Exam 11: Determinants

Solve by determinants: $x+2 y+z=5$ -2x+y=-5 y-z=-6

Calculate the currents $I_{1}$ and $I_{2}: 3.00 I_{1}+5.00\left(I_{1}-I_{2}\right)=12.0$ $10.0 I_{2}+5.00\left(I_{2}-I_{1}\right)=6.00$

Evaluate the determinant: $\left|\begin{array}{ccc}1.25 & 2.10 & 1.55 \\ 0.00 & -1.10 & 2.00 \\ 3.50 & 4.11 & 1.35\end{array}\right|$

Evaluate the determinant: $\left|\begin{array}{ccc}1.5 & 1.8 & -2.1 \\ -1.9 & 2.2 & 1.6 \\ -2.3 & 2.0 & 1.7\end{array}\right|$

Evaluate the determinant: $\left|\begin{array}{cccc}2 & -1 & 7 & 1 \\ 0 & 4 & -3 & 4 \\ -1 & 6 & -2 & 1 \\ 3 & 1 & 0 & -1\end{array}\right|$

Find the value of the second-order determinant: $\left|\begin{array}{cc}\frac{2}{3} & \frac{1}{3} \\ \frac{-1}{3} & \frac{-2}{3}\end{array}\right|$

Solve by determinants: $5 x+2 z=1$ 3y-2z=4 2x+3y=-1

Solve for $x$ and $y$ by determinants: $4 m x-2 n y=2$ $3 m x+5 n y=6$

Evaluate the determinant: $\left|\begin{array}{ccc}2 & 1 & -4 \\ -3 & 8 & 1 \\ 5 & 3 & 4\end{array}\right|$

Solve by determinants: $\frac{x}{4}+\frac{3 y}{5}=1$ $\frac{x}{2}-\frac{2 y}{3}=2$

Solve by determinants: $x+2 z-w+u=1$ $3 y+z+2 w+u=-2$ -x-2y-2z+3w+u=15 x+3z-w+u=2.5 $x+y+z+w+u=1$

Evaluate the determinant: $\left|\begin{array}{cccc}2 & 5 & 3 & 4 \\ 0 & 1 & 2 & -1 \\ 4 & 3 & 2 & 2 \\ 7 & 3 & 5 & 1\end{array}\right|$

Solve by determinants: $a+b-c-d+\mathrm{e}=-3$ a-b+d-e=3 a-c-d-e=2 b+c+d-e=4 a-b-c+d=2

Solve by determinants: $2 a+5 b-3 c-4 d+e=2$ 7a-3b+d-2e =2 a-3c-d-e =-6 3b+4c+5d-2e =-11 a-3b-4c+d =-6

Find the value of the second-order determinant: $\left|\begin{array}{cc}(x-2)^{2} & x^{2}+4 x+4 \\ x-2 & x+2\end{array}\right|$

Evaluate the determinant: $\left|\begin{array}{cccc}1 & -4 & 2 & 12 \\ 3 & 1 & 0 & 2 \\ 0 & 2 & 0 & 6 \\ 6 & 3 & 0 & 10\end{array}\right|$

Determine the coefficients $a, b$ , and $c$ which balance the chemical equation $a \mathrm{Fe}_{2} \mathrm{O}_{3}+b \mathrm{Al} \rightarrow c \mathrm{Al}_{2} \mathrm{O}_{3}+$ $10 \mathrm{Fe}$ . In other words, solve this system: Fe: $\quad 2 a+0 b=10$ O: $\quad 3 a+0 b=3 c$ Al: $\quad 0 a+b=2 c$

Solve by determinants: $\frac{1}{x}-\frac{2}{y}+\frac{1}{z}=3$ +-=9 -+=-9

Find the value of the second-order determinant: $\left|\begin{array}{cc}-4 & 3 \\ 6 & -2\end{array}\right|$

Solve by determinants: $\frac{6}{x}+\frac{4}{y}=3$ $\frac{-9}{x}+\frac{16}{y}=1$