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Calculus 3 : Vectors and Vector Operations

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Example Questions

Example Question #751 : Vectors And Vector Operations

Find the determinant of the 3x3 matrix $\begin{bmatrix} 1&0 &1 \\ 12&2 &4 \\ 5&10 &3 \end{bmatrix}$ $\displaystyle \begin{bmatrix} 1&0 &1 \\ 12&2 &4 \\ 5&10 &3 \end{bmatrix}$

Possible Answers:

100 $\displaystyle 100$

$\displaystyle 76$

$\displaystyle 95$

$\displaystyle 82$

Correct answer:

$\displaystyle 76$

Explanation:

To find the determinant of a 3x3 matrix $M=\begin{bmatrix} x&y &z \\ a_1&a_2 &a_3 \\ b_1& b_2&b_3 \end{bmatrix}$ $\displaystyle M=\begin{bmatrix} x&y &z \\ a_1&a_2 &a_3 \\ b_1& b_2&b_3 \end{bmatrix}$ , you use the formula

$\displaystyle determinant(M)=x(a_2b_3-b_2a_3)-y(a_1b_3-b_1a_3)+z(a_1b_2-b_1a_2)$

Using the matrix from the problem statement, we get

$\displaystyle 1(6-40)-0(36-20)+1(120-10)=76$

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Example Question #752 : Vectors And Vector Operations

Find the determinant of the matrix $\begin{bmatrix} \hat{i}&\hat{j} &\hat{k} \\ 1&-5 &3 \\ 2& 7& 0 \end{bmatrix}$ $\displaystyle \begin{bmatrix} \hat{i}&\hat{j} &\hat{k} \\ 1&-5 &3 \\ 2& 7& 0 \end{bmatrix}$

Possible Answers:

$\left \langle -21,6,17\right \rangle$ $\displaystyle \left \langle -21,6,17\right \rangle$

$\left \langle -20,6,14\right \rangle$ $\displaystyle \left \langle -20,6,14\right \rangle$

$\left \langle -21,9,17\right \rangle$ $\displaystyle \left \langle -21,9,17\right \rangle$

$\left \langle -21,-6,17\right \rangle$ $\displaystyle \left \langle -21,-6,17\right \rangle$

Correct answer:

$\left \langle -21,6,17\right \rangle$ $\displaystyle \left \langle -21,6,17\right \rangle$

Explanation:

To find the determinant of a 3x3 matrix $m=\begin{bmatrix} x& y & z\\ a_1&a_2 &a_3 \\ b_1& b_2& b_3 \end{bmatrix}$ $\displaystyle m=\begin{bmatrix} x& y & z\\ a_1&a_2 &a_3 \\ b_1& b_2& b_3 \end{bmatrix}$ , we use the formula

$\displaystyle determinant(m)=x(a_2b_3-b_2a_3)-y(a_1b_3-b_1a_3)+z(a_1b_2-b_1a_2)$

Using the matrix from the problem statement, we get

$\hat{i}(0-21)-\hat{j}(0-6)+\hat{k}(7+10)=\left \langle -21,6,17\right \rangle$ $\displaystyle \hat{i}(0-21)-\hat{j}(0-6)+\hat{k}(7+10)=\left \langle -21,6,17\right \rangle$

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Example Question #751 : Vectors And Vector Operations

Calculate the determinant of Matrix $\displaystyle A$ .

$A=\begin{bmatrix} 1 & 2&3 \\ 0& 2& 1\\ 3& 4& 5 \end{bmatrix}$ $\displaystyle A=\begin{bmatrix} 1 & 2&3 \\ 0& 2& 1\\ 3& 4& 5 \end{bmatrix}$

Possible Answers:

$\det(A)=-2$ $\displaystyle \det(A)=-2$

$\det(A)=1$ $\displaystyle \det(A)=1$

$\det(A)=6$ $\displaystyle \det(A)=6$

$\det(A)=-6$ $\displaystyle \det(A)=-6$

$\det(A)=0$ $\displaystyle \det(A)=0$

Correct answer:

$\det(A)=-6$ $\displaystyle \det(A)=-6$

Explanation:

In order to find the determinant of $\displaystyle A$ , we first need to copy down the first two columns into columns 4 and 5.

