# Matrix Problems and Solutions (Olympiad Level)

The following matrix problems are provided along with the solutions. The problems are in high school olympiad level, so you need higher thinking skills to tackle them. It may not easy, but keep learning and you can.

### Quote by Arthur Ashe

Success is a journey, not a destination. The doing is often more important than the outcome.

Multiple Choice Section

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

Solution

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

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

Solution

Given that $M \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & b \\ a- c & b- d \end{pmatrix}$
Therefore, we will have
\begin{aligned} |M| \cdot \begin{vmatrix} a & b \\ c & d \end{vmatrix} & = \begin{vmatrix} a & b \\ a- c & b- d \end{vmatrix} \\ |M| \cdot (ad- bc) & = a(b- d)- b(a- c) \\ |M| \cdot (ad- bc) & = \cancel{ab}- ad \cancel{- ab} + bc \\ |M| \cdot (ad- bc) & =-(ad- bc) \\ |M| & = \dfrac{-(ad-bc)}{ad-bc} =-1 \end{aligned}
Thus, the determinant of matrix $M$ equals $\boxed{-1}$

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

Solution

It is given that
$\begin{vmatrix} a & b & c\\ d & e & f \\ g & h & i \end{vmatrix} = 3$
By swapping the first and second row entries, causing the determinant to be negative, we have
$\begin{vmatrix} d & e & f \\ a & b & c \\ g & h & i \end{vmatrix} =-3$
Transpose the matrix, so we have
$\begin{vmatrix} d & a & g \\ e & b & h \\ f & c & i \end{vmatrix} =-3$
(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).
$\begin{vmatrix} 2a+d & a & g \\ 2b + e & b & h \\ 2c + f & c & i \end{vmatrix} =-3$
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).
$\begin{vmatrix} 2a+d & a & 4a+2d+g\\ 2b + e & b & 4b+2e+h \\ 2c + f & c & 4c+2f+i \end{vmatrix} =-3$
Thus, the value of
$\boxed{\begin{vmatrix} 2a+d & a & 4a+2d+g\\ 2b + e & b & 4b+2e+h \\ 2c + f & c & 4c+2f+i \end{vmatrix} =-3}$

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

Solution

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

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Problem Number 5
If a constant $k$ satisfies the matrix equation
$\begin{pmatrix} k & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x-1 \\ y-1 \end{pmatrix} = \begin{pmatrix} 0 \\ k \end{pmatrix}$,
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{aligned} \begin{pmatrix} k & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x- 1 \\ y- 1 \end{pmatrix} & = \begin{pmatrix} 0 \\ k \end{pmatrix} \\ \begin{pmatrix} k(x-1) + 1(y-1) \\ 1(x-1) + 0(y-1) \end{pmatrix} & = \begin{pmatrix} 0 \\ k \end{pmatrix} \\ \begin{pmatrix} k(x-1)+(y-1) \\ x-1 \end{pmatrix} & = \begin{pmatrix} 0 \\ k \end{pmatrix} \end{aligned}
Hence, we have the following equation system.
$\begin{cases} k(x-1) + (y-1) = 0 & (\cdots 1) \\ x- 1 = k & (\cdots 2) \end{cases}$
Substitute equation $(2)$ into equation $(1)$.
$k(k) + y- 1 = 0 \Leftrightarrow y = 1- k^2$
Therefore, we have
\begin{aligned}x+y & =(k+1)+(1-k^2) \\ & =(k+1)+(k+1)(1-k) \\ & = (k+1)(1+(1-k)) \\ & = (k+1)(2-k) \end{aligned}
So, the value of $x+y$ is $(k+1)(2-k)$ or can be rewritten into $\boxed{(2-k) (1+k)}$

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Problem Number 6
If $x : y = 5 : 4$, then the value of $x$ and $y$ that satisfy
$\begin{pmatrix} 2 & 10 & 1 \end{pmatrix} \begin{pmatrix} x & y \\ 4 & 5 \\ 30 & 25 \end{pmatrix}\begin{pmatrix} 5 \\ 10 \end{pmatrix} = 1.360$
is $\cdots \cdot$
A. $x = 1$ and $y = \dfrac45$
B. $x=\dfrac45$ and $y=1$
C. $x=5$ and $y=4$
D. $x=-10$ and $y=-8$
E. $x=10$ and $y=8$

Solution

Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$.
Hence, we have
\begin{aligned} \begin{pmatrix} 2 & 10 & 1 \end{pmatrix} \begin{pmatrix} 5k & 4k \\ 4 & 5 \\ 30 & 25 \end{pmatrix}\begin{pmatrix} 5 \\ 10 \end{pmatrix} & = 1.360 \\ \begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \end{pmatrix} \begin{pmatrix} 5 \\ 10 \end{pmatrix} & = 1.360 \\ \begin{pmatrix} 10k + 70 & 8k + 75 \end{pmatrix} \begin{pmatrix} 5 \\ 10 \end{pmatrix} & =1.360 \\ 5(10k+70) + 10(8k+75) & = 1.360 \\ \text{Divide each side by}~& 5 \\10k+70 + 2(8k+75) & = 272 \\ 26k + 220 & = 272 \\ 26k & = 52 \\ k & = 2 \end{aligned}Thus, we have
\boxed{\begin{aligned} x & = 5k = 5(2) = 10 \\ y & =4k=4(2)=8 \end{aligned}}

