Question

║⊕QUESTION⊕║
Math Class Is Sometimes Like Watching A Foreign Movie without subtitles
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CLASS 12
MATRICES
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Using Elementary transformations, find the inverse of the matrice
$\left[\begin{array}{ccc}-1&1&2\\1&2&3\\3&1&1\end{array}\right]$

Answer 1

Given:

A matrix

$\left[\begin{array}{ccc}-1&1&2\\1&2&3\\3&1&1\end{array}\right]$

To Find:

Inverse of the given matrix by elementary transformations

Solution:

Let the given matrix be A

So, A can be written as

$\longrightarrow\rm{A=IA}$

where, I is the identity matrix

On applying elementary row transformations on given matrix, we get

$\longrightarrow \left[\begin{array}{ccc}-1&1&2\\1&2&3\\3&1&1\end{array}\right]=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]A$

$\rm{R_{2}\longrightarrow R_{1}+R_{2}}$

$\longrightarrow\left[\begin{array}{ccc}-1&1&2\\0&3&5\\3&1&1\end{array}\right]=\left[\begin{array}{ccc}1&0&0\\1&1&0\\0&0&1\end{array}\right]\rm{A}$

$\rm{R_{3}\longrightarrow 3R_{1}+R_{3}}$

$\longrightarrow\left[\begin{array}{ccc}-1&1&2\\0&3&5\\0&4&7\end{array}\right]=\left[\begin{array}{ccc}1&0&0\\1&1&0\\3&0&1\end{array}\right]\rm{A}$

$\rm{R_{1}\longrightarrow (-1)R_{1}}$

$\longrightarrow\left[\begin{array}{ccc}1&-1&2\\0&3&5\\0&4&7\end{array}\right]=\left[\begin{array}{ccc}-1&0&0\\1&1&0\\3&0&1\end{array}\right]\rm{A}$

$\rm{R_{2}\longrightarrow \bigg(\dfrac{1}{3}\bigg)R_{2}}$

$\longrightarrow\left[\begin{array}{ccc}1&-1&2\\0&1&\frac{5}{3} \\0&4&7\end{array}\right]=\left[\begin{array}{ccc}-1&0&0\\\frac{1}{3} &\frac{1}{3}&0\\3&0&1\end{array}\right]\rm{A}$

$\rm{R_{1}\longrightarrow R_{1}+R_{2}}$

$\longrightarrow\left[\begin{array}{ccc}1&0&-\frac{1}{3}\\0&1&\frac{5}{3} \\0&4&7\end{array}\right]=\left[\begin{array}{ccc}-\frac{2}{3}&\frac{1}{3}&0\\\frac{1}{3} &\frac{1}{3}&0\\3&0&1\end{array}\right]\rm{A}$

$\rm{R_{3}\longrightarrow R_{3}-4R_{2}}$

$\longrightarrow\left[\begin{array}{ccc}1&0&-\frac{1}{3}\\0&1&\frac{5}{3} \\0&0&\frac{1}{3}\end{array}\right]=\left[\begin{array}{ccc}-\frac{2}{3}&\frac{1}{3}&0\\\frac{1}{3} &\frac{1}{3}&0\\\frac{5}{3}&-\frac{4}{3}&1\end{array}\right]\rm{A}$

$\rm{R_{1}\longrightarrow R_{1}+R_{3}}$

$\longrightarrow\left[\begin{array}{ccc}1&0&0\\0&1&\frac{5}{3} \\0&0&\frac{1}{3}\end{array}\right]=\left[\begin{array}{ccc}1&-1&1\\\frac{1}{3} &\frac{1}{3}&0\\\frac{5}{3}&-\frac{4}{3}&1\end{array}\right]\rm{A}$

$\rm{R_{2}\longrightarrow R_{2}-5R_{3}}$

$\longrightarrow\left[\begin{array}{ccc}1&0&0\\0&1&0 \\0&0&\frac{1}{3}\end{array}\right]=\left[\begin{array}{ccc}1&-1&1\\-8&7&-5\\\frac{5}{3}&-\frac{4}{3}&1\end{array}\right]\rm{A}$

$\rm{R_{3}\longrightarrow 3R_{3}}$

$\longrightarrow\left[\begin{array}{ccc}1&0&0\\0&1&0 \\0&0&1\end{array}\right]=\left[\begin{array}{ccc}1&-1&1\\-8&7&-5\\5&-4&3\end{array}\right]\rm{A}$

Since, the product of obtained matrix and matrix A is an identity matrix.

So, the obtained matrix is the the required inverse matrix

$\therefore \green{\rm{A^{-1}}=\left[\begin{array}{ccc}1&-1&1\\-8&7&-5\\5&-4&3\end{array}\right]}$

Answer 2

Answer:

Let f:Z↦Z be defined as f(x)=x

2

,x∈Z.

We know that the square of an integer is always a unique integer.

So, ''f'' is a function.

Now, since f(−2)=f(2)=4, ''f'' is not an injection.

There is no integer x∈Z:f(x)=−1.

Hence , ''f'' is not a surjection.

Since ''f'' is neither one-one nor onto, ''f'' is not a bijection.