Question

6 Solve the set of linear equations by Guassian elimination method : a+2b+3c=5, 3a-b+2c=8, 4a-6b-4c=-2. Find c
4
5
9
10

7 Solve the set of linear equations by the matrix method : a+3b+2c=3 , 2a-b-3c= -8, 5a+2b+c=9. Sove for b
9
-3
5
-4

8 Solve the set of linear equations by Guassian elimination method : a+2b+3c=5, 3a-b+2c=8, 4a-6b-4c=-2. Find b
4
-5
-3
5

9 Solve the set of linear equations by Guassian elimination method : x+2y+3z=5, 3x-y+2z=8, 4x-6y-4=-2. Find a
-1
4
5
-11

10 Solve the set of linear equations by the matrix method : a+3b+2c=3 , 2a-b-3c= -8, 5a+2b+c=9. Sove for a
2
4
7
3

Answer 1

Answer on Question #55839 – Math – Linear Algebra

6. Solve the set of linear equations by Gaussian elimination method: $a + 2b + 3c = 5$, $3a - b + 2c = 8$, $4a - 6b - 4c = -2$. Find c

4

5

9

10

Solution

$$\left\{ \begin{array}{l} a + 2b + 3c = 5 \\ 3a - b + 2c = 8 \\ 4a - 6b - 4c = -2 \end{array} \right.$$

In matrix form:

$$\left( \begin{array}{rrr|r} 1 & 2 & 3 & 5 \\ 3 & -1 & 2 & 8 \\ 4 & -6 & -4 & -2 \end{array} \right)$$

$1^{\mathrm{st}}$ row is multiplied by 3 and subtracted from $2^{\mathrm{nd}}$ row; $1^{\mathrm{st}}$ row is multiplied by 4 and subtracted from $3^{\mathrm{rd}}$ row

$$\left( \begin{array}{rrr|r} 1 & 2 & 3 & 5 \\ 0 & -7 & -7 & -7 \\ 0 & -14 & -16 & -22 \end{array} \right)$$

$2^{\mathrm{nd}}$ row is divided by 7

$$\left( \begin{array}{rrr|r} 1 & 2 & 3 & 5 \\ 0 & 1 & 1 & 1 \\ 0 & -14 & -16 & -22 \end{array} \right)$$

$2^{\mathrm{nd}}$ row is multiplied by 2 and subtracted from $1^{\mathrm{st}}$ row; $2^{\mathrm{nd}}$ row is multiplied by 14 and added to $3^{\mathrm{rd}}$ row

$$\left( \begin{array}{rrr|r} 1 & 0 & 1 & 3 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & -2 & -8 \end{array} \right)$$

$3^{\mathrm{rd}}$ row is divided by -2

$$\left( \begin{array}{ccc|c} 1 & 0 & 1 & 3 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 4 \end{array} \right)$$

$3^{\mathrm{rd}}$ row is subtracted from $2^{\mathrm{nd}}$ row

$$\left( \begin{array}{ccc|c} 1 & 0 & 1 & 3 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 4 \end{array} \right)$$

$3^{\mathrm{rd}}$ row is subtracted from $1^{\mathrm{st}}$ row

$$\left( \begin{array}{ccc|c} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 4 \end{array} \right)$$ $$\left\{ \begin{array}{l} a = -1 \\ b = -3 \\ c = 4 \end{array} \right.$$

Answer: $c = 4$

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

9

-3

5

-4

Solution

$$\left\{ \begin{array}{l} a + 3b + 2c = 3 \\ 2a - b - 3c = -8 \\ 5a + 2b + 1c = 9 \end{array} \right.$$ $$A = \left( \begin{array}{ccc} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{array} \right); \ B = \left( \begin{array}{c} 3 \\ -8 \\ 9 \end{array} \right); \ X = \left( \begin{array}{c} a \\ b \\ c \end{array} \right);$$ $$A \cdot X = B;$$ $$X = A^{-1} \cdot B$$

Let's find the inverse of A.