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

Solution

Given that $A = \begin{pmatrix} 2 & 1 \\ 3 & 5 \end{pmatrix}$ and $B =\begin{pmatrix}-5 & 3 \\ 1 &-2 \end{pmatrix}$
Observe the matrices’ entries and you will spot its relationship.
Two entries of the main diagonal in matrix $A$ are swapped and multiplied by $-1$ to yield the entries of the main diagonal in matrix $B$.
Two entries of the antidiagonal in matrix $A$ are swapped to yield the entries of the antidiagonal in matrix $B$.
Mathematically, we write as
$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \iff B = \begin{pmatrix}-d & c \\ b &-a \end{pmatrix}$
Since $C = \begin{pmatrix} 3 & 2 \\ 1 &-5 \end{pmatrix}$, then by following the relationship pattern, we have
$D = \begin{pmatrix} 5 & 1 \\ 2 &-3 \end{pmatrix}$
Hence,
\begin{aligned} C + D & = \begin{pmatrix} 3 & 2 \\ 1 &-5 \end{pmatrix} + \begin{pmatrix} 5 & 1 \\ 2 &-3 \end{pmatrix} \\ & = \begin{pmatrix} 8 & 3 \\ 3 &-8 \end{pmatrix} \end{aligned}

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Problem Number 8
If $A$ be a matrix with an order of $2 \times 2$ that has an inverse and $A^2-2A-I = 0$, then $A-2I = \cdots \cdot$
A. $(2A)^{-1}$                   D. $A^2-2A$
B. $A^2+2A$                      E. $A^{-1}$
C. $2I-A$

Solution

Start from $A^2-2A-I=0$ with $I$ denotes identity matrix and $0$ denotes zero matrix, we have
\begin{aligned} A^2-2A & =I \\ A^2-2IA & = I \\ A(A- 2I) & = I \end{aligned}
Based on the definition of an inverse of the matrix, $B$ is an inverse of matrix $A$ if it satisfies $AB=BA=I$
Thus, we have $\boxed{A- 2I= A^{-1}}$

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Problem Number 9
Suppose there are matrices $A$ and $B$ with an order of $2 \times 2$ and each of them has an inverse. Given that $AB = \begin{pmatrix} 4 & 2 \\ 3 & 2 \end{pmatrix}$.$AB(B^{-1}+A)A^{-1}$ equals to $\cdots \cdot$
A. $\begin{pmatrix} 4 & 2 \\ 3 & 2 \end{pmatrix}$            D. $\begin{pmatrix} 4 & 3 \\ 4 & 2 \end{pmatrix}$
B. $\begin{pmatrix} 5 & 2 \\ 3 & 3 \end{pmatrix}$            E. $\begin{pmatrix} 1 & 9 \\ 6 & 1 \end{pmatrix}$
C. $\begin{pmatrix} 5 & 2 \\ 2 & 3 \end{pmatrix}$

Solution

Notice that $AB(B^{-1}+A)A^{-1}$ can be simplified furthermore by applying some matrix operation rules.
\begin{aligned} & AB(B^{-1}+A)A^{-1} \\ & = AB(B^{-1}A^{-1} + AA^{-1}) && (\text{Sifat Dis}\text{tributif}) \\ & = AB((AB)^{-1} + I) && (\text{Inve}\text{rse rules}) \\ & = (AB)(AB^{-1}) + AB(I) && (\text{Distributi}\text{ve Property}) \\ & = I + AB && (\text{Inve}\text{rse rules}) \\ & = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 4 & 2 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 5 & 2 \\ 3 & 3 \end{pmatrix} \end{aligned}Thus,
$\boxed{AB(B^{-1}+A)A^{-1} = \begin{pmatrix} 5 & 2 \\ 3 & 3 \end{pmatrix}}$

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Problem Number 10
If $A = \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix}$ and $B = \begin{pmatrix}-2 & 5 \\ 1 &-3 \end{pmatrix}$, then the matrix expressing
$\left(\left(\left(\left(\left(\left(\left(\left(AB\right)^T\right)^{-1}\right)^T\right)^{-1}\right)^T\right)^{-1}\right)^T\right)^{-1}$
is $\cdots \cdot$
A. $\begin{pmatrix}-4 &-5 \\ 9 & 11 \end{pmatrix}$          D. $\begin{pmatrix} 11 & 5 \\-9 &-4 \end{pmatrix}$
B. $\begin{pmatrix}-4 & 9 \\-5 & 11 \end{pmatrix}$          E. $\begin{pmatrix}-9 & 11 \\ 5 &-4 \end{pmatrix}$
C. $\begin{pmatrix} 11 &-9 \\ 5 &-4 \end{pmatrix}$

Solution

Suppose $A$ expressing a non-singular matrix.
\begin{aligned} (A^T)^{-1} & = (A^{-1})^T \\ (A^{-1})^{-1} & = A \\ (A^T)^T & = A \end{aligned}
You may see that there are an even number times of the transpose and inverse, and it will yield the original matrix. We have
\begin{aligned} & \left(\left(\left(\left(\left(\left(\left(\left(AB\right)^T\right)^{-1}\right)^T\right)^{-1}\right)^T\right)^{-1}\right)^T\right)^{-1} \\ & = AB = \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix} \begin{pmatrix}-2 & 5 \\ 1 &-3 \end{pmatrix} \\ & = \begin{pmatrix}-4 & 9 \\-5 & 11 \end{pmatrix} \end{aligned}(Answer B)

