Compute the determinant:
[tex]\[
\left[\begin{array}{ccc}
1-2 & 4 & 2 \\
1 & 1+1 & -5 \\
0 & 1 & 1-14
\end{array}\right]
\][/tex]



Answer :

Alright! Let's calculate the determinant of the given 3x3 matrix step-by-step. The matrix is:

[tex]\[ \begin{pmatrix} 1-2 & 4 & 2 \\ 1 & 1+1 & -5 \\ 0 & 1 & 1-14 \end{pmatrix} \][/tex]

First, we simplify each entry:

[tex]\[ \begin{pmatrix} -1 & 4 & 2 \\ 1 & 2 & -5 \\ 0 & 1 & -13 \end{pmatrix} \][/tex]

Now, to find the determinant, we can use the formula for the determinant of a 3x3 matrix:

[tex]\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \][/tex]

where the matrix [tex]\( A \)[/tex] is:

[tex]\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \][/tex]

For our matrix:

[tex]\[ \begin{pmatrix} -1 & 4 & 2 \\ 1 & 2 & -5 \\ 0 & 1 & -13 \end{pmatrix} \][/tex]

we can identify the variables as follows:
- [tex]\(a = -1\)[/tex]
- [tex]\(b = 4\)[/tex]
- [tex]\(c = 2\)[/tex]
- [tex]\(d = 1\)[/tex]
- [tex]\(e = 2\)[/tex]
- [tex]\(f = -5\)[/tex]
- [tex]\(g = 0\)[/tex]
- [tex]\(h = 1\)[/tex]
- [tex]\(i = -13\)[/tex]

Now we substitute these into the determinant formula:

[tex]\[ \text{det}(A) = -1 \cdot (2 \cdot -13 - (-5) \cdot 1) - 4 \cdot (1 \cdot -13 - (-5) \cdot 0) + 2 \cdot (1 \cdot 1 - 2 \cdot 0) \][/tex]

We simplify inside the parentheses first:

[tex]\[ 2 \cdot -13 - (-5) \cdot 1 = -26 + 5 = -21 \][/tex]

[tex]\[ 1 \cdot -13 - (-5) \cdot 0 = -13 \][/tex]

[tex]\[ 1 \cdot 1 - 2 \cdot 0 = 1 \][/tex]

Now substitute these values back into the equation:

[tex]\[ \text{det}(A) = -1 \cdot (-21) - 4 \cdot (-13) + 2 \cdot 1 \][/tex]

Now, calculate each term:

[tex]\[ -1 \cdot (-21) = 21 \][/tex]

[tex]\[ -4 \cdot (-13) = 52 \][/tex]

[tex]\[ 2 \cdot 1 = 2 \][/tex]

Add these results together:

[tex]\[ 21 + 52 + 2 = 75 \][/tex]

Hence, the determinant of the matrix is:

[tex]\[ \boxed{75} \][/tex]