Math, asked by sravani17032001, 5 months ago

[ 1 2 1
Show that A=
3 2 3
is non - singular
1 1 2
find A-1
(MARCH-1​

Answers

Answered by ak4733
0

Answer:

unable to understand the question please rewrite it

Answered by hukam0685
1

Step-by-step explanation:

Given:

A=\left[\begin{array}{ccc}1&2&1\\3&2&3\\1&1&2\end{array}\right]

To find:Show that matrix A is non singular and find A^-1.

Solution:

Tip:

1) For non-singular matrix:

\bf |A|\neq 0\\

2)Inverse of a matrix:

\bf A^{-1}=\frac{Adj.(A)}{|A|}\\

Step 1: Calculate determinant of A.

|A|=\left|\begin{array}{ccc}1&2&1\\3&2&3\\1&1&2\end{array}\right|\\

 |A|  = 1(4 - 3) - 2(6 - 3) + 1(3 - 2) \\

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

 |A|  = 1 - 6 + 1

 |A|  =  - 4

 |A|  \neq0

Thus, A is non-singular matrix.

Step 2: Find Minor matrix M_{ij} for every element of matrix A.

Minor of a element of can be calculated by hiding that row and column.

M=\left[\begin{array}{ccc}1&3&1\\3&1&-1\\4&0&-4\end{array}\right]\\

Step 3: Find Co-factor matrix C_{ij}

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

C=\left[\begin{array}{ccc}1&-3&1\\-3&1&1\\4&0&-4\end{array}\right]\\

Step 4: Find Adj.(A)

Adj.(A) is transpose of Co-factor matrix.

Adj.(A)=\left[\begin{array}{ccc}1&-3&4\\-3&1&0\\1&1&-4\end{array}\right]\\

Step 5: Write inverse of A.

A^{-1}=\frac{Adj.(A)}{|A|}\\

A^{-1}=\frac{-1}{4} \left[\begin{array}{ccc}1&-3&4\\-3&1&0\\1&1&-4\end{array}\right]\\

or

A^{-1}=\frac{1}{4} \left[\begin{array}{ccc}-1&3&-4\\3&-1&0\\-1&-1&4\end{array}\right]\\

or

A^{-1}= \left[\begin{array}{ccc}-\frac{1}{4}&\frac{3}{4}&-1\\\\\frac{3}{4}&-\frac{1}{4}&0\\\\-\frac{1}{4}&-\frac{1}{4}&1\end{array}\right]\\

Final answer:

Yes, Matrix A is non singular.

\bf A^{-1}= \left[\begin{array}{ccc}-\frac{1}{4}&\frac{3}{4}&-1\\\\\frac{3}{4}&-\frac{1}{4}&0\\\\-\frac{1}{4}&-\frac{1}{4}&1\end{array}\right]\\

Hope it helps you.

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