MATRIKS MATERI KE 2

 

If given 

$A=\left(\begin{array}{ccc}6& -3& 2\\ 1& 0& 4\\ 5& 7& 6\end{array}\right)$ and $B=\left(\begin{array}{ccc}1& -9& 4\\ 2& 2& -3\\ 3& 13& -10\end{array}\right)$, then what is the value of $det\left(B^{-1}\right(A^{-1}{B}^{-1}{)}^{-1}{A}^{-1})$?
A. $-1$               C. $1$                E. $5$
B. $0$                  D. $3$

Solution

Use the following matrix inversing theorem.
$\begin{array}{cc}(A^{-1}{B}^{-1}{)}^{-1}& =BA\\ A^{-1}\cdot A& =A\cdot A^{-1}=I\\ I\cdot I& =I\end{array}$
and note that the determinant of identity matrix $I$ always equals $1$.
Therefore, we have
$\begin{array}{cc}& det\left(B^{-1}\right(A^{-1}{B}^{-1}{)}^{-1}{A}^{-1})\\ & =det\left(B^{-1}\right(BA\left)A^{-1}\right)\\ & =det\left(\right(B^{-1}B\left)\right(A{A}^{-1}\left)\right)\\ & =det(I\cdot I)\\ & =det\left(I\right)=1\end{array}$
Thus, the determinant of $B^{-1}(A^{-1}{B}^{-1}{)}^{-1}{A}^{-1}$ equals $1$
(Answer C)

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Problem Number 2
Let $M$ be a matrix such that
$M\cdot \left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}a& b\\ a-c& b-d\end{array}\right)$. The determinant of matrix $M$ equals $\cdots \cdot$
A. $-1$                 C. $1$               E. $3$
B. $0$                    D. $2$

Solution

Given that $M\cdot \left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}a& b\\ a-c& b-d\end{array}\right)$
Therefore, we will have
$\begin{array}{cc}|M|\cdot \left|\begin{array}{cc}a& b\\ c& d\end{array}\right|& =\left|\begin{array}{cc}a& b\\ a-c& b-d\end{array}\right|\\ |M|\cdot (ad-bc)& =a(b-d)-b(a-c)\\ |M|\cdot (ad-bc)& =ab-ad-ab+bc\\ |M|\cdot (ad-bc)& =-(ad-bc)\\ |M|& =\frac{-(ad-bc)}{ad-bc}=-1\end{array}$
Thus, the determinant of matrix $M$ equals $-1$
(Answer A)

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Problem Number 3
If given $\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|=3$, then $\left|\begin{array}{ccc}2a+d& a& 4a+2d+g\\ 2b+e& b& 4b+2e+h\\ 2c+f& c& 4c+2f+i\end{array}\right|=\cdots \cdot$
A. $-3$               C. $0$              E. $3$
B. $-2$               D. $2$

Solution

It is given that
$\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|=3$
By swapping the first and second row entries, causing the determinant to be negative, we have
$\left|\begin{array}{ccc}d& e& f\\ a& b& c\\ g& h& i\end{array}\right|=-3$
Transpose the matrix, so we have
$\left|\begin{array}{ccc}d& a& g\\ e& b& h\\ f& c& i\end{array}\right|=-3$
(Transposing a matrix doesn’t affect the determinant)
After that, add correspondingly each entry in the first column to two times of each entry in the second column (it also won’t change the determinant).
$\left|\begin{array}{ccc}2a+d& a& g\\ 2b+e& b& h\\ 2c+f& c& i\end{array}\right|=-3$
Finally, add correspondingly each entry in the third column to two times of each entry in the first column (it also won’t change the determinant).
$\left|\begin{array}{ccc}2a+d& a& 4a+2d+g\\ 2b+e& b& 4b+2e+h\\ 2c+f& c& 4c+2f+i\end{array}\right|=-3$
Thus, the value of
$\left|\begin{array}{ccc}2a+d& a& 4a+2d+g\\ 2b+e& b& 4b+2e+h\\ 2c+f& c& 4c+2f+i\end{array}\right|=-3$
(Answer A)

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Problem Number 4
If matrix $A=\left(\begin{array}{cc}3& 7\\ -1& -2\end{array}\right)$, then $A^{27}+A^{31}+A^{40}$ equals $\cdots \cdot$
A. $\left(\begin{array}{cc}5& 1\\ 3& -4\end{array}\right)$
B. $\left(\begin{array}{cc}-7& 14\\ -2& 3\end{array}\right)$
C. $\left(\begin{array}{cc}7& -14\\ 3& 2\end{array}\right)$
D. $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$
E. $\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)$

