No, $\newline$ Tugas $\newline$ Date $\newline$ Tentukan determinan dari matriks berileut ini menggunakan cara ckspansi $\newline$ $A=\left[\begin{array}{cccc} 2 & 0 & -3 & 6 \\ 0 & 5 & 2 & 3 \\ -2 & 3 & -1 & 4 \\ 3 & 6 & 0 & 5 \end{array}\right]$

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Mean, median, mode, and range: find the missing number

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Q. No,

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Tugas

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Date

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Tentukan determinan dari matriks berileut ini menggunakan cara ckspansi

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A=\left[\begin{array}{cccc} 2 & 0 & -3 & 6 \\ 0 & 5 & 2 & 3 \\ -2 & 3 & -1 & 4 \\ 3 & 6 & 0 & 5 \end{array}\right]

Write Matrix A: First, let's write down the matrix $A$ for clarity: $\newline$ $A = \begin{bmatrix} 2 & 0 & -3 & 6 \ 0 & 5 & 2 & 3 \ -2 & 3 & -1 & 4 \ 3 & 6 & 0 & 5 \end{bmatrix}$ $\newline$ To find the determinant of a $4 \times 4$ matrix, we can use the expansion method by expanding along the first row (or any row/column that we find convenient). In this case, the first row has a zero which will make calculations a bit easier.
Expand Along First Row: We will expand along the first row. The determinant of $A$ , denoted as $\text{det}(A)$ , is given by: $\newline$ $\text{det}(A) = a_{11} \times C_{11} - a_{12} \times C_{12} + a_{13} \times C_{13} - a_{14} \times C_{14}$ $\newline$ where $a_{ij}$ is the element in the $i$ th row and $j$ th column, and $C_{ij}$ is the cofactor of $a_{ij}$ .
Calculate Cofactors: Calculating the cofactors for the first row: $\newline$ $C_{11}$ is the determinant of the $3 \times 3$ matrix obtained by removing the first row and first column from $A$ : $\newline$ $C_{11} = \begin{vmatrix} 5 & 2 & 3 \ 3 & -1 & 4 \ 6 & 0 & 5 \end{vmatrix}$ $\newline$ $C_{12}$ is zero since $a_{12}$ is zero, so we don't need to calculate it. $\newline$ $C_{13}$ is the determinant of the $3 \times 3$ matrix obtained by removing the first row and third column from $A$ : $\newline$ $C_{13} = \begin{vmatrix} 0 & 5 & 3 \ -2 & 3 & 4 \ 3 & 6 & 5 \end{vmatrix}$ $\newline$ $C_{14}$ is the determinant of the $3 \times 3$ matrix obtained by removing the first row and fourth column from $A$ : $\newline$ $C_{14} = \begin{vmatrix} 0 & 5 & 2 \ -2 & 3 & -1 \ 3 & 6 & 0 \end{vmatrix}$
Calculate Determinants: Now we calculate the determinant of each $3$ x $3$ matrix: $\newline$ For $C_{11}$ : $\newline$ $\text{det}(C_{11}) = 5 \times ((-1)\times5 - 0\times4) - 2 \times (3\times5 - 0\times4) + 3 \times (3\times0 - (-1)\times6)$ $\newline$ $\text{det}(C_{11}) = 5 \times (-5) - 2 \times 15 + 3 \times 6$ $\newline$ $\text{det}(C_{11}) = -25 - 30 + 18$ $\newline$ $\text{det}(C_{11}) = -37$ $\newline$ For $C_{13}$ (we only need to calculate this if necessary, depending on the value of $a_{13}$ ): $\newline$ $\text{det}(C_{13}) = 0 \times ((3)\times5 - (6)\times4) - 5 \times ((-2)\times5 - (3)\times4) + 3 \times ((-2)\times6 - (3)\times3)$ $\newline$ $\text{det}(C_{13}) = 0 - 5 \times (-10 - 12) + 3 \times (-12 - 9)$ $\newline$ $\text{det}(C_{13}) = 0 + 5 \times (-22) - 3 \times 21$ $\newline$ $\text{det}(C_{11}) = 5 \times ((-1)\times5 - 0\times4) - 2 \times (3\times5 - 0\times4) + 3 \times (3\times0 - (-1)\times6)$ $0$ $\newline$ $\text{det}(C_{11}) = 5 \times ((-1)\times5 - 0\times4) - 2 \times (3\times5 - 0\times4) + 3 \times (3\times0 - (-1)\times6)$ $1$ $\newline$ For $\text{det}(C_{11}) = 5 \times ((-1)\times5 - 0\times4) - 2 \times (3\times5 - 0\times4) + 3 \times (3\times0 - (-1)\times6)$ $2$ (we only need to calculate this if necessary, depending on the value of $\text{det}(C_{11}) = 5 \times ((-1)\times5 - 0\times4) - 2 \times (3\times5 - 0\times4) + 3 \times (3\times0 - (-1)\times6)$ $3$ ): $\newline$ $\text{det}(C_{11}) = 5 \times ((-1)\times5 - 0\times4) - 2 \times (3\times5 - 0\times4) + 3 \times (3\times0 - (-1)\times6)$ $4$ $\newline$ $\text{det}(C_{11}) = 5 \times ((-1)\times5 - 0\times4) - 2 \times (3\times5 - 0\times4) + 3 \times (3\times0 - (-1)\times6)$ $5$ $\newline$ $\text{det}(C_{11}) = 5 \times ((-1)\times5 - 0\times4) - 2 \times (3\times5 - 0\times4) + 3 \times (3\times0 - (-1)\times6)$ $6$ $\newline$ $\text{det}(C_{11}) = 5 \times ((-1)\times5 - 0\times4) - 2 \times (3\times5 - 0\times4) + 3 \times (3\times0 - (-1)\times6)$ $7$ $\newline$ $\text{det}(C_{$14$}) = $-87$
Combine Results: Now we can combine these results to find the determinant of A:$\newline$det(A) = $2 \times (-37) - 0 \times (C_{12}$, which we don't need to calculate) $- 3 \times (-173) + 6 \times (-87)$$\newline$det(A) = $-74 + 0 + 519 - 522$$\newline$det(A) = $-74 + 519 - 522$$\newline$det(A) = $445 - 522$$\newline$det(A) = $-77$