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Problem Number 11
Given that $A = \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\-2 &-1 &-3 \end{pmatrix}$.
The value of $A^{2017} + 2017A^{2018} + 2I^{2018}$ is $\cdots \cdot$
A. $O$                D. $2017A+2I$
B. $2I$               E. $A+2I$
C. $A$

Solution

Notice the pattern.
\begin{aligned} A^2 & = A \times A \\ & = \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\-2 &-1 &-3 \end{pmatrix} \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\-2 &-1 &-3 \end{pmatrix} \\ & = \begin{pmatrix} 0 & 0 & 0 \\ 3 & 3 & 9 \\-1 &-1 &-3 \end{pmatrix} \\ A^3 & = A^2 \times A \\ & = \begin{pmatrix} 0 & 0 & 0 \\ 3 & 3 & 9 \\-1 &-1 &-3 \end{pmatrix} \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\-2 &-1 &-3 \end{pmatrix} \\ & = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \end{aligned}Since the product of two matrices involving zero matrix yields a zero matrix, while any power of the identity matrix yields a same identity matrix, then we have
\begin{aligned} & A^{2017} + 2017A^{2018} + 2I^{2018} \\ & = O + 2017O + 2I = 2I \end{aligned}
Thus, $\boxed{A^{2017} + 2017A^{2018} + 2I^{2018} = 2I}$

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Problem Number 12
If $A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, then $A^{2019} + B^{2020} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is satisfied by $a + b + c + d = \cdots \cdot$
A. $2020$                 D. $4040$
B. $2036$                 E. $4043$
C. $4038$

Solution

Given
$A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}~~~~~B = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$
We will discover the pattern entries of $A^{2019}$.
\begin{aligned} A^2 & = \begin{pmatrix} 1 & 0 \\ 1&1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1&1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \\ A^3 & = A^2 \cdot A = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} \\ A^4 & = A^3 \cdot A =\begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix} \end{aligned}See that only the entry in the second row and first column of this matrix changed according to the power of $A$. This implies that
$A^{2019} = \begin{pmatrix} 1 & 0 \\ 2019 & 1 \end{pmatrix}$
By applying the same observation and principle, we have
$B^{2020} = \begin{pmatrix} 1 & 2020 \\ 0 & 1 \end{pmatrix}$
Next,
\begin{aligned} A^{2019} + B^{2020} & = \begin{pmatrix} 1 & 0 \\ 2019 & 1 \end{pmatrix} + \begin{pmatrix} 1 & 2020 \\ 0 & 1 \end{pmatrix} \\ & = \begin{pmatrix} 2 & 2020 \\ 2019 & 2 \end{pmatrix} \end{aligned}Lastly, we have
$a = d = 2; b = 2020; c = 2019$
Thus, the value of $\boxed{a+b+c+d=4043}$

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Problem Number 13
If $A = \begin{pmatrix} 0 & a \\ 1 & 0 \end{pmatrix}$, then $A^{2009} = \cdots \cdot$
A. $\begin{pmatrix} a^{1004} & 0 \\ 0 & a^{1004} \end{pmatrix}$        D. $\begin{pmatrix} a^{2008} & 0 \\ 0 & a^{1004} \end{pmatrix}$
B. $\begin{pmatrix} 0 & a^{1005} \\ a^{1004} & 0 \end{pmatrix}$        E. $\begin{pmatrix} 0 & a^{1004} \\ a^{1004} & 0 \end{pmatrix}$
C. $\begin{pmatrix} 0 & a^{1005} \\ a^{1005} & 0 \end{pmatrix}$

Solution

Given $A = \begin{pmatrix} 0 & a \\ 1 & 0 \end{pmatrix}$
Spot the pattern by finding the matrix expressing the power of $A$.
\begin{aligned} A^2 & = A \times A = \begin{pmatrix} 0 & a \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & a \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} \\ A^3 & = A^2 \times A = \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} \begin{pmatrix} 0 & a \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & a^2 \\ 0 & a \end{pmatrix} \\ A^4 & = A^3 \times A = \begin{pmatrix} 0 & a^2 \\ 0 & a \end{pmatrix} \begin{pmatrix} 0 & a \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} a^2 & 0 \\ 0 & a^2 \end{pmatrix} \\ A^5 & = A^4 \times A = \begin{pmatrix} a^2 & 0 \\ 0 & a^2 \end{pmatrix} \begin{pmatrix} 0 & a \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & a^3 \\ a^2 & 0 \end{pmatrix} \\ A^6 & = A^5 \times A = \begin{pmatrix} 0 & a^3 \\ a^2 & 0 \end{pmatrix}\begin{pmatrix} 0 & a \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} a^3 & 0 \\ 0 & a^3 \end{pmatrix} \end{aligned}Noticing the pattern, we will have
$A^n = \begin{pmatrix} a^{\frac{n} {2}} & 0 \\ 0 & a^{\frac{n} {2}} \end{pmatrix}$
for every even number $n$, or
$A^n = \begin{pmatrix} 0 & a^{\frac{n+1} {2}} \\ a^{\frac{n-1} {2}} & 0\end{pmatrix}$
for every odd number $n$.
Since $2009$ is an odd number, then
$$A^{2009} = \begin{pmatrix} 0 & a^{\frac{2009+1} {2}} \\ a^{\frac{2009-1} {2}} & 0\end{pmatrix} = \begin{pmatrix} 0 & a^{1005} \\ a^{1004} & 0\end{pmatrix}$$(Answer B)

