Question

The standard ordered basis of R³ is {e₁, e₂, e₃} Let T : R³ → R³ be the linear transformation such that T(e₁) = 7e₁ - 5e₃, T (e₂) = -2e₂ + 9e₃, T(e₃) = e₁ + e₂ + e₃. The standard matrix of T is:
1. \(\left( {\begin{array}{*{20}{c}} 7&0&1\\ 0&{ - 2}&1\\ { - 5}&9&1 \end{array}} \right)\)
2. \(\left( {\begin{array}{*{20}{c}} 7&-2&1\\ -5&{ 9}&1\\ { 0}&0&1 \end{array}} \right)\)
3. \(\left( {\begin{array}{*{20}{c}} 7&0&-5\\ 0&{ - 2}&9\\ 1&1&1 \end{array}} \right)\)
4. \(\left( {\begin{array}{*{20}{c}} 7&-5&0\\ -2&{ 9}&1\\ { 1}&1&1 \end{array}} \right)\)

Answer 1

Correct Answer - Option 1 : \(\left( {\begin{array}{*{20}{c}} 7&0&1\\ 0&{ - 2}&1\\ { - 5}&9&1 \end{array}} \right)\)

Concept:

Matrix transformations:

Theorem: Suppose L: Rⁿ → R^m is a linear map. Then there exists an m×n matrix A such that L(x) = Ax for all x ∈ Rⁿ. Columns of A are vectors L(e₁), L(e₂), . . . , L(e_n), where e₁, e₂, . . . , e_n is the standard basis for R_n.

Calculation:

Given linear transformation are:

T(e1) = 7e1 - 5e3,

T(e2) = -2e2 + 9e3,

T(e3) = e1 + e2 + e3

Let the standard matrix be A with respect to the basis e1, e2, e3, 

Now T(e1) = 7e1 + 0e₂ - 5e3,

T(e2) = 0e₁ -2e2 + 9e3,

T(e3) = e1 + e2 + e3.

The standard matrix will be (transpose of linear combinations)

\(\left( {\begin{array}{*{20}{c}} 7&0&1\\ 0&{ - 2}&1\\ { - 5}&9&1 \end{array}} \right)\)