Question 8
GCE_MODEL_QUESTIONS_VOL1
•
2024
•
FURTHER_MATHEMATICS
•
objective
Let \(T:\mathbb{R}^{2}\to\mathbb{R}^{2}\) be a linear transformation such that \(T(1,2)=(3,5)\) and \(T(2,-1)=(4,-2)\).\n\n(a) Find the matrix \(A\) of \(T\) with respect to the standard basis.\n(b) Compute the determinant and trace of \(A\).\n(c) State whether \(T\) is invertible and, if it is, give the matrix \(A^{-1}\).
Answer Options
Correct Answer
A.
A=\begin{pmatrix}\frac{11}{5}&\frac{2}{5}\\[2pt]\frac{1}{5}&\frac{12}{5}\end{pmatrix}, \;\det A=\frac{26}{5}, \;\operatorname{tr}A=\frac{23}{5}, \;A^{-1}=\begin{pmatrix}\frac{6}{13}&-\frac{1}{13}\\-\frac{1}{26}&\frac{11}{26}\end{pmatrix}
B.
A=\begin{pmatrix}\frac{11}{5}&\frac{1}{5}\\[2pt]\frac{2}{5}&\frac{12}{5}\end{pmatrix}, \;\det A=\frac{27}{5}, \;\operatorname{tr}A=\frac{24}{5}, \;A^{-1}=\begin{pmatrix}\frac{6}{13}&-\frac{2}{13}\\-\frac{1}{26}&\frac{11}{26}\end{pmatrix}
C.
A=\begin{pmatrix}\frac{10}{5}&\frac{2}{5}\\[2pt]\frac{1}{5}&\frac{13}{5}\end{pmatrix}, \;\det A=\frac{128}{25}, \;\operatorname{tr}A=\frac{23}{5}, \;A^{-1}=\begin{pmatrix}\frac{5}{12}&-\frac{1}{12}\\-\frac{1}{24}&\frac{9}{24}\end{pmatrix}
D.
A=\begin{pmatrix}\frac{11}{5}&\frac{2}{5}\\[2pt]\frac{1}{5}&\frac{12}{5}\end{pmatrix}, \;\det A=\frac{25}{5}, \;\operatorname{tr}A=\frac{22}{5}, \;A^{-1}=\begin{pmatrix}\frac{5}{13}&-\frac{2}{13}\\-\frac{1}{26}&\frac{10}{26}\end{pmatrix}
Correct Answer
A
Correct Option:
A=\begin{pmatrix}\frac{11}{5}&\frac{2}{5}\\[2pt]\frac{1}{5}&\frac{12}{5}\end{pmatrix}, \;\det A=\frac{26}{5}, \;\operatorname{tr}A=\frac{23}{5}, \;A^{-1}=\begin{pmatrix}\frac{6}{13}&-\frac{1}{13}\\-\frac{1}{26}&\frac{11}{26}\end{pmatrix}
About This Question
This is
Gce Model Questions Vol1
2024
Further Mathematics
Question 8.
It is one of the
objective
questions from the
2024
Gce Model Questions Vol1
Further Mathematics
examination.
Difficulty level:
Hard
.