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Problem Number 14
If $A = \begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{pmatrix}$, then $\det(A) = \cdots \cdot$
A. $(a-b)(b-c)(c-a)(a+b+c)$
B. $(a-b)(b-c)(c-a)(a+b-c)$
C. $(a-b)(b-c)(c-a)(a-b+c)$
D. $(a-b)(b-c)(c+a)(a-b-c)$
E. $(a-b)(b-c)(c+a)(a-b+c)$

Solution

\begin{aligned} \det A & = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} \\ & = 1(bc^3-b^3c)- 1(ac^3-a^3c) + 1(ab^3-a^3b) \\ & = ab^3-bc^3 + bc^3-a^3b + ac^3-a^3c \\ & = (a-c)b^3 + (c^3-a^3)b + ac(c^2-a^2) \\ & = (a-c)b^3-(a-c)[(a^2+ac+c^2)b + ac(a+c)] \\ & = (a-c)\color{red}{[b^3- (a^2+ac+c^2)b + ac(a+c)]} \end{aligned}Next, we simplify the algebraic expression in red color above.
\begin{aligned} & b^3- (a^2+ac+c^2)b + ac(a+c) \\ & = b^3- a^2b-abc- bc^2 + a^2c + ac^2 \\ & = b^3-bc^2-a^2b + a^2c-abc + ac^2 \\ & = b(b^2-c^2)-(b-c)a^2- ac(b-c) \\ & = b(b-c)(b+c)- (b-c)a^2- ac(b-c) \\ & = (b-c)\color{blue}{[b(b+c)-a^2-ac]} \end{aligned}After that, we simplify the algebraic expression in blue color above.
\begin{aligned} b(b+c)- a^2-ac & = b^2+bc-a^2-ac \\ & = b^2-a^2+bc-ac \\ & = (b-a)(b+a)+c(b-a) \\ & = (b-a)(a+b+c) \end{aligned}Hence, we write as
\begin{aligned} \det A & = (a-c)(b-c)(b-a)(a+b+c) \\ & = (a-b)(b-c)(c-a)(a+b+c) \end{aligned}
Thus, the determinant of matrix $A$ equals $\boxed{(a-b)(b-c)(c-a)(a+b+c)}$

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Problem Number 15
Given that $A = \begin{pmatrix} b &-\sin x \\ \sin x & b \end{pmatrix}$. The product of all possible values of $b$ that satisfy the equation $A^{-1} = A^T$ is $\cdots \cdot$
A. $-\sin^2 x$              D. $\cos x$
B. $-\cos^2 x$             E. $\cos^2 x$
C. $-\cos x$

Solution

Given
$A = \begin{pmatrix} b &-\sin x \\ \sin x & b \end{pmatrix}$ $A^T = \begin{pmatrix} b & \sin x \\-\sin x & b \end{pmatrix}$
Of the equation $A^{-1} = A^T$, multiply both side by $A$, and we have
\begin{aligned} A^{-1}A & = A^TA \\ I & = A^TA \\ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} & = \begin{pmatrix} b & \sin x \\-\sin x & b \end{pmatrix} \begin{pmatrix} b &-\sin x \\ \sin x & b \end{pmatrix} \\ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} & = \begin{pmatrix} b^2 + \sin^2 x & 0 \\ 0 & \sin^2 x + b^2 \end{pmatrix} \end{aligned}From the equivalence of two matrices, we have
\begin{aligned} \sin^2 x + b^2 & = 1 \\ b^2 & = 1- \sin^2 x \\ b^2 & = \cos^2 x \\ b & = \pm \cos x \end{aligned}
The product of all possible values of $b$ is
$\boxed{b_1b_2 = \cos x(-\cos x) =-\cos^2 x}$

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Problem Number 16
Which one of the following are true statements about matrix?

1. If $A$ and $B$ both are square matrices, then $(A+B)(A-B) = A^2-B^2$
2. If $AB = C$ and $C$ has $2$ columns, then $A$ also has $2$ columns.
3. If $BC = D$, then $B^{-1}C^{-1} = D^{-1}$
4. If $AC=O$, then it holds $A=O$ or $C=O$
5. If $A$ and $B$ both are $m \times n$ matrices, then $AB^T$ and $A^TB$ both are defined.

Solution

We will check the truth of the given options.
Check option A: False
Since $A, B$ are two square matrices, then by applying the distributive property, it holds
\begin{aligned} (A+B)(A-B) &= A^2-AB+BA-B^2 \\ & \neq A^2-B^2 \end{aligned}
The inequality occured because $AB \neq BA$ (recall that multiplication of matrices is not commutative).
Check option B: False
Given $AB=C$ and $C$ has two columns. Based on the definition of multiplication of matrices, the number of row in $A$ must be equal to the number of row in $C$, while the number of column in $B$ must be equal to the number of column in $C$. Hence, the said statement isn’t always true that $A$ has two columns.
Check option C: False
Given $BC = D$. If we take the inverse to the both sides of equation, we have
\begin{aligned} (BC)^{-1} & = D^{-1} \\ C^{-1}B^{-1} & = D^{-1} \end{aligned}
Thus, the statement is absolutely false.
Check option D: False
Given $AC = O$. Besides $A = O$ or $C = O$, there are other possibilities that the product of two matrices yields a zero matrix as follows.
$ACC^{-1} = OC^{-1} \Leftrightarrow A = C^{-1}$
Thus, the matrix that satisfies $A=C^{-1}$ is a solution to that matrix equation.
Check option E: True
Given $A$ and $B$ are $m \times n$ matrices. Therefore, both $A^T$ and $B^T$ have an order of $n \times m$. Hence, $AB^T$ is a product of two matrices with an order of $m \times n$ and $n \times m$, so it holds the matrix multiplication rule (it is defined). The same also holds for $A^TB$.

