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

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

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

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

$$Rank(A)=2$$

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

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