Answer :

|A| = 3(-2)-1(3) +2(-4)

= -6 -3 -8

= -17

∴ As, |A| is not equal to zero.

So, A^{-1} exists.

Element of a co-factor matrix is given by,

C_{ij} = (-1)^{i+j}(M_{ij})

Where M_{ij} is minor of A_{ij}.

M_{ij} is equal to the determinant of the matrix obtained after removing i^{th} row and j^{th} column.

So, C_{11} = (-1)^{1+1}

⇒ C_{11} = -2 + 0

⇒ C_{11} = -2

C_{12} = (-1)^{1+2}

⇒ C_{12} = -1(-3 + 6)

⇒ C_{12} = -3

C_{13} = (-1)^{1+3}

⇒ C_{13} = 0 – 4

⇒ C_{13} = -4

C_{21} = (-1)^{2+1}

⇒ C_{21} = (-1)(-1 + 0)

⇒ C_{21} = 1

C_{22} = (-1)^{2+2}

⇒ C_{22} = -3 – 4

⇒ C_{22} = -7

C_{23} = (-1)^{2+3}

⇒ C_{23} = (-1)(0 – (-2))

⇒ C_{23} = (-1)(2)

⇒ C_{23} = 2

C_{31} = (-1)^{3+1}

⇒ C_{31} = -3 – 4

⇒ C_{31} = -7

C_{32} = (-1)^{3+2}

⇒ C_{32} = (-1)(-9 – 6)

⇒ C_{32} = (-1)(-15)

⇒ C_{32} = 15

C_{33} = (-1)^{3+3}

⇒ C_{33} = 6 – 3

⇒ C_{33} = 3

So, co-factor matrix of A =

Now for given system of equations.

Clearly, (A^{t})X = B

⇒ X = (A^{t})^{-1} B

As, (A^{t})^{-1} = (A^{-1})^{t}

∴ X = (A^{-1})^{t} B

Therefore, x = 2, y = 1, and z = –4

**OR**

Let,

As A = IA

So,

R_{1}→R_{1}+R_{3}

R_{1}→ (-1) R_{1}

R_{2}→ R_{2} + 5R_{1}

R_{3}→ R_{3} + 3R_{1}

Interchanging R_{2} and R_{3}

R_{2}→ (-1) R_{2}

R_{1}→ R_{1} + R_{2}

R_{3}→ R_{3} + 2R_{2}

R_{1}→ R_{1} - 9R_{3}

R_{2}→ R_{2} - 152R_{3}

Therefore,

Rate this question :

<span lang="EN-USMathematics - Exemplar

<span lang="EN-USMathematics - Exemplar

Using matrices soMathematics - Board Papers

On her birthday SMathematics - Board Papers

Using matrices soMathematics - Board Papers

<span lang="EN-USRS Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal - Mathematics

Using elementary RS Aggarwal - Mathematics

Using elementary RS Aggarwal - Mathematics