Question

6 Solve for x and y: 3x + 4y = 9, 2x + 3y =8
x = 7.5, y = - 4. 5
x = 7.0, y = - 4. 5
x = 4.5, y = - 7. 5
x = 7.5, y = - 4. 1

7 If A.x=,where A=[Math Processing Error],determine the eigen values of the matrix A, and an eigen vector corresponding to each eigen value. If [Math Processing Error],what is b
{0,1,0}
{3,0,2}
{2,0,1}
{0,1,1}

8 If A.x=[Math Processing Error],where A=[Math Processing Error],determine the eigen values of the matrix A, and an eigen vector corresponding to each eigen value. If [Math Processing Error],what is c
{2,3,0}
{-2,1,1}
{2,1,1}
{3,2,6}

9 Solve the set of linear equations by the matrix method : a+3b+2c=3 , 2a-b-3c= -8, 5a+2b+c=9. Sove for c
2
1
5
7

10 If A.x=[Math Processing Error],where A=[Math Processing Error],determine the eigen values of the matrix A, and an eigen vector corresponding to each eigen value. If [Math Processing Error],what is a
{-2,1,0}
{3,5,2}
{1,0,0}
{2,1,4}

Answer 1

Answer on Question #58059 – Math – Linear Algebra

Question

6. Solve for $x$ and $y$: $3x + 4y = 9$, $2x + 3y = 8$

$$x = 7.5, y = -4.5$$ $$x = 7.0, y = -4.5$$ $$x = 4.5, y = -7.5 \quad x = 7.5, y = -4.1$$

Solution

$$\left\{ \begin{array}{l} 3x + 4y = 9 \\ 2x + 3y = 8 \end{array} \right.$$ $$3(3x + 4y) - 4(2x + 3y) = 3(9) - 4(8)$$ $$x = -5 \rightarrow y = \frac{9 - 3 \cdot (-5)}{4} = 6$$

Answer: $x = -5, y = 6$.

Question

9. Solve the set of linear equations by the matrix method: $a + 3b + 2c = 3$, $2a - b - 3c = -8$, $5a + 2b + c = 9$. Solve for $c$

2

1

5

7

Solution

$$\left\{ \begin{array}{l} a + 3b + 2c = 3 \\ 2a - b - 3c = -8 \\ 5a + 2b + c = 9 \end{array} \right. \to \left( \begin{array}{rrr} 1 & 3 & 2 & 3 \\ 2 & -1 & -3 - 8 \\ 5 & 2 & 1 & 9 \end{array} \right)$$ $$\left( \begin{array}{rrr} 1 & 3 & 2 & 3 \\ 2 & -1 & -3 -8 \\ 5 & 2 & 1 & 9 \end{array} \right) R_2 = r_2 - 2r_1 \left( \begin{array}{rrr} 1 & 3 & 2 & 3 \\ 0 & -7 & -7 -14 \\ 5 & 2 & 1 & 9 \end{array} \right) R_3 = r_3 - 5r_1 \left( \begin{array}{rrr} 1 & 3 & 2 & 3 \\ 0 & -7 & -7 -14 \\ 0 & -13 & -9 -6 \end{array} \right)$$ $$\left( \begin{array}{rrr} 1 & 3 & 2 & 3 \\ 0 & -7 & -7 -14 \\ 0 & -13 & -9 -6 \end{array} \right) R_3 = 7r_3 - 13r_2 \left( \begin{array}{rrr} 1 & 3 & 2 & 3 \\ 0 & -7 & -7 -14 \\ 0 & 0 & 28140 \end{array} \right)$$ $$28c = 140 \rightarrow c = \frac{140}{28} = 5$$

Answer: 5.

Remark. Parts 7, 8, 10 are not properly written, therefore we can't solve them.

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