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If \mathrm{A}=\left[\begin{array}{lll} {1} & {3} & {4} \\ {2} & {1} & {2} \\ {5} & {1} & {1} \end{array}\right] find A^{-1}

Hence solve the system of equations

x+3y+4z=8

2x+y+2z=5

and 5x+y+z=7

 

 

 

 
 
 
 
 

Answers (1)

If \mathrm{A}=\left[\begin{array}{lll} {1} & {3} & {4} \\ {2} & {1} & {2} \\ {5} & {1} & {1} \end{array}\right]      \left | A \right |=1(1-2)-3(2-10)+4(2-5)

=11\neq 0

\therefore A^{-1}exists.

Then cofactors of A are 

A_{11}=-1, A_{12}=8, A_{13}=-3, A_{21}=1, A_{22}=-19, A_{23}=14, A_{31}=2, A_{32}=6, A_{33}=-5

Then adj A = \begin{bmatrix} A_{11} &A_{12} &A_{12} \\ A_{21}&A_{22} &A_{23} \\ A_{31}&A_{32} &A_{33} \end{bmatrix}^T

=\begin{bmatrix} -1 &8 & -3\\ 1& -19 & 14\\ 2& 6 & -5 \end{bmatrix}^T

adjA=\begin{bmatrix} -1 &1 & 2\\ 8& -19 & 6\\ -3 & 14 & -5 \end{bmatrix}

Now, A^{-1}=\frac{1}{|A|}adjA

=\frac{1}{11}\begin{bmatrix} -1 &1 &2 \\ 8& -19 & 6\\ -3& 14 &-5 \end{bmatrix}\: \: \: -(ii)

Given system of equations are 

x+3y+4z=8

2x+y+2z=5

5x+y+z=7

Let 

A=\begin{bmatrix} 1 & 3 & 4\\ 2& 1 & 2\\ 5& 1 & 1 \end{bmatrix}\: \: \: \: B=\begin{bmatrix} 8\\5 \\7 \end{bmatrix}\: \: \: and\: \: \: X=\begin{bmatrix} x\\y \\z \end{bmatrix}

Then   AX=B

A=\begin{bmatrix} 1 & 3 & 4\\ 2& 1 & 2\\ 5& 1 & 1 \end{bmatrix}X=\begin{bmatrix} 8\\5 \\7 \end{bmatrix}

X=\begin{bmatrix} 1 & 3 & 4\\ 2& 1 & 2\\ 5& 1 & 1 \end{bmatrix}^{-1}\begin{bmatrix} 8\\5 \\7 \end{bmatrix}

 =\frac{1}{11}\begin{bmatrix} -1 & 1 & 2\\ 8& -19 & 6\\ -3& 14 & -5 \end{bmatrix}\begin{bmatrix} 8\\5 \\7 \end{bmatrix}

Using (ii) equation 

=\frac{1}{11}\begin{bmatrix} -8 +5 +14\\ 64-95 +42\\ -24 +70 -35 \end{bmatrix}=\frac{1}{11}\begin{bmatrix} 11\\11 \\11 \end{bmatrix}

X=\frac{1}{11}\begin{bmatrix} 11\\11 \\11 \end{bmatrix}

\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 1\\1 \\1 \end{bmatrix}

Hence x=1,y=1,z=1

Posted by

Ravindra Pindel

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