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Question: Find a basis for the set of all vectors of the form

\(\left[ {\begin{array}{*{20}{c}}{a - {\bf{2}}b + {\bf{5}}c}\\{{\bf{2}}a + {\bf{5}}b - {\bf{8}}c}\\{ - a - {\bf{4}}b + {\bf{7}}c}\\{{\bf{3}}a + b + c}\end{array}} \right]\).

Short Answer

Expert verified

The basis of the matrix is \(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\2\\{ - 1}\\3\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}{ - 2}\\5\\{ - 4}\\1\end{array}} \right]} \right\}\).

Step by step solution

01

Step 1:Form the matrix using the vectors

The matrix using the set of vectors can be written in the following manner:

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&5\\2&5&{ - 8}\\{ - 1}&{ - 4}&7\\3&1&1\end{array}} \right]\)

02

Row reduce the matrix

Use the row operationson the above matrix as shown below:

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}1&{ - 2}&5\\2&5&{ - 8}\\{ - 1}&{ - 4}&7\\3&1&1\end{array}} \right] = \,\,\left[ {\begin{array}{*{20}{c}}1&{ - 2}&5\\0&9&{ - 18}\\0&{ - 6}&{12}\\0&7&{ - 14}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}{R_2} \to {R_2} - 2{R_1}\\{R_3} \to {R_3} + {R_1}\\{R_4} \to {R_4} - 3{R_1}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&5\\0&1&{ - 2}\\0&{ - 6}&{12}\\0&7&{ - 14}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_2} \to \frac{1}{9}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&0&1\\0&1&{ - 2}\\0&0&0\\0&0&0\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}{R_1} \to {R_1} + 2{R_2}\\{R_3} \to {R_3} + 6{R_2}\end{array} \right\}\end{array}\)

From the row reduced form, it can be observed that column 1 and column 2 are the pivot columns.

So, the basis of the matrix is \(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\2\\{ - 1}\\3\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}{ - 2}\\5\\{ - 4}\\1\end{array}} \right]} \right\}\).

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Most popular questions from this chapter

Consider the following two systems of equations:

\(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 1\\4{x_1} + {x_2} - 6{x_3} = 9\end{array}\) \(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 5\\4{x_1} + {x_2} - 6{x_3} = 45\end{array}\)

It can be shown that the first system of a solution. Use this fact and the theory from this section to explain why the second system must also have a solution. (Make no row operations.)

(M) Determine whether w is in the column space of \(A\), the null space of \(A\), or both, where

\({\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}1\\2\\1\\0\end{array}} \right),A = \left( {\begin{array}{*{20}{c}}{ - 8}&5&{ - 2}&0\\{ - 5}&2&1&{ - 2}\\{10}&{ - 8}&6&{ - 3}\\3&{ - 2}&1&0\end{array}} \right)\)

If a\({\bf{6}} \times {\bf{3}}\)matrix A has a rank 3, find dim Nul A, dim Row A, and rank\({A^T}\).

Define a linear transformation by \(T\left( {\mathop{\rm p}\nolimits} \right) = \left( {\begin{array}{*{20}{c}}{{\mathop{\rm p}\nolimits} \left( 0 \right)}\\{{\mathop{\rm p}\nolimits} \left( 0 \right)}\end{array}} \right)\). Find \(T:{{\mathop{\rm P}\nolimits} _2} \to {\mathbb{R}^2}\)polynomials \({{\mathop{\rm p}\nolimits} _1}\) and \({{\mathop{\rm p}\nolimits} _2}\) in \({{\mathop{\rm P}\nolimits} _2}\) that span the kernel of T, and describe the range of T.

Suppose a nonhomogeneous system of six linear equations in eight unknowns has a solution, with two free variables. Is it possible to change some constants on the equations’ right sides to make the new system inconsistent? Explain.

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