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Problem Number 17
The determinant of $\begin{pmatrix} 3+2\sqrt5+\sqrt7 & 2\sqrt5 \\ 3-\sqrt5-\sqrt7 & 3-\sqrt7 \end{pmatrix}$ equals to $\cdots \cdot$
A. $-8$             C. $2\sqrt35$               E. $7\sqrt15$
B. $2$                 D. $12$

Solution

Notice that the determinant of $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ equals to the determinant of $\begin{pmatrix} a-b & b \\ c-d & d \end{pmatrix}$.
By using this fact, we have
\begin{aligned} & \det \begin{pmatrix} 3+2\sqrt5+\sqrt7 & 2\sqrt5 \\ 3-\sqrt5-\sqrt7 & 3-\sqrt7 \end{pmatrix} \\ & = \det \begin{pmatrix} 3+2\sqrt5+\sqrt7-\color{red}{2\sqrt5} & 2\sqrt5 \\ 3-\sqrt5-\sqrt7-(\color{red}{3-\sqrt7}) & 3-\sqrt7 \end{pmatrix} \\ & = \det \begin{pmatrix} 3+\sqrt7 & 2\sqrt5 \\ -\sqrt5 & 3-\sqrt7 \end{pmatrix} \\ & = (3+\sqrt7)(3-\sqrt7)-(-\sqrt5)(2\sqrt5) \\ & = (9-7)-(-2 \cdot 5) \\ & = 2-(-10) = 12 \end{aligned}Thus, the determinant of said matrix equals to $\boxed{12}$

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Problem Number 18
For every natural number $n$, it is defined a matrix $A_n = \begin{pmatrix} 3n & n \\ 4n & 2n \end{pmatrix}$. If $\det(A_1 + A_2 + \cdots +A_t) = 882$, then the value of $\det(A_{2t}) = \cdots \cdot$
A. $288$               C. $122$                E. $70$
B. $144$               D. $72$

Solution

Given that $A_n = \begin{pmatrix} 3n & n \\ 4n & 2n \end{pmatrix}$ with $n \in \mathbb{N}$. The determinant of this matrix is
\begin{aligned} \det(A_n) & = 3n(2n)-4n(n) \\ & = 6n^2-4n^2=2n^2 \end{aligned}
Notice that $A_1+A_2+\cdots+A_t = A(1+2+\cdots+t)$ with $A = \begin{pmatrix} 3 & 1 \\ 4 & 2 \end{pmatrix}$ and $\det(A) = 3(2)-1(4)=2$ so that if we apply the theorem $\det(n \cdot A) = n^2 \det(A)$ with $A$ is a square matrix with an order of $2$ along with the determination equation, we have
\begin{aligned} \det(A_1 + A_2 + \cdots +A_t) & = 882 \\ \det(A(1+2+\cdots t)) & = 882 \\ \det(A) \cdot (1+2+\cdots + t)^2 & = 882 \\ 2(1+2+\cdots + t)^2 & = 882 \\ (1+2+\cdots+t)^2 & = 441 \\ 1+2+\cdots+t & = 21 \\ \dfrac{t}{2}(t+1) & = 21 \\ t^2+t-42 & = 0 \\ (t+7)(t-6) & = 0 \end{aligned}
We obtain $t = -7$ or $t = 6$ (It satisfies since $t \in \mathbb{N}$).
Thus, the value of $\boxed{\det(A_{2 \cdot 6}) = A_{12} = 2 \cdot 12^2 = 288}$

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Problem Number 19
If $A = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix}$, then $A^{2013} \cdot \begin{pmatrix}2 \\ 1 \end{pmatrix} + B^{2014} \cdot \begin{pmatrix} 1 \\ -3 \end{pmatrix}$ equals $\cdots \cdot$
A. $\begin{pmatrix} 2023 \\ -6044 \end{pmatrix}$             D. $\begin{pmatrix} -4023 \\ -6044 \end{pmatrix}$
B. $\begin{pmatrix} -4023 \\ 6044 \end{pmatrix}$             E. $\begin{pmatrix} -6044 \\ -4023 \end{pmatrix}$
C. $\begin{pmatrix} 4023 \\ 6044 \end{pmatrix}$

