Science:Math Exam Resources/Courses/MATH307/December 2006/Question 02 (c)

MATH307 December 2006

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Question 02 (c)
Consider the matrix $A={\begin{bmatrix}1&2&0&3&4\\2&4&3&0&5\\1&2&-1&5&5\end{bmatrix}}$ (c) Find a basis for the left nullspace of A.

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Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. To find the basis of left nullspace, $N(A^{T})$, we solve for $\displaystyle A^{T}{\vec {x}}={\vec {0}}$ This can be done by finding the row echelon form of $A^{T}$ $A^{T}={\begin{bmatrix}1&2&1\\2&4&2\\0&3&-1\\3&0&5\\4&5&5\end{bmatrix}}$ Perform the following row operations: row 2 = row 2 - row 1 row 4 = row 4 $-3\times$ row 1 row 5 = row 5 $-4\times$ row 1 and we get the matrix : ${\begin{bmatrix}1&2&1\\0&0&0\\0&3&-1\\0&-6&2\\0&-3&1\end{bmatrix}}$ Further row operations, row 4 = row 4 $+2\times$ row3 row 5 = row 5 $+$ row 3 and interchanging row 2 and 3 gives us the row echelon form of the matrix as: ${\begin{bmatrix}1&2&1\\0&-3&1\\0&0&0\\0&0&0\\0&0&0\end{bmatrix}}$ From this we can see that only $x_{3}$ is a free variable. Solving this row echelon for homogeneous solution gives us : ${\begin{aligned}3x_{2}=x_{3}&\Rightarrow x_{2}={\frac {1}{3}}x_{3}\\x_{1}=-2x_{2}-x_{3}&\Rightarrow x_{1}=-{\frac {2}{3}}x_{3}-x_{3}=-{\frac {5}{3}}x_{3}\end{aligned}}$ ${\vec {x}}={\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}=t{\begin{bmatrix}-5\\1\\3\end{bmatrix}},t\in \mathbb {R}$ Thus a basis of $N(A^{T})$ is given by $\left\{{\begin{bmatrix}-5\\1\\3\end{bmatrix}}\right\}$