A $$ =\left(\begin{array}{lll}2 & 1 & 0 \ 1 & 2 & 1 \ 0 & 1 & 2\end{array} ight) \quad B=\left(\begin{array}{cccc}2 & -3 & -1 & 0 \ 3 & 0 & 0 & 0 \ 4 & 3 & -6 & 4 \ 5 & -1 & 9 & 1\end{array} ight) $$ 1. Tentukan determinan matriks $$ A $$ dengan metode Sarrus. 2. Tentukan determinan matriks A dengan ekspansi kofaktor terhadap baris ke-3 3. Tentukan determinan matriks A dengan ekspansi kofaktor terhadap kolom ke-2 4. Tentukan determinan transpose matriks A , ditulis: $$ \left|\mathrm{A}^{ op} ight| $$ 5. Tentukan determinan matriks B ordo $$ 4 imes 4 $$ (metode atau cara bebas)

Question

A $$ =\left(\begin{array}{lll}2 & 1 & 0 \ 1 & 2 & 1 \ 0 & 1 & 2\end{array}ight) \quad B=\left(\begin{array}{cccc}2 & -3 & -1 & 0 \ 3 & 0 & 0 & 0 \ 4 & 3 & -6 & 4 \ 5 & -1 & 9 & 1\end{array}ight) $$ 1. Tentukan determinan matriks $$ A $$ dengan metode Sarrus. 2. Tentukan determinan matriks A dengan ekspansi kofaktor terhadap baris ke-3 3. Tentukan determinan matriks A dengan ekspansi kofaktor terhadap kolom ke-2 4. Tentukan determinan transpose matriks A , ditulis: $$ \left|\mathrm{A}^{	op}ight| $$ 5. Tentukan determinan matriks B ordo $$ 4 	imes 4 $$ (metode atau cara bebas)

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1. We have
$$
A=\begin{pmatrix}2 & 1 & 0 \ 1 & 2 & 1 \ 0 & 1 & 2\end{pmatrix}.
$$
Using the Sarrus rule we extend the first two columns to the right:
$$
\begin{array}{ccc|cc}
2 & 1 & 0 & 2 & 1\[4mm]
1 & 2 & 1 & 1 & 2\[4mm]
0 & 1 & 2 & 0 & 1
\end{array}.
$$
The positive diagonals are:
$$
2\cdot2\cdot2,\quad 1\cdot1\cdot0,\quad 0\cdot1\cdot1,
$$
so
$$
P=2\cdot2\cdot2+1\cdot1\cdot0+0\cdot1\cdot1=8+0+0=8.
$$
The negative diagonals are:
$$
0\cdot2\cdot0,\quad 1\cdot1\cdot2,\quad 2\cdot1\cdot1,
$$
so
$$
N=0\cdot2\cdot0+1\cdot1\cdot2+2\cdot1\cdot1=0+2+2=4.
$$
Thus, the determinant is
$$
\det A=P-N=8-4=4.
$$

2. We now compute $$\det A$$ using cofactor expansion along the third row. The third row is $$[0,\ 1,\ 2]$$.

- For element $$a_{31}=0$$ (at position $$(3,1)$$):
  - The sign is $$(-1)^{3+1}=1$$.
  - The minor is
    $$
    \begin{pmatrix}1 & 0 \ 2 & 1\end{pmatrix},
    $$
    with $$\det = 1\cdot1-0\cdot2=1$$.
  - Contribution: $$0\cdot1\cdot1=0$$.

- For element $$a_{32}=1$$ (at position $$(3,2)$$):
  - The sign is $$(-1)^{3+2}=-1$$.
  - The minor is
    $$
    \begin{pmatrix}2 & 0 \ 1 & 1\end{pmatrix},
    $$
    with $$\det =2\cdot1-0\cdot1=2$$.
  - Contribution: $$1\cdot(-1)\cdot2=-2$$.

- For element $$a_{33}=2$$ (at position $$(3,3)$$):
  - The sign is $$(-1)^{3+3}=1$$.
  - The minor is
    $$
    \begin{pmatrix}2 & 1 \ 1 & 2\end{pmatrix},
    $$
    with $$\det =2\cdot2-1\cdot1=4-1=3$$.
  - Contribution: $$2\cdot1\cdot3=6$$.

