}{ccc}{-1} & {1} & {1}\end{array}\right) \quad X=\left(\begin{array}{c}{-1} \\ {3} \\ {-1}\end{array}\right)$

$\left(\begin{array}{ccc|c}{1} & {0} & {-1} & {3} \\ {1} & {1} & {1} & {2}\end{array}\right) R_{2} \rightarrow R_{2}-R_{1}$

$\left(\begin{array}{ccc|c}{1} & {0} & {-1} & {3} \\ {0} & {1} & {2} & {-1}\end{array}\right)$

$\Rightarrow$ the system has Infinite No. of sol.

$\left(\begin{array}{l}{3} \\ {2}\end{array}\right) \in span\left(\left(\begin{array}{l}{1} \\ {1}\end{array}\right),\left(\begin{array}{l}{0} \\ {1}\end{array}\right),\left(\begin{array}{c}{-1} \\ {1}\end{array}\right)\right)$

$\left(\begin{array}{l}{3} \\ {2}\end{array}\right) \in col(A)$

$\left[A^{T} | w^{T}\right]=\left(\begin{array}{cc|c}{1} & {1} & {-1} \\ {0} & {1} & {1} \\ {-1} & {1} & {1}\end{array}\right) R_{3}\rightarrow R_{3}+R_{1}$

$\left(\begin{array}{cc|c}{1} & {1} & {-1} \\ {0} & {1} & {1} \\ {0} & {2} & {0}\end{array}\right) {R_{3}} \rightarrow R_{3} - 2 R_{2}$

$\left(\begin{array}{ll|l}{1} & {1} & {-1} \\ {0} & {1} & {1} \\ {0} & {0} & {-2}\end{array}\right)$

$0=-2$

The system has no sol.

$\left(\begin{array}{c}{-1} \\ {1} \\ {1}\end{array}\right) \notin span \left(\left(\begin{array}{c}{1} \\ {0} \\ {-1}\end{array}\right)\left(\begin{array}{l}{1} \\ {1} \\ {1}\end{array}\right)\right)$

$w^{T} \notin col \left(A^{T}\right) \Rightarrow w \notin Row(A)$

$A X=0$

$\left(\begin{array}{ccc}{1} & {0} & {-1} \\ {1} & {1} & {1}\end{array}\right)\left(\begin{array}{c}{-1} \\ {3} \\ {-1}\end{array}\right)=\left(\begin{array}{c}{0} \\ {1}\end{array}\right) \neq\left(\begin{array}{l}{0} \\ {0}\end{array}\right) \Rightarrow \left(\begin{array}{c}{0} \\ {1}\end{array}\right) \notin nul(A)$

$x \notin null(A)$

Find basis for $\text {col}(A)$ and null(A) of the given matrices and then compute the rank and
the nullity for the given matrices:

$A=\left(\begin{array}{ccc}{1} & {0} & {-1} \\ {1} & {1} & {1}\end{array}\right)$

Rank(A)+nallity(A)=No. of col.

= No. of var.

$[A | 0]\left(\begin{array}{ccc|c}{1} & {0} & {-1} & {0} \\ {1} & {1} & {1} & {0}\end{array}\right) R_{2} \rightarrow R_{2} - R_{1}$

$\left(\begin{array}{ccc|c}{1} & {0} & {-1} & {0} \\ {0} & {1} & {2} & {0}\end{array}\right)$

$Col(A)=span \left(\left(\begin{array}{l}{1} \\ {1}\end{array}\right)\left(\begin{array}{l}{0} \\ {1}\end{array}\right)\right)$

$\Rightarrow\text {Basis of col}(A)=\left\{\left(\begin{array}{ll}{1} \\ {1}\end{array}\right),\left(\begin{array}{l}{0} \\ {1}\end{array}\right)\right\}$

$AX=0$

Let $c_{3}=t, c_{1}=t, c_{2}=-2 t$

$\left(\begin{array}{l}{c_{1}} \\ {c_{2}} \\ {c_{3}}\end{array}\right)=\left(\begin{array}{c}{t} \\ {-2 t} \\ {t}\end{array}\right)=\left(\begin{array}{c}{1} \\ {-2} \\ {1}\end{array}\right) t$

$\text {Null}(A)=\text {span}\left(\left(\begin{array}{l} {1} \\ {-2} \\ {1} \end{array}\right)\right)$

$\text {Basis of null} (A)=\left\{\left(\begin{array}{c}{1} \\ {-2} \\ {1}\end{array}\right)\right\}$

$Rank(A)=2$

$Rank(A)+nullity(A)=\text {No. of col.}$

$2+nullity(A)=3$

$nullity(A)=1$

$A=\left(\begin{array}{ccccc}{2} & {-4} & {0} & {2} & {1} \\ {-1} & {2} & {1} & {2} & {3} \\ {1} & {-2} & {1} & {4} & {4}\end{array}\right)$

$\Rightarrow A x=0$

$\left(\begin{array}{rrrrr|r}{2} & {-4} & {0} & {2} & {1} & {0} \\ {-1} & {2} & {1} & {2} & {3}& {0} \\ {1} & {-2} & {1} & {4} & {4} &{0}\end{array}\right) \Rightarrow\left(\begin{array}{rrrrr|r}{1} & {-2} & {0} & {1} & {0.5} & {0} \\ {0} & {0} & {1} & {3} & {3.5} & {0} \\ {0} & {0} & {0} & {0} & {0} & {0}\end{array}\right)$

Basis of col.(A) $=\left\{c_{1}, c_{3}\right\}=\left\{\left(\begin{array}{c}{2} \\ {-1} \\ {1}\end{array}\right),\left(\begin{array}{l}{0} \\ {1} \\ {1}\end{array}\right)\right\}$

Let $c_{2}=r$
$c_{4}=s$
$c_{5}=t$

$c_{1}-2 r+s+0.5 t=0$
$c_{3}+3 s+3.5 t=0$

$c_{1}=2 r-s-0.5 t$
$c_{3}=-3 s-3.5 t$

$\left[\begin{array}{l}{c_{1}} \\ {c_{2}} \\ {c_{3}} \\ {c_{4}} \\ {c_{5}}\end{array}\right]=\left[\begin{array}{c}{2} \\ {1} \\ {0} \\ {0} \\ {0}\end{array}\right] r+\left[\begin{array}{c}{-1} \\ {0} \\ {-3} \\ {1} \\ {0}\end{array}\right] s+\left[\begin{array}{c}{0.5} \\ {0} \\ {-3.5} \\ {0} \\ {1}\end{array}\right] t$

$Rank(A)=2$

$\text {nullity}(A)=3$