$\left[\begin{array}{ccc}0 & 1 & -3 \\ 4 & 1 & 0 \\ 1 & 0 & 5\end{array}\right]$

View step-by-step help

Home

Math Problems

Algebra 1

Domain and range of relations

Full solution

\left[\begin{array}{ccc}0 & 1 & -3 \\ 4 & 1 & 0 \\ 1 & 0 & 5\end{array}\right]

Use Sarrus Rule: To find the determinant of a $3\times3$ matrix, we use the rule of Sarrus or the cofactor expansion method. Let's use the rule of Sarrus.
Write Extended Matrix: The matrix is $\begin{bmatrix} 0 & 1 & -3 \ 4 & 1 & 0 \ 1 & 0 & 5 \end{bmatrix}$ . Write down the first two columns of the matrix to the right to apply the rule of Sarrus.
Calculate Diagonal Products: The extended matrix looks like this: $\left[\begin{array}{ccccc} 0 & 1 & -3 & 0 & 1 \ 4 & 1 & 0 & 4 & 1 \ 1 & 0 & 5 & 1 & 0 \end{array}\right]$ .
Add Diagonal Products: Now, add the products of the diagonals going from the top left to the bottom right: $0\times 1\times 5$ + $1\times 0\times 1$ + \-3\times 4\times 0.
Subtract Diagonal Products: The sum of the products is: $(0) + (0) + (0) = 0$ .
Calculate Determinant: Next, subtract the products of the diagonals going from the bottom left to the top right: $(1 \times 1 \times 1) + (0 \times 0 \times 5) + (-3 \times 4 \times 0)$ .
Calculate Determinant: Next, subtract the products of the diagonals going from the bottom left to the top right: $(1 \times 1 \times 1) + (0 \times 0 \times 5) + (-3 \times 4 \times 0)$ .The sum of these products is: $(1) + (0) + (0) = 1$ .
Calculate Determinant: Next, subtract the products of the diagonals going from the bottom left to the top right: $(1\times1\times1) + (0\times0\times5) + (-3\times4\times0)$ .The sum of these products is: $(1) + (0) + (0) = 1$ .Subtract the second sum from the first sum to get the determinant: $0 - 1$ .
Calculate Determinant: Next, subtract the products of the diagonals going from the bottom left to the top right: $1\times 1\times 1$ + $0\times 0\times 5$ + \-3\times 4\times 0. The sum of these products is: $1$ + $0$ + $0$ = $1$ . Subtract the second sum from the first sum to get the determinant: $0 - 1$ . The determinant of the matrix is \-1.