Summing the contributions:
$$
\det A=0-2+6=4.
$$

3. Next, we expand along the second column. The entries in column 2 are $$[1,\ 2,\ 1]$$.

- For element $$a_{12}=1$$ (at position $$(1,2)$$):
  - The sign is $$(-1)^{1+2}=-1$$.
  - The minor is obtained by deleting row 1 and column 2:
    $$
    \begin{pmatrix}1 & 1 \ 0 & 2\end{pmatrix},
    $$
    with $$\det =1\cdot2-1\cdot0=2$$.
  - Contribution: $$1\cdot(-1)\cdot2=-2$$.

- For element $$a_{22}=2$$ (at position $$(2,2)$$):
  - The sign is $$(-1)^{2+2}=1$$.
  - The minor is obtained by deleting row 2 and column 2:
    $$
    \begin{pmatrix}2 & 0 \ 0 & 2\end{pmatrix},
    $$
    with $$\det =2\cdot2-0\cdot0=4$$.
  - Contribution: $$2\cdot1\cdot4=8$$.

- For element $$a_{32}=1$$ (at position $$(3,2)$$):
  - The sign is $$(-1)^{3+2}=-1$$.
  - The minor is obtained by deleting row 3 and column 2:
    $$
    \begin{pmatrix}2 & 0 \ 1 & 1\end{pmatrix},
    $$
    with $$\det =2\cdot1-0\cdot1=2$$.
  - Contribution: $$1\cdot(-1)\cdot2=-2$$.

Summing:
$$
\det A=-2+8-2=4.
$$

4. Since the determinant of a transpose equals the determinant of the matrix, we have:
$$
\left|\mathrm{A}^{\top}\right|=\det A=4.
$$

5. We now compute the determinant of the $$4\times 4$$ matrix
$$
B=\begin{pmatrix}
2 & -3 & -1 & 0 \
3 & 0 & 0 & 0 \
4 & 3 & -6 & 4 \
5 & -1 & 9 & 1
\end{pmatrix}.
$$
Notice that row 2 contains a single nonzero element (in the first column). Expanding along row 2:

- The only nonzero element is $$a_{21}=3$$ at position $$(2,1)$$. Its sign is $$(-1)^{2+1}=-1$$.

- The corresponding minor $$M_{21}$$ is obtained by deleting row 2 and column 1:
  $$
  M_{21}=\begin{pmatrix}
  -3 & -1 & 0 \
  3 & -6 & 4 \
  -1 & 9 & 1
  \end{pmatrix}.
  $$
  
We now compute $$\det M_{21}$$ using the formula for a $$3\times3$$ determinant:
$$
\det M_{21} = a_{11}(a_{22}a_{33}-a_{23}a_{32}) - a_{12}(a_{21}a_{33}-a_{23}a_{31}) + a_{13}(a_{21}a_{32}-a_{22}a_{31}),
$$
where the entries are:
$$
a_{11}=-3,\quad a_{12}=-1,\quad a_{13}=0,
$$
$$
a_{21}=3,\quad a_{22}=-6,\quad a_{23}=4,
$$
$$
a_{31}=-1,\quad a_{32}=9,\quad a_{33}=1.
$$
Calculating each part:

- First term:
  $$
  -3\big[(-6)(1)-4\cdot9\big] = -3\big[-6-36\big] = -3(-42)=126.
  $$
- Second term:
  $$
  -(-1)\big[3\cdot1-4\cdot(-1)\big] = 1\big[3+4\big]=7.
  $$
- Third term is zero since $$a_{13}=0$$.

Thus,
$$
\det M_{21} = 126+7 = 133.
$$
The contribution from row 2 is then:
$$
3\cdot(-1)\cdot133 = -399.
$$
So,
$$
\det B=-399.
$$

---

Final answers:

1. $$\det A=4$$.

2. Using cofactor expansion along the third row, $$\det A=4$$.

3. Using cofactor expansion along the second column, $$\det A=4$$.

4. $$\left|\mathrm{A}^{\top}\right|=4$$.

5. $$\det B=-399$$.
1. Determinant of matrix $$ A $$ is 4. 2. Determinant of matrix $$ A $$ using cofactor expansion along the third row is 4. 3. Determinant of matrix $$ A $$ using cofactor expansion along the second column is 4. 4. Determinant of the transpose of matrix $$ A $$ is 4. 5. Determinant of matrix $$ B $$ is -399.

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