Solution

Given that $A = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}$.
We will find the matrix’s entry pattern for $A^n$.
You should note that
\begin{aligned} A^2 & = \begin{pmatrix} 1 & \color{red}{-2} \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & \color{red}{-2} \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & \color{red}{-4} \\ 0 & 1 \end{pmatrix} \\ A^3 & = A^2 \cdot A = \begin{pmatrix} 1 & \color{red}{-4} \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & \color{red}{-2} \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & \color{red}{-6} \\ 0 & 1 \end{pmatrix} \end{aligned}From here, we found that only an entry of the first row second column changed and it’s a multiple of $2$ for corresponding power of $A$.
We can write $A^n = \begin{pmatrix} 1 & \color{red}{-2n} \\ 0 & 1 \end{pmatrix}$.
Given that $B = \begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix}$.
We will find the matrix’s entry pattern for $B^n$.
You should note that
\begin{aligned} B^2 & = \begin{pmatrix} 1 & 0 \\ \color{blue}{-3} & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ \color{blue}{-3} & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ \color{blue}{-6} & 1 \end{pmatrix} \\ B^3 & = B^2 \cdot B = \begin{pmatrix} 1 & 0 \\ \color{blue}{-6} & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ \color{blue}{-3} & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ \color{blue}{-9} & 1 \end{pmatrix} \end{aligned}From here, we found that only an entry of the second row first column changed and it’s a multiple of $3$ for corresponding power of $B$.
We can write $B^n = \begin{pmatrix} 1 & 0 \\ -3n & 1 \end{pmatrix}$.
Therefore,
\begin{aligned} & A^{2013} \cdot \begin{pmatrix}2 \\ 1 \end{pmatrix} + B^{2014} \cdot \begin{pmatrix} 1 \\ -3 \end{pmatrix} \\ & = \begin{pmatrix} 1 & -4026 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ -3 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ -6042 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ -3 \end{pmatrix} \\ & = \begin{pmatrix} -4024 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ -6039 \end{pmatrix} = \begin{pmatrix} -4023 \\ -6044 \end{pmatrix} \end{aligned}Thus, it is evaluated that $\boxed{\begin{pmatrix} -4023 \\ -6044 \end{pmatrix}}$

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Problem Number 20
Given a matrix $A = \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix}$ and $a_{ij}$ expresses the entry of matrix $A^{10}$ lies in the row number $i$ and column number $j$. If $a_{21}=p^2-q$ with $p, q$ be the positive integers, then the value of $p+q=\cdots \cdot$.
A. $32$                C. $35$                 E. $39$
B. $33$                D. $38$

Solution

Given $A = \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix}$.
Notice that
\begin{aligned} A^2 & = A \cdot A = \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 3 & 4 \end{pmatrix} \\ A^3 & = A^2 \cdot A = \begin{pmatrix} 1 & 0 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 7 & 8 \end{pmatrix} \end{aligned}Start from here, we already discover the pattern as follows.
$A^n = \begin{pmatrix} 1 & 0 \\ 2^n-1 & 2^n \end{pmatrix}$
so that
$A^{10} = \begin{pmatrix} 1 & 0 \\ 2^{10}-1 & 2^{10} \end{pmatrix}$
$a_{21}$ denotes the entry lies in the second row and first column, that is $a_{21} = 2^{10}-1$.
Hence, we will have
\begin{aligned} a_{21} & = p^2-q \\ 2^{10}-1 & = p^2-q \\ (2^5)^2-1 & = p^2-q \\ 32^2-1 & = p^2-q \end{aligned}
The last equation implies that $p=32$ and $q=1$.
Thus, the value of $\boxed{p+q=32+1=33}$

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Problem Number 21
If $A = \begin{pmatrix} 2a & 2 \\ -4 & a \end{pmatrix}$ and $B = \begin{pmatrix} 2b & b \\ -4 & b \end{pmatrix}$ are both invertible, then all values of $b$ that satisfy the inequation $\det(ABA^{-1}B^{-1}) > 0$ are $\cdots \cdot$
A. $b < 0$
B. $b > 0$
C. $b > -2$
D. $b < -2$ or $b > 0$
E. $b \neq -2$ or $b \neq 0$

Solution

Given:
\begin{aligned} A & = \begin{pmatrix} 2a & 2 \\ -4 & a \end{pmatrix} \\ B & = \begin{pmatrix} 2b & b \\ -4 & b \end{pmatrix} \end{aligned}
Let’s examine matrix $B$ first.
Since $B$ is invertible (does have an inverse), then its determinant must not be zero.
\begin{aligned} \det(B) & \neq 0 \\ 2b(b)-b(-4) & \neq 0 \\ 2b^2+4b & \neq 0 \\ 2b(b+2) & \neq 0 \end{aligned}
We have $2b \neq 0$ or $b+2 \neq 0$, which means $b \neq 0$ or $b \neq -2$.
In matrix, the following determinant properties hold.
\boxed{\begin{aligned} \det(A^{-1}) & = \dfrac{1}{\det(A)} \\ \det(AB) & = \det A \cdot \det B \end{aligned}}
Therefore, $\det(ABA^{-1}B^{-1}) > 0$ can be rewritten into
\begin{aligned} \det(A) \cdot \det(B) \cdot \det(A^{-1}) \cdot \det(B^{-1}) & > 0 \\ \Leftrightarrow \cancel{\det(A)} \cdot \bcancel{\det(B)} \cdot \dfrac{1}{\cancel{\det(A)}} \cdot \dfrac{1}{\bcancel{\det(B)}} & > 0 \\ \Leftrightarrow 1 & > 0 \end{aligned}The last proposition is true in value that $1 > 0$.
Thus, real numbers $b$ except $0$ or $-2$ always satisfy the inequation. In other words, all real numbers $b$ that satisfy $\det(ABA^{-1}B^{-1}) > 0$ are $b \neq -2$ or $b \neq 0$.

