ONE-EIGHTH STAGE 2025
Prempeh College: 62 points
Suhum SHTS: 37 points
Benkum SHS: 13 points
Mfantsipim School: 53 points
Anglican SHS, Kumasi: 37 points
Sogakope SHS: 24 points
Adisadel College: 82 points
St. Hubert Seminary SHS: 24 points
Serwaa Nyarko Girls' SHS: 24 points
ROUND 4 - TRUE OR FALSE
PREAMBLE
The given linear transformation has an inverse.
1. $T(x,y)\to(x-y,\ x+y)$
ANSWER: TRUE
EXPLANATION:
Check if the determinant is not zero; if it is zero, the transformation has no inverse:
$\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$
$\begin{pmatrix}1&-1\\1&1\end{pmatrix}$
$\det=(1)(1)-(-1)(1)$
$\det=1-(-1)$
$\det=2$
Since the determinant is not zero, T has an inverse.
2. $T(x,y)\to(x-y,\ -x+y)$
ANSWER: FALSE
EXPLANATION:
Check if the determinant is not zero; if it is zero, the transformation has no inverse:
$\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$
$\begin{pmatrix}1&-1\\-1&1\end{pmatrix}$
$\det=(1)(1)-(-1)(-1)$
$\det=1-(1)$
$\det=0$
Since the determinant is zero, T has no inverse.
3. $T(x,y)\to(2x+y,\ x-2y)$
ANSWER: TRUE
EXPLANATION:
Check if the determinant is not zero; if it is zero, the transformation has no inverse:
$\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$
$\begin{pmatrix}2&1\\1&-2\end{pmatrix}$
$\det=(2)(-2)-(1)(1)$
$\det=-4-(1)$
$\det=-5$
Since the determinant is not zero, T has an inverse.
---
PRACTICE QUESTIONS
PREAMBLE 1
The given linear transformation has an inverse:
1. $T(x, y) \to (2x,\ 3y)$
ANSWER: TRUE
EXPLANATION:
$\det\begin{pmatrix}2&0\\0&3\end{pmatrix} = (2)(3) - (0)(0) = 6 \ne 0$.
2. $T(x, y) \to (x + 2y,\ 2x + 4y)$
ANSWER: FALSE
EXPLANATION:
$\det\begin{pmatrix}1&2\\2&4\end{pmatrix} = (1)(4) - (2)(2) = 0$.
3. $T(x, y) \to (y,\ x)$
ANSWER: TRUE
EXPLANATION:
$\det\begin{pmatrix}0&1\\1&0\end{pmatrix} = 0 - 1 = -1 \ne 0$.
PREAMBLE 2
For the matrix $M = \begin{pmatrix}3&1\\2&1\end{pmatrix}$:
1. $\det M = 1$
ANSWER: TRUE
EXPLANATION:
$(3)(1) - (1)(2) = 3 - 2 = 1$.
2. $M$ has an inverse.
ANSWER: TRUE
EXPLANATION:
$\det M \ne 0$.
3. $M^{-1} = \begin{pmatrix}1&1\\2&3\end{pmatrix}$
ANSWER: FALSE
EXPLANATION:
$M^{-1} = \dfrac{1}{1}\begin{pmatrix}1&-1\\-2&3\end{pmatrix}$; the signs of the other diagonal must change.
PREAMBLE 3
A linear transformation with matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$:
1. Has an inverse when $ad - bc \ne 0$.
ANSWER: TRUE
EXPLANATION:
A non-zero determinant is needed.
2. Multiplies areas by $|ad - bc|$.
ANSWER: TRUE
EXPLANATION:
The determinant is the area scale factor.
3. Always has an inverse when $a = d$.
ANSWER: FALSE
EXPLANATION:
$\begin{pmatrix}1&1\\1&1\end{pmatrix}$ has $a = d$ but determinant $1 - 1 = 0$.