Evaluate the determinant of the matrix. Expand by minors along the row or column that appears to make the computation easiest.
$[−10−101012−2]\left[ \begin{array} { c c c } - 1 & 0 & - 1 \\0 & 1 & 0 \\1 & 2 & - 2\end{array} \right]$

A) 3
B) $- 1$
C) $- 3$
D) $2$
E) $1$

Question

Evaluate the determinant of the matrix. Expand by minors along the row or column that appears to make the computation easiest.

[−10−101012−2]\left[ \begin{array} { c c c } - 1 & 0 & - 1 \\0 & 1 & 0 \\1 & 2 & - 2\end{array} \right]

A) 3
B)

- 1

C)

- 3

D)

2

E)

1

McClain Laster · Accepted Answer

Final Answer : A, Explanation :   Expanding along the second row, which contains two zeros, simplifies the calculation. The determinant is   0 ∗ ( − 1 ) 2 + 1 ∗ ∣ 0 − 1 2 − 2 ∣ + 1 ∗ ( − 1 ) 2 + 2 ∗ ∣ − 1 − 1 1 − 2 ∣ = 0 + 1 ∗ ( ( − 1 ) ∗ ( − 2 ) − ( − 1 ) ∗ 1 ) = 1 ∗ ( 2 + 1 ) = 3. 0*(-1)^{2+1}*\left|\begin{array}{cc}0  &  -1 \ 2  &  -2\end{array}ight| + 1*(-1)^{2+2}*\left|\begin{array}{cc}-1  &  -1 \ 1  &  -2\end{array}ight| = 0 + 1*((-1)*(-2) - (-1)*1) = 1*(2+1) = 3. 0 ∗ ( − 1 ) 2 + 1 ∗ ​ 0 2 ​ − 1 − 2 ​ ​ + 1 ∗ ( − 1 ) 2 + 2 ∗ ​ − 1 1 ​ − 1 − 2 ​ ​ = 0 + 1 ∗ (( − 1 ) ∗ ( − 2 ) − ( − 1 ) ∗ 1 ) = 1 ∗ ( 2 + 1 ) = 3.

Evaluate the determinant of the matrix. Expand by minors along the row or column that appears to make the computation easiest.
$[−10−101012−2]\left[ \begin{array} { c c c } - 1 & 0 & - 1 \\0 & 1 & 0 \\1 & 2 & - 2\end{array} \right]$

A) 3
B) $- 1$
C) $- 3$
D) $2$
E) $1$

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