Question

1. calculate the determinant.

\\(\

$$\begin{bmatrix}5 & 2 \\\\ 3 & 1\\end{bmatrix}$$

\\)
\\(-11\\)
\\(1\\)
\\(-1\\)
\\(11\\)

1. calculate the determinant.

\\(\

$$\begin{bmatrix}0 & 2 & -1 \\\\ -4 & 1 & 0 \\\\ -2 & 3 & 4\\end{bmatrix}$$

\\)
\\(42\\)
\\(46\\)
\\(-42\\)
\\(-22\\)

Explanation:

Question 6

Step1: Recall determinant formula for 2x2 matrix

For a matrix

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

, the determinant is $ad - bc$.

Step2: Apply formula to given matrix

Given matrix

$$\begin{bmatrix}5&2\\3&1\end{bmatrix}$$

, $a = 5$, $b = 2$, $c = 3$, $d = 1$.
Determinant = $5\times1 - 2\times3 = 5 - 6 = -1$.

Step1: Recall determinant formula for 3x3 matrix

For a matrix

$$\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}$$

, the determinant is $a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31})$.

Step2: Apply formula to given matrix

Given matrix

$$\begin{bmatrix}0&2&-1\\-4&1&0\\-2&3&4\end{bmatrix}$$

, $a_{11} = 0$, $a_{12} = 2$, $a_{13} = -1$, $a_{21} = -4$, $a_{22} = 1$, $a_{23} = 0$, $a_{31} = -2$, $a_{32} = 3$, $a_{33} = 4$.
Determinant = $0\times(1\times4 - 0\times3)-2\times((-4)\times4 - 0\times(-2))+(-1)\times((-4)\times3 - 1\times(-2))$
= $0 - 2\times(-16 - 0)+(-1)\times(-12 + 2)$
= $0 + 32+(-1)\times(-10)$
= $32 + 10 = 42$. Wait, no, wait: Wait, let's recalculate:

Wait, step by step:

First term: $a_{11}(a_{22}a_{33}-a_{23}a_{32}) = 0\times(1\times4 - 0\times3) = 0$.

Second term: $-a_{12}(a_{21}a_{33}-a_{23}a_{31}) = -2\times((-4)\times4 - 0\times(-2)) = -2\times(-16 - 0) = -2\times(-16) = 32$.

Third term: $a_{13}(a_{21}a_{32}-a_{22}a_{31}) = -1\times((-4)\times3 - 1\times(-2)) = -1\times(-12 + 2) = -1\times(-10) = 10$.

Now sum all terms: $0 + 32 + 10 = 42$. Wait, but the options have 42, 46, -42, -22. Wait, maybe I made a mistake. Wait, let's check again.

Wait, the formula is $a_{11}(a_{22}a_{33}-a_{23}a_{32}) - a_{12}(a_{21}a_{33}-a_{23}a_{31}) + a_{13}(a_{21}a_{32}-a_{22}a_{31})$.

So for $a_{12}$, it's $-a_{12}(...)$, so:

$a_{11}=0$, $a_{12}=2$, $a_{13}=-1$

$a_{21}=-4$, $a_{22}=1$, $a_{23}=0$

$a_{31}=-2$, $a_{32}=3$, $a_{33}=4$

First part: $0\times(1\times4 - 0\times3) = 0$

Second part: $-2\times((-4)\times4 - 0\times(-2)) = -2\times(-16 - 0) = -2\times(-16) = 32$

Third part: $-1\times((-4)\times3 - 1\times(-2)) = -1\times(-12 + 2) = -1\times(-10) = 10$

Total: $0 + 32 + 10 = 42$. So the determinant is 42.

Answer:

-1 (Option: -1)

Question 7