Question

show that
[121
323
112]. is non singular and find A^-1​

Answer 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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