There are two matrix methods. I’ve shown both methods, one for each of the questions.

(i)(a) Matrix row reduction method to find the inverse matrix

The following row operations are used to convert the identity matrix to its inverse:

- R1←R1-R2
- R2←R2+2R3
- R1←R1-R3
- R2←R2+R3
- R3←3R3-R1
- R2←4R2-R3
- R2←7R3-3R2
- R1←28R1+R3
- R1←R1/84, R2←R2/42, R3←-R3/28

When applied to the original coefficient matrix we get the identity matrix, and when applied to the identity matrix we get the inverse.

Original matrix is

( 1 1 2 )

( -3 -1 2 )

( 1 2 -1 )

The inverse is

( 3 -5 -4 )

( 1 3 8 ) × 1/14

( 5 1 -2 )

When the inverse matrix is applied to the constants we get the variable values

(3×0-5×1-(-4)(-4))/14=(-5+16)/14=11/14=x

(1×0+3×1-8(-4))/14=(3-32)/14=-29/14=y

(5×0+1×1-2(-4))/14=(1+8)/14=9/14=z

(i)(b) Cramer’s Rule

The determinant ∆ for the original matrix evaluates to (1)(1-4)-(1)(3-2)+2(-6+1)=-3-1-10=-14.

x=

| 0 1 2 |

| 1 -1 2 | ÷ ∆

| -4 2 -1 |

which evaluates to ((-1)(-1+8)+2(2-4))/(-14)=(-7-4)/(-14)=11/14

y=

| 1 0 2 |

| -3 1 2 | ÷ ∆

| 1 -4 -1 |=

(1(-1+8)+2(12-1))/(-14)=(7+22)/(-14)=-29/14

z=

| 1 1 0 |

| -3 -1 1 | ÷ ∆

| 1 2 -4 |=

((4-2)-(12-1))/(-14)=(2-11)/(-14)=9/14

(ii)(a) Matrix method to find the inverse—another method

∆=

| 1 -1 1 |

| 1 2 3 |

| 1 -3 -1 |=-2

To create the inverse matrix we first swap rows and columns:

( 1 1 1 )

( -1 2 -3 )

( 1 -3 -1 )

Replace each element by its subdeterminant:

( 7 4 -5 )

( -4 -2 2 )

( -5 -2 3)

Then we reverse the sign of every other element and apply 1/∆:

( 7 -4 -5 )

( 4 -2 -2 ) × -½

( -5 2 3 )

Finally, we apply the inverse matrix to the constants:

x=((7)(2)-(4)(6)-(5)(-4))/(-2)=10/(-2)=-5

y=((4)(2)-(2)(6)-(2)(-4))/(-2)=4/(-2)=-2

z=((-5)(2)+(2)(6)+(3)(-4))/(-2)=(-10)/(-2)=5

(ii)(b) Cramer’s Rule

x=(2(-2+9)+1(-6+12)+1(-18+8))/(-2)=(14+6-10)/(-2)=10/(-2)=-5

y=((1)(-6+12)-(2)(-1-3)+1(-4-6))/(-2)=(6+8-10)/(-2)=4/(-2)=-2

z=((1)(-8+18)+(1)(-4-6)+(2)(-3-2))/(-2)=(-10)/(-2)=5