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In Exercise 1-10, assume that\(T\)is a linear transformation. Find the standard matrix of\(T\).

\(T:{\mathbb{R}^3} \to {\mathbb{R}^2}\), \(T\left( {{e_1}} \right) = \left( {1,3} \right)\) and \(T\left( {{e_2}} \right) = \left( {4, - 7} \right) - \) and \(T\left( {{e_3}} \right) = \left( { - 5,4} \right)\)where \({e_1}\), \({e_2}\), and \({e_3}\).

Short Answer

Expert verified

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

Step by step solution

01

Find the value of \(T\) using linear transformation

Using linear transformation,

\(\begin{aligned}{c}T &= T\left( {{x_1}{e_1} + {x_2}{e_2} + {x_3}{e_3}} \right)\\ &= {x_1}T\left( {{e_1}} \right) + {x_2}T\left( {{e_2}} \right) + {x_3}T\left( {{e_3}} \right)\\ &= \left[ {\begin{array}{*{20}{c}}{T\left( {{e_1}} \right)}&{T\left( {{e_2}} \right)}&{T\left( {{e_3}} \right)}\end{array}} \right]x\end{aligned}\)

02

Substitute values of \(T\left( {{e_1}} \right)\), \(T\left( {{e_2}} \right)\), and \(T\left( {{e_3}} \right)\)

By the equation \(T = \left[ {\begin{array}{*{20}{c}}{T\left( {{e_1}} \right)}&{T\left( {{e_2}} \right)}&{T\left( {{e_3}} \right)}\end{array}} \right]x\),

\(T = \left[ {\begin{array}{*{20}{c}}1&4&{ - 5}\\3&{ - 7}&4\end{array}} \right]x\).

03

Find the standard matrix \(T\) for linear transformation

In the equation \(T = Ax\), the matrix \(A\) is the matrix for linear transformation \(T\).

By the equation \(T = \left[ {\begin{array}{*{20}{c}}1&4&{ - 5}\\3&{ - 7}&4\end{array}} \right]x\), \(A = \left[ {\begin{array}{*{20}{c}}1&4&{ - 5}\\3&{ - 7}&4\end{array}} \right]\).

So, the linear transformation matrix is \(\left[ {\begin{array}{*{20}{c}}1&4&{ - 5}\\3&{ - 7}&4\end{array}} \right]\).

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Most popular questions from this chapter

If Ais an \(n \times n\) matrix and the transformation \({\bf{x}}| \to A{\bf{x}}\) is one-to-one, what else can you say about this transformation? Justify your answer.

Suppose Tand Ssatisfy the invertibility equations (1) and (2), where T is a linear transformation. Show directly that Sis a linear transformation. (Hint: Given u, v in \({\mathbb{R}^n}\), let \({\mathop{\rm x}\nolimits} = S\left( {\mathop{\rm u}\nolimits} \right),{\mathop{\rm y}\nolimits} = S\left( {\mathop{\rm v}\nolimits} \right)\). Then \(T\left( {\mathop{\rm x}\nolimits} \right) = {\mathop{\rm u}\nolimits} \), \(T\left( {\mathop{\rm y}\nolimits} \right) = {\mathop{\rm v}\nolimits} \). Why? Apply Sto both sides of the equation \(T\left( {\mathop{\rm x}\nolimits} \right) + T\left( {\mathop{\rm y}\nolimits} \right) = T\left( {{\mathop{\rm x}\nolimits} + y} \right)\). Also, consider \(T\left( {cx} \right) = cT\left( x \right)\).)

Describe the possible echelon forms of the matrix A. Use the notation of Example 1 in Section 1.2.

a. A is a \({\bf{2}} \times {\bf{3}}\) matrix whose columns span \({\mathbb{R}^{\bf{2}}}\).

b. A is a \({\bf{3}} \times {\bf{3}}\) matrix whose columns span \({\mathbb{R}^{\bf{3}}}\).

In Exercise 23 and 24, make each statement True or False. Justify each answer.

23.

a. Another notation for the vector \(\left[ {\begin{array}{*{20}{c}}{ - 4}\\3\end{array}} \right]\) is \(\left[ {\begin{array}{*{20}{c}}{ - 4}&3\end{array}} \right]\).

b. The points in the plane corresponding to \(\left[ {\begin{array}{*{20}{c}}{ - 2}\\5\end{array}} \right]\) and \(\left[ {\begin{array}{*{20}{c}}{ - 5}\\2\end{array}} \right]\) lie on a line through the origin.

c. An example of a linear combination of vectors \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\) is the vector \(\frac{1}{2}{{\mathop{\rm v}\nolimits} _1}\).

d. The solution set of the linear system whose augmented matrix is \(\left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}&b\end{array}} \right]\) is the same as the solution set of the equation\({{\mathop{\rm x}\nolimits} _1}{a_1} + {x_2}{a_2} + {x_3}{a_3} = b\).

e. The set Span \(\left\{ {u,v} \right\}\) is always visualized as a plane through the origin.

Find the general solutions of the systems whose augmented matrices are given as

14. \(\left[ {\begin{array}{*{20}{c}}1&2&{ - 5}&{ - 6}&0&{ - 5}\\0&1&{ - 6}&{ - 3}&0&2\\0&0&0&0&1&0\\0&0&0&0&0&0\end{array}} \right]\).

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