Science:Math Exam Resources/Courses/MATH221/December 2008/Question 01 (b)

MATH221 December 2008

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Question 01 (b)
Let $A$ denote the matrix $A={\begin{bmatrix}3&0&-2\\1&-2&3\\4&-2&1\end{bmatrix}}$ Determine whether the vector $b={\begin{pmatrix}1\\1\\1\end{pmatrix}}$ is in the column space of the matrix $A$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1

For $b$ to be in the column space of $A$, it must be a linear combination of the basis vectors. From part (a) we know that a basis for the column space of $A$ consists of the first two column vectors of $A$. Hence, check if there are constants $\alpha _{1}$, $\alpha _{2}$ such that ${\begin{pmatrix}1\\1\\1\end{pmatrix}}=\alpha _{1}{\begin{pmatrix}3\\1\\4\end{pmatrix}}+\alpha _{2}{\begin{pmatrix}0\\-2\\-2\end{pmatrix}}$

Hint 2

Rewriting the equation in the hint above: If $b$ was in the column space of $A$, how many solutions are there to the equation $0=\beta _{1}{\begin{pmatrix}3\\1\\4\end{pmatrix}}+\beta _{2}{\begin{pmatrix}0\\-2\\-2\end{pmatrix}}+\beta _{3}{\begin{pmatrix}1\\1\\1\end{pmatrix}}$ If $\beta _{3}=0$ we know from part (a) that there is only the trivial solution. Otherwise, divide this equation by $\beta _{3}$ and set $\alpha _{1}=\beta _{1}/\beta _{3}$, $\alpha _{2}=\beta _{2}/\beta _{3}$ to get back to the equation on the first hint.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

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 determine if b is in the column space of A, we have to make sure that b is a linear combination of the column space of A $A{\vec {x}}=b$ If there is a solution for ${\vec {x}}$, then we can write b as a linear combination of the columns of A We can do this by row reduce the matrix [A|b] ${\begin{bmatrix}3&0&1\\1&-2&1\\4&-2&1\end{bmatrix}}{\xrightarrow[{R2\rightarrow R1-3R2}]{R3\rightarrow R3-4R2}}{\begin{bmatrix}1&0&{\frac {1}{3}}\\0&6&-2\\0&6&-3\end{bmatrix}}{\xrightarrow[{R3\rightarrow R2-R3}]{R2\rightarrow R2/6}}{\begin{bmatrix}1&0&{\frac {1}{3}}\\0&1&{\frac {-1}{3}}\\0&0&1\end{bmatrix}}$ The last row of the reduced row form shows that there is no solution for this augmented matrix. Therefore, vector b = ${\begin{pmatrix}1\\1\\1\end{pmatrix}}$ is not in the column space of matrix A = ${\begin{pmatrix}3&0&-2\\1&-2&3\\4&-2&1\end{pmatrix}}$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The fastest way is to check if the matrix, whose first two columns are the basis of the column space of $A$ and the third column is the vector $b$, has determinante $0$. Thereby we check if these three vectors are linear dependent. If yes, then $b$ lies in the column space of $A$. ${\text{det}}\left({\begin{pmatrix}3&0&1\\1&-2&1\\4&-2&1\end{pmatrix}}\right)=3*(-2)*1+0*1*4+1*1*(-2)-4*(-2)*1-(-2)*1*3+1*1*0=6\neq 0$ Hence, the vectors are linear independent and $b$ does not lies in the column space of $A$.