$A=\begin{bmatrix} 1 & 2&3 & 1 &2\\ 0& 2& 1&0 &2\\ 3& 4& 5 &3 &4\end{bmatrix}$ $\displaystyle A=\begin{bmatrix} 1 & 2&3 & 1 &2\\ 0& 2& 1&0 &2\\ 3& 4& 5 &3 &4\end{bmatrix}$

The next step is to multiply the down diagonals.

$A_d=1\cdot2\cdot5+2\cdot1\cdot 3+3\cdot0\cdot4=10+6+0=16$ $\displaystyle A_d=1\cdot2\cdot5+2\cdot1\cdot 3+3\cdot0\cdot4=10+6+0=16$

The next step is to multiply the up diagonals.

$A_u=3\cdot2\cdot3+4\cdot1\cdot1+5\cdot0\cdot2=18+4=22$ $\displaystyle A_u=3\cdot2\cdot3+4\cdot1\cdot1+5\cdot0\cdot2=18+4=22$

The last step is to substract A_u $\displaystyle A_u$ from A_d $\displaystyle A_d$ .

$\det(A)=A_d-A_u=16-22=-6$ $\displaystyle \det(A)=A_d-A_u=16-22=-6$

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Example Question #753 : Vectors And Vector Operations

Calculate the determinant of Matrix $\displaystyle A$ .

$A=\begin{bmatrix} 1 & 2&3 \\ 0& 2& 1\\ 3& 4& 5 \end{bmatrix}$ $\displaystyle A=\begin{bmatrix} 1 & 2&3 \\ 0& 2& 1\\ 3& 4& 5 \end{bmatrix}$

Possible Answers:

$\det(A)=1$ $\displaystyle \det(A)=1$

$\det(A)=-2$ $\displaystyle \det(A)=-2$

$\det(A)=0$ $\displaystyle \det(A)=0$

$\det(A)=-6$ $\displaystyle \det(A)=-6$

$\det(A)=6$ $\displaystyle \det(A)=6$

Correct answer:

$\det(A)=-6$ $\displaystyle \det(A)=-6$

Explanation:

In order to find the determinant of $\displaystyle A$ , we first need to copy down the first two columns into columns 4 and 5.

$A=\begin{bmatrix} 1 & 2&3 & 1 &2\\ 0& 2& 1&0 &2\\ 3& 4& 5 &3 &4\end{bmatrix}$ $\displaystyle A=\begin{bmatrix} 1 & 2&3 & 1 &2\\ 0& 2& 1&0 &2\\ 3& 4& 5 &3 &4\end{bmatrix}$

The next step is to multiply the down diagonals.

$A_d=1\cdot2\cdot5+2\cdot1\cdot 3+3\cdot0\cdot4=10+6+0=16$ $\displaystyle A_d=1\cdot2\cdot5+2\cdot1\cdot 3+3\cdot0\cdot4=10+6+0=16$

The next step is to multiply the up diagonals.

$A_u=3\cdot2\cdot3+4\cdot1\cdot1+5\cdot0\cdot2=18+4=22$ $\displaystyle A_u=3\cdot2\cdot3+4\cdot1\cdot1+5\cdot0\cdot2=18+4=22$

The last step is to substract A_u $\displaystyle A_u$ from A_d $\displaystyle A_d$ .

$\det(A)=A_d-A_u=16-22=-6$ $\displaystyle \det(A)=A_d-A_u=16-22=-6$

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Calculus 3 : Vectors and Vector Operations

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #751 : Vectors And Vector Operations

Example Question #752 : Vectors And Vector Operations

Example Question #751 : Vectors And Vector Operations

Example Question #753 : Vectors And Vector Operations

All Calculus 3 Resources

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