$$\begin{array}{l} \det(A) = \left| \begin{array}{ccc} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{array} \right| = 1 \cdot \left| \begin{array}{cc} -1 & -3 \\ 2 & 1 \end{array} \right| - 3 \cdot \left| \begin{array}{cc} 2 & -3 \\ 5 & 1 \end{array} \right| + 2 \cdot \left| \begin{array}{cc} 2 & -1 \\ 5 & 2 \end{array} \right| = \\ = (-1) \cdot 1 - 2 \cdot (-3) - 3[2 \cdot 1 - 5 \cdot (-3)] + 2 \cdot [2 \cdot 2 - 5 \cdot (-1)] = -1 + 6 - 3[2 + 15] + 2 \cdot [4 + 5] = 23 - 51 = -28 \\ \end{array}$$ $$\begin{array}{l} M_{11} = \left| \begin{array}{cc} -1 & -3 \\ 2 & 1 \end{array} \right| = -1 \cdot 1 - 2 \cdot (-3) = 5; \\ A_{11} = (-1)^{1+1} M_{11} = 5; \\ \end{array}$$ $$\begin{array}{l} M_{12} = \left| \begin{array}{cc} 2 & -3 \\ 5 & 1 \end{array} \right| = 2 \cdot 1 - 5 \cdot (-3) = 17; \\ A_{12} = (-1)^{1+2} M_{12} = -17; \\ \end{array}$$ $$\begin{array}{l} M_{13} = \left| \begin{array}{cc} 2 & -1 \\ 5 & 2 \end{array} \right| = 2 \cdot 2 - 5 \cdot (-1) = 9; \\ A_{13} = (-1)^{1+3} M_{13} = 9; \\ \end{array}$$ $$\begin{array}{l} M_{21} = \left| \begin{array}{cc} 3 & 2 \\ 2 & 1 \end{array} \right| = 3 \cdot 1 - 2 \cdot 2 = -1; \\ A_{21} = (-1)^{2+1} M_{21} = 1; \\ \end{array}$$ $$\begin{array}{l} M_{22} = \left| \begin{array}{cc} 1 & 2 \\ 5 & 1 \end{array} \right| = 1 \cdot 1 - 5 \cdot 2 = -9; \\ A_{22} = (-1)^{2+2} M_{22} = -9; \\ \end{array}$$ $$\begin{array}{l} M_{23} = \left| \begin{array}{cc} 1 & 3 \\ 5 & 2 \end{array} \right| = 1 \cdot 2 - 5 \cdot 3 = -13; \\ A_{23} = (-1)^{2+3} M_{23} = 13; \\ \end{array}$$ $$\begin{array}{l} M_{31} = \left| \begin{array}{cc} 3 & 2 \\ -1 & -3 \end{array} \right| = 3 \cdot (-3) - (-1) \cdot 2 = -7; \\ A_{31} = (-1)^{3+1} M_{31} = 7; \\ \end{array}$$ $$M_{32} = \left| \begin{array}{cc} 1 & 2 \\ 2 & -3 \end{array} \right| = 1 \cdot (-3) - 2 \cdot 2 = -7;$$ $$A_{32} = (-1)^{3+2} M_{32} = 7;$$ $$M_{33} = \left| \begin{array}{cc} 1 & 3 \\ 2 & -1 \end{array} \right| = 1 \cdot (-1) - 2 \cdot 3 = -7;$$ $$A_{33} = (-1)^{3+3} M_{33} = -7;$$

Cofactor matrix:

$$C^* = \begin{pmatrix} 5 & -17 & 9 \\ 1 & -9 & 13 \\ 7 & 7 & -7 \end{pmatrix}$$

Adjugate:

$$C^{*T} = \begin{pmatrix} 5 & 1 & 7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{pmatrix}$$

Inverse of A:

$$A^{-1} = \frac{C^{*T}}{\det A} = \begin{pmatrix} -\frac{5}{28} & -\frac{1}{28} & \frac{1}{4} \\ \frac{17}{28} & \frac{9}{28} & -\frac{1}{4} \\ -\frac{9}{28} & -\frac{13}{28} & \frac{1}{4} \end{pmatrix}$$

The solution:

$$X = A^{-1} \cdot B = \begin{pmatrix} -\frac{5}{28} & -\frac{1}{28} & \frac{1}{4} \\ \frac{17}{28} & \frac{9}{28} & -\frac{1}{4} \\ -\frac{9}{28} & -\frac{13}{28} & \frac{1}{4} \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -8 \\ 9 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix};$$ $$\left\{ \begin{array}{l} a = 2 \\ b = -3 \\ c = 5 \end{array} \right.$$

Answer: $b = -3$

8. Solve the set of linear equations by Gaussian elimination method: $a + 2b + 3c = 5$ , $3a - b + 2c = 8$ , $4a - 6b - 4c = -2$ . Find b

4

-5

-3

5

Solution

$$\left\{ \begin{array}{l} a + 2 b + 3 c = 5 \\ 3 a - b + 2 c = 8 \\ 4 a - 6 b - 4 c = - 2 \end{array} \right.$$