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Problem Number 22
If $B = \begin{pmatrix} 1 & 9 & 7 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$, then $B^{12345678765432} = \cdots \cdot$
A. $\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}$
B. $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$
C. $\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$
D. $\begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix}$
E. $\begin{pmatrix} 1 & 9 & 7 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$

Solution

Try to observe the second power of $B$.
\begin{aligned} B^2 & = B \cdot B \\ & = \begin{pmatrix} 1 & 9 & 7 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 9 & 7 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \\ & = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I \end{aligned}
We got an identity matrix $I$.
This makes $A^3 = A^2 \cdot A = I \cdot A = A$.
Hence, we can conclude that $A^{n} = A$ for every odd number $n$ and $A^n = I$ for every even number $n$.
Since $12345678765432$ is an even number, then obviously $B^{12345678765432} = I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$.

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Essay Section

Problem Number 1
Evaluate all the values of $a, b, c$ if it is given that $A$ be a symmetric matrix with
$A = \begin{pmatrix} 2 & a- 2b + 2c & 2a+b+c \\ 3 & 5 & a + c \\ 0 &-2 & 7 \end{pmatrix}$.

Solution

Since $A$ be a symmetric matrix, then $A$ will be equal to its transpose.
\begin{aligned} A & = A^T \\ \begin{pmatrix} 2 & a- 2b + 2c & 2a+b+c \\ 3 & 5 & a + c \\ 0 &-2 & 7 \end{pmatrix} & = \begin{pmatrix} 2 & 3 & 0 \\ a-2b + 2c & 5 &-2 \\ 2a+b+c & a+c & 7\end{pmatrix} \end{aligned}Hence, we obtain a system of linear equations as follows.
$\begin{cases} a- 2b + 2c & = 3 && (1) \\ 2a + b + c &= 0 && (2) \\ a + c & =-2 && (-3) \end{cases}$
Eliminate $b$ from the equation $1$ and $2$:
\begin{aligned} \! \begin{aligned} a-2b+2c & = 3 \\ 2a+b+c & = 0 \end{aligned} \left| \! \begin{aligned} \times 1 \\ \times 2 \end{aligned} \right| & \! \begin{aligned} a-2b+2c & = 3 \\~4a +2b+2c & = 0 \end{aligned} \\ & \rule{3.5 cm}{0.6pt} + \\ & \! \begin{aligned} 5a + 4c & = 3~~~~~~(4) \end{aligned} \end{aligned}
Find the value of $a$ and $c$ by using equations $3$ and $4$.
\begin{aligned} \! \begin{aligned} a + c& =-2 \\ 5a + 4c & = 3\end{aligned} \left| \! \begin{aligned} \times 4 \\ \times 1 \end{aligned} \right| & \! \begin{aligned}~4a + 4c & =-8 \\~ 5a + 4c & = 3 \end{aligned} \\ & \rule{2.5 cm}{0.6pt}- \\ & \! \begin{aligned} a & = 11 \end{aligned} \end{aligned}
For $a = 11$, we have
\begin{aligned} a + c & =-2 \\ 11 + c & =-2 \\ c & =-13 \end{aligned}
Substitute the value of $a$ and $c$ to one of three initial equations, for instance, equation $1$.
\begin{aligned} a-2b + 2c & = 3 \\ 11- 2b + 2(-13) & = 3 \\-15-2b & = 3 \\-2b & = 18 \\ b & =-9 \end{aligned}
Thus, the values of $a, b$, and $c$ are $\color{red}{a = 11, b =-9}$, and $\color{red}{c =-13}$, respectively.

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Problem Number 2
Show that the determinant of matrix
$\begin{pmatrix} \sin \theta & \cos \theta & 0 \\-\cos \theta & \sin \theta & 0 \\ \sin \theta \cos \theta & \sin \theta + \cos \theta & 1 \end{pmatrix}$
doesn’t depend on the value of $\theta$.

Solution

Let
$X = \begin{pmatrix} \sin \theta & \cos \theta & 0 \\-\cos \theta & \sin \theta & 0 \\ \sin \theta- \cos \theta & \sin \theta + \cos \theta & 1 \end{pmatrix}$
By applying cofactor expansion along the third column, we have
\begin{aligned} \det(X) & = 0 \begin{vmatrix}-cos \theta & \sin \theta \\ \sin \theta- \cos \theta & \sin \theta + \cos \theta \end{vmatrix} \\ &- 0 \begin{vmatrix} \sin \theta & \cos \theta \\ \sin \theta- \cos \theta & \sin \theta + \cos \theta \end{vmatrix} + 1 \begin{vmatrix} \sin \theta & \cos \theta \\-\cos \theta & \sin \theta \end{vmatrix} \\ & = 0- 0 + (\sin^2 \theta- (-\cos^2 \theta)) \\ & = \sin^2 \theta + \cos^2 \theta = 1 \end{aligned}We found that the determinant is always to $1$. In other words, the determinant doesn’t depend on the value of $\theta$. Any $\theta$ will yield the same value.
Note:
Recall the Pythagorean Identity in trigonometry as follows.
$\boxed{\sin^2 \theta + \cos^2 \theta = 1}$

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Problem Number 3
Show that $(ABCD)^{-1} = D^{-1} \cdot C^{-1} \cdot B^{-1} \cdot A^{-1}$.