Solution

Given $A=\left(\begin{array}{cc}3& 7\\ -1& -2\end{array}\right)$
Try to spot the pattern by finding the power of matrix $A$.
$$\begin{array}{cc}{A}^2& =\left(\begin{array}{cc}3& 7\\ -1& -2\end{array}\right)\left(\begin{array}{cc}3& 7\\ -1& -2\end{array}\right)=\left(\begin{array}{cc}2& 7\\ -1& -3\end{array}\right)\\ A^3& =A^2\cdot A=\left(\begin{array}{cc}2& 7\\ -1& -3\end{array}\right)\left(\begin{array}{cc}3& 7\\ -1& -2\end{array}\right)\\ & =\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)=-\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=-I\end{array}$$with $I$ denotes the identity matrix.
After then,
$\begin{array}{cc}& A^{27}+A^{31}+A^{40}\\ & =A^{27}(I+A^4+A^{13})\\ & =\left(A^3{)}^9\right(I+A^{3}A+\left(A^3{)}^{4}A\right)\\ & =(-I{)}^9(I-IA+(-I{)}^{4}A)\\ & =-I(I-A+A)\\ & =-I=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\end{array}$
Thus, $A^{27}+A^{31}+A^{40}=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)$
Note: Any power of a identity matrix always yields a same identity matrix.
(Answer E)

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Problem Number 5
If a constant $k$ satisfies the matrix equation
$\left(\begin{array}{cc}k& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x-1\\ y-1\end{array}\right)=\left(\begin{array}{c}0\\ k\end{array}\right)$,
then the value of $x+y=\cdots \cdot$
A. $(2+k)(1+k)$
B. $(2-k)(1-k)$
C. $(2+k)(1-k)$
D. $(1-k)(1+k)$
E. $(2-k)(1+k)$

Solution

By applying matrix multiplication rule, we have
$\begin{array}{cc}\left(\begin{array}{cc}k& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x-1\\ y-1\end{array}\right)& =\left(\begin{array}{c}0\\ k\end{array}\right)\\ \left(\begin{array}{c}k(x-1)+1(y-1)\\ 1(x-1)+0(y-1)\end{array}\right)& =\left(\begin{array}{c}0\\ k\end{array}\right)\\ \left(\begin{array}{c}k(x-1)+(y-1)\\ x-1\end{array}\right)& =\left(\begin{array}{c}0\\ k\end{array}\right)\end{array}$
Hence, we have the following equation system.
$\{\begin{array}{cc}k(x-1)+(y-1)=0& (\cdots 1)\\ x-1=k& (\cdots 2)\end{array}$
Substitute equation $\left(2\right)$ into equation $\left(1\right)$.
$k\left(k\right)+y-1=0\iff y=1-k^2$
Therefore, we have
$\begin{array}{cc}x+y& =(k+1)+(1-k^2)\\ & =(k+1)+(k+1)(1-k)\\ & =(k+1)(1+(1-k\left)\right)\\ & =(k+1)(2-k)\end{array}$
So, the value of $x+y$ is $(k+1)(2-k)$ or can be rewritten into $(2-k)(1+k)$
(Answer E)

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Problem Number 6
If $x:y=5:4$, then the value of $x$ and $y$ that satisfy
$\left(\begin{array}{ccc}2& 10& 1\end{array}\right)\left(\begin{array}{cc}x& y\\ 4& 5\\ 30& 25\end{array}\right)\left(\begin{array}{c}5\\ 10\end{array}\right)=1.360$
is $\cdots \cdot$
A. $x=1$ and $y=\frac{4}{5}$
B. $x=\frac{4}{5}$ and $y=1$
C. $x=5$ and $y=4$
D. $x=-10$ and $y=-8$
E. $x=10$ and $y=8$

Solution

Problem Number 7
It is given that $A=\left(\begin{array}{cc}2& 1\\ 3& 5\end{array}\right)$ holds a certain relationship to matrix $B=\left(\begin{array}{cc}-5& 3\\ 1& -2\end{array}\right)$. If matrix $C=\left(\begin{array}{cc}3& 2\\ 1& -5\end{array}\right)$ and matrix $D$ have the same relationship, then the value of $C+D=\cdots \cdot$
A. $\left(\begin{array}{cc}8& 3\\ 3& 8\end{array}\right)$             D. $\left(\begin{array}{cc}3& 8\\ 8& 3\end{array}\right)$
B. $\left(\begin{array}{cc}8& 3\\ 3& -8\end{array}\right)$             E. $\left(\begin{array}{cc}-8& -3\\ -3& 8\end{array}\right)$
C. $\left(\begin{array}{cc}3& 8\\ -8& 3\end{array}\right)$