In matrix form:

$$\left( \begin{array}{c c c c} 1 & 2 & 3 & 5 \\ 3 & - 1 & 2 & 8 \\ 4 & - 6 & - 4 & - 2 \end{array} \right)$$

$1^{\mathrm{st}}$ row is multiplied by 3 and subtracted from $2^{\mathrm{nd}}$ row; $1^{\mathrm{st}}$ row is multiplied by 4 and subtracted from $3^{\mathrm{rd}}$ row

$$\left( \begin{array}{c c c c} 1 & 2 & 3 & 5 \\ 0 & - 7 & - 7 & - 7 \\ 0 & - 14 & - 16 & - 22 \end{array} \right)$$

$2^{\mathrm{nd}}$ row is divided by -7

$$\left( \begin{array}{c c c c} 1 & 2 & 3 & 5 \\ 0 & 1 & 1 & 1 \\ 0 & - 14 & - 16 & - 22 \end{array} \right)$$

$2^{\mathrm{nd}}$ row is multiplied by 2 and subtracted from $1^{\mathrm{st}}$ row; $2^{\mathrm{nd}}$ row is multiplied by 14 and added to $3^{\mathrm{rd}}$ row

$$\left( \begin{array}{c c c c} 1 & 0 & 1 & 3 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & - 2 & - 8 \end{array} \right)$$

$3^{\mathrm{rd}}$ row is divided by -2

$$\left( \begin{array}{ccc|c} 1 & 0 & 1 & 3 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 4 \end{array} \right)$$

$3^{\mathrm{rd}}$ row is subtracted from $2^{\mathrm{nd}}$ row

$$\left( \begin{array}{ccc|c} 1 & 0 & 1 & 3 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 4 \end{array} \right)$$

$3^{\mathrm{rd}}$ row is subtracted from $1^{\mathrm{st}}$ row

$$\left( \begin{array}{ccc|c} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 4 \end{array} \right)$$ $$\left\{ \begin{array}{l} a = -1 \\ b = -3 \\ c = 4 \end{array} \right.$$

Answer: $b = -3$

9. Solve the set of linear equations by Gaussian elimination method: $x + 2y + 3z = 5$, $3x - y + 2z = 8$, $4x - 6y - 4 = -2$. Find a

-1

4

5

-11

Solution

$$\left\{ \begin{array}{l} a + 2b + 3c = 5 \\ 3a - b + 2c = 8 \\ 4a - 6b - 4c = -2 \end{array} \right.$$

In matrix form:

$$\left( \begin{array}{ccc|c} 1 & 2 & 3 & 5 \\ 3 & -1 & 2 & 8 \\ 4 & -6 & -4 & -2 \end{array} \right)$$

$1^{\mathrm{st}}$ row is multiplied by 3 and subtracted from $2^{\mathrm{nd}}$ row; $1^{\mathrm{st}}$ row is multiplied by 4 and subtracted from $3^{\mathrm{rd}}$ row

$2^{\mathrm{nd}}$ row is divided by -7

$2^{\mathrm{nd}}$ row is multiplied by 2 and subtracted from $1^{\mathrm{st}}$ row; $2^{\mathrm{nd}}$ row is multiplied by 14 and added to $3^{\mathrm{rd}}$ row

$3^{\mathrm{rd}}$ row is divided by -2

$3^{\mathrm{rd}}$ row is subtracted from $2^{\mathrm{nd}}$ row

$3^{\mathrm{rd}}$ row is subtracted from $1^{\mathrm{st}}$ row

Answer: $a = -1$ .

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

2

4

7

3

Solution

$$\left\{ \begin{array}{l} a + 3 b + 2 c = 3 \\ 2 a - b - 3 c = - 8 \\ 5 a + 2 b + 1 c = 9 \end{array} \right.$$ $$A = \left( \begin{array}{ccc} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{array} \right); B = \left( \begin{array}{c} 3 \\ -8 \\ 9 \end{array} \right); X = \left( \begin{array}{c} a \\ b \\ c \end{array} \right);$$ $$A \cdot X = B;$$ $$X = A ^ {- 1} \cdot B$$

Let's find the inverse of A.