Solution

You should use the definition of the inverse of a matrix as follows.
$\boxed{AA^{-1} = I}$
with $I$ denotes an identity matrix.
Following this, we have
$(ABCD)^{-1}ABCD = I$
Multiply both side by $D^{-1}$:
\begin{aligned} (ABCD)^{-1}ABCDD^{-1} & = ID^{-1} \\ (ABCD)^{-1}ABC & = D^{-1} \end{aligned}
Multiply both side by $C^{-1}$:
\begin{aligned} (ABCD)^{-1}ABCC^{-1} & = D^{-1}C^{-1} \\ (ABCD)^{-1}AB & = D^{-1}C^{-1} \end{aligned}
Multiply both side by $B^{-1}$:
\begin{aligned} (ABCD)^{-1}ABB^{-1} & = D^{-1}C^{-1}B^{-1} \\ (ABCD)^{-1}A & = D^{-1}C^{-1}B^{-1} \end{aligned}
Finally, multiply both side by $A^{-1}$:
\begin{aligned} (ABCD)^{-1}AA^{-1} & = D^{-1}C^{-1}B^{-1}A^{-1} \\ (ABCD)^{-1} & = D^{-1}C^{-1}B^{-1}A^{-1} \end{aligned}
Thus, it is showed that $(ABCD)^{-1} = D^{-1} \cdot C^{-1} \cdot B^{-1} \cdot A^{-1}$.

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Problem Number 4
Evaluate the determinant of the following matrix.
$A = \begin{pmatrix} n & n + 1 & n + 2 \\ n + 1 & n + 2 & n + 3 \\ n + 2 & n+3 & n+4 \end{pmatrix}$

Solution

We will apply Elementary Matrix Operations, specifically Row Elementary Operations to track the determinant.
Given $A = \begin{pmatrix} n & n + 1 & n + 2 \\ n + 1 & n + 2 & n + 3 \\ n + 2 & n+3 & n+4 \end{pmatrix}$.
Obeying the elementary matrix operations, the steps are as follows.
$|A| = \begin{vmatrix} n & n + 1 & n + 2 \\ n + 1 & n + 2 & n + 3 \\ n + 2 & n+3 & n+4 \end{vmatrix}$
Multiply the elements in the second row by $-1$ then add that result to the third row. Also multiply the elements in the first row by $-1$ then add that result to the second row.

$|A| = \begin{vmatrix} n & n+1 & n+2 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix}$
Multiply the elements in the second row by $-1$ then add that result to the third row.
$|A| = \begin{vmatrix} n & n+1 & n+2 \\ 1 & 1 & 1 \\ 0 & 0 & 0 \end{vmatrix}$
We found that all entries in one row are being zero. It can be cocluded that the determinant of the matrix is $\boxed{|A| = 0}$

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Problem Number 5
Evaluate the determinant of the following matrix.
$A = \begin{pmatrix} n^2 & (n + 1)^2 & (n + 2)^2 \\ (n + 1)^2 & (n + 2)^2 & (n + 3)^2 \\ (n + 2)^2 & (n+3)^2 & (n+4)^2 \end{pmatrix}$

Solution

We will apply Elementary Matrix Operations to track the determinant.
Given $A = \begin{pmatrix} n^2 & (n + 1)^2 & (n + 2)^2 \\ (n + 1)^2 & (n + 2)^2 & (n + 3)^2 \\ (n + 2)^2 & (n+3)^2 & (n+4)^2 \end{pmatrix}$.
Obeying the elementary matrix operations, the steps are as follows.
$|A| = \begin{vmatrix} n^2 & (n + 1)^2 & (n + 2)^2 \\ (n + 1)^2 & (n + 2)^2 & (n + 3)^2 \\ (n + 2)^2 & (n+3)^2 & (n+4)^2 \end{vmatrix}$.
Multiply the elements in the first column by $-1$ then add that result to the second column and also the third column.
$|A| = \begin{vmatrix} n^2 & 2n+1 & 4n+4 \\ (n + 1)^2 & 2n+3 & 4n + 8 \\ (n + 2)^2 & 2n+5 & 4n+12 \end{vmatrix}$
Multiply the elements in the first row by $-1$ then add that result to the second and third rows.
$|A| = \begin{vmatrix} n^2 & 2n+1 & 4n+4 \\ 2n+1 & 2 & 4 \\ 4n+4 & 4 & 8 \end{vmatrix}$
Factorize out $4$ from the third column.
$|A| = 4\begin{vmatrix} n^2 & 2n+1 & n+1 \\ 2n+1 & 2 & 1 \\ 4n+4 & 4 & 2 \end{vmatrix}$
Factorize out $2$ from the third row.
$|A| = 8\begin{vmatrix} n^2 & 2n+1 & n+1 \\ 2n+1 & 2 & 1 \\ 2n+2 & 2 & 1 \end{vmatrix}$
Multiply the elements in the second row by $-1$ then add them to the third row.
$|A| = 8\begin{vmatrix} n^2 & 2n+1 & n+1 \\ 2n+1 & 2 & 1 \\ 1 & 0 & 0 \end{vmatrix}$
Apply a cofactor expansion along the third row.
\begin{aligned}|A| & = 8 \cdot 1 \cdot \begin{vmatrix} 2n+1 & n+1 \\ 2 & 1 \end{vmatrix} \\ & = 8 \cdot [(2n+1)-2(n+1)] \\ & = 8(-1) = -8 \end{aligned}
Thus, the determinant of matrix $A$ is $\boxed{|A| = -8}$

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