$$\begin{array}{l} \det (A) = \left| \begin{array}{ccc} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{array} \right| = 1 \cdot \left| \begin{array}{cc} -1 & -3 \\ 2 & 1 \end{array} \right| - 3 \cdot \left| \begin{array}{cc} 2 & -3 \\ 5 & 1 \end{array} \right| + 2 \cdot \left| \begin{array}{cc} 2 & -1 \\ 5 & 2 \end{array} \right| = \\ = (- 1) \cdot 1 - 2 \cdot (- 3) - 3 [ 2 \cdot 1 - 5 \cdot (- 3) ] + 2 \cdot [ 2 \cdot 2 - 5 \cdot (- 1) ] = - 1 + 6 - 3 [ 2 + 15 ] + 2 \cdot [ 4 + 5 ] = 23 - 51 = - 28 \\ \end{array}$$ $$det (A) = - 28$$ $$M _ {1 1} = \left| \begin{array}{cc} - 1 & - 3 \\ 2 & 1 \end{array} \right| = - 1 \cdot 1 - 2 \cdot (- 3) = 5;$$ $$A _ {1 1} = (- 1) ^ {1 + 1} M _ {1 1} = 5;$$ $$M _ {1 2} = \left| \begin{array}{cc} 2 & -3 \\ 5 & 1 \end{array} \right| = 2 \cdot 1 - 5 \cdot (- 3) = 17;$$ $$A _ {1 2} = (- 1) ^ {1 + 2} M _ {1 2} = - 17;$$ $$M_{13} = \begin{vmatrix} 2 & -1 \\ 5 & 2 \end{vmatrix} = 2 \cdot 2 - 5 \cdot (-1) = 9;$$ $$A_{13} = (-1)^{1 + 3} M_{13} = 9;$$ $$M_{21} = \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 - 2 \cdot 2 = -1;$$ $$A_{21} = (-1)^{2 + 1} M_{21} = 1;$$ $$M_{22} = \begin{vmatrix} 1 & 2 \\ 5 & 1 \end{vmatrix} = 1 \cdot 1 - 5 \cdot 2 = -9;$$ $$A_{22} = (-1)^{2 + 2} M_{22} = -9;$$ $$M_{23} = \begin{vmatrix} 1 & 3 \\ 5 & 2 \end{vmatrix} = 1 \cdot 2 - 5 \cdot 3 = -13;$$ $$A_{23} = (-1)^{2 + 3} M_{23} = 13;$$ $$M_{31} = \begin{vmatrix} 3 & 2 \\ -1 & -3 \end{vmatrix} = 3 \cdot (-3) - (-1) \cdot 2 = -7;$$ $$A_{31} = (-1)^{3 + 1} M_{31} = 7;$$ $$M_{32} = \begin{vmatrix} 1 & 2 \\ 2 & -3 \end{vmatrix} = 1 \cdot (-3) - 2 \cdot 2 = -7;$$ $$A_{32} = (-1)^{3 + 2} M_{32} = 7;$$ $$M_{33} = \begin{vmatrix} 1 & 3 \\ 2 & -1 \end{vmatrix} = 1 \cdot (-1) - 2 \cdot 3 = -7;$$ $$A_{33} = (-1)^{3 + 3} M_{33} = -7;$$

Cofactor matrix:

$$C^* = \begin{pmatrix} 5 & -17 & 9 \\ 1 & -9 & 13 \\ 7 & 7 & -7 \end{pmatrix}$$

Adjugate:

$$C^{*T} = \begin{pmatrix} 5 & 1 & 7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{pmatrix}$$

Inverse of A:

$$A ^ {- 1} = \frac {C ^ {* T}}{\det A} = \left( \begin{array}{c c c} - \frac {5}{2 8} & - \frac {1}{2 8} & \frac {1}{4} \\ \frac {1 7}{2 8} & \frac {9}{2 8} & - \frac {1}{4} \\ - \frac {9}{2 8} & - \frac {1 3}{2 8} & \frac {1}{4} \end{array} \right)$$

The solution:

$$X = A ^ {- 1} \cdot B = \left( \begin{array}{c c c} - \frac {5}{2 8} & - \frac {1}{2 8} & \frac {1}{4} \\ \frac {1 7}{2 8} & \frac {9}{2 8} & - \frac {1}{4} \\ - \frac {9}{2 8} & - \frac {1 3}{2 8} & \frac {1}{4} \end{array} \right) \cdot \left( \begin{array}{c} 3 \\ - 8 \\ 9 \end{array} \right) = \left( \begin{array}{c} 2 \\ - 3 \\ 5 \end{array} \right);$$ $$\left\{ \begin{array}{l} a = 2 \\ b = - 3 \\ c = 5 \end{array} \right.$$

Answer: $a = 2$ .

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