Exercise 3.4 class 12 maths solutions Matrices

Using elementary transformations, find the inverse of each of the matrices, if it exists in questions 1 to 17.

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\2&3\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\2&3\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\2&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{ – 2}&1\end{array}} \right]A $$

Applying $R_2 \to \frac{1}{5}R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{\frac{{ – 2}}{5}}&{\frac{1}{5}}\end{array}} \right]A $$

Applying $R_1 \to R_1 + R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{3}{5}}&{\frac{1}{5}}\\{\frac{{ – 2}}{5}}&{\frac{1}{5}}\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}{\frac{3}{5}}&{\frac{1}{5}}\\{\frac{{ – 2}}{5}}&{\frac{1}{5}}\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 1}&2\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 1}&2\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}1&3\\2&7\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}1&3\\2&7\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}1&3\\2&7\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&3\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{ – 2}&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – 3R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}7&{ – 3}\\{ – 2}&1\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}7&{ – 3}\\{ – 2}&1\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}2&3\\5&7\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}2&3\\5&7\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}2&3\\5&7\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_1$

$$ \left[ {\begin{array}{*{20}{c}}2&3\\1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{ – 2}&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&2\\1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&{ – 1}\\{ – 2}&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&2\\0&{ – 1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&{ – 1}\\{ – 5}&2\end{array}} \right]A $$

Applying $R_2 \to -R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&{ – 1}\\5&{ – 2}\end{array}} \right]A $$

Applying $R_1 \to R_1 – 2R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 7}&3\\5&{ – 2}\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}{ – 7}&3\\5&{ – 2}\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}2&1\\7&4\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}2&1\\7&4\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}2&1\\7&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 3R_1$

$$ \left[ {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{ – 3}&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4&{ – 1}\\{ – 3}&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4&{ – 1}\\{ – 7}&2\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}4&{ – 1}\\{ – 7}&2\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}2&5\\1&3\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}2&5\\1&3\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}2&5\\1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&2\\1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 1}&2\end{array}} \right]A $$

Applying $R_1 \to R_1 – 2R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&{ – 5}\\{ – 1}&2\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}3&{ – 5}\\{ – 1}&2\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}3&1\\5&2\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}3&1\\5&2\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}3&1\\5&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – R_1$

$$ \left[ {\begin{array}{*{20}{c}}3&1\\2&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{ – 1}&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\2&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&{ – 1}\\{ – 1}&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&{ – 1}\\{ – 5}&3\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}2&{ – 1}\\{ – 5}&3\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}4&5\\3&4\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}4&5\\3&4\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}4&5\\3&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&1\\3&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 3R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 3}&4\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4&{ – 5}\\{ – 3}&4\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}4&{ – 5}\\{ – 3}&4\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}3&{10}\\2&7\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}3&{10}\\2&7\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}3&{10}\\2&7\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&3\\2&7\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&3\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 2}&3\end{array}} \right]A $$

Applying $R_1 \to R_1 – 3R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}7&{ – 10}\\{ – 2}&3\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}7&{ – 10}\\{ – 2}&3\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}3&{ – 1}\\{ – 4}&2\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}3&{ – 1}\\{ – 4}&2\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}3&{ – 1}\\{ – 4}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 + R_2$

$$ \left[ {\begin{array}{*{20}{c}}{ – 1}&1\\{ – 4}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]A $$

Applying $R_1 \to -R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 4}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 1}&{ – 1}\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 + 4R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&{ – 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 1}&{ – 1}\\{ – 4}&{ – 3}\end{array}} \right]A $$

Applying $R_2 \to -R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 1}&{ – 1}\\4&3\end{array}} \right]A $$

Applying $R_2 \to \frac{1}{2}R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 1}&{ – 1}\\2&{\frac{3}{2}}\end{array}} \right]A $$

Applying $R_1 \to R_1 + R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\2&{\frac{3}{2}}\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}1&{\frac{1}{2}}\\2&{\frac{3}{2}}\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}2&{ – 6}\\1&{ – 2}\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}2&{ – 6}\\1&{ – 2}\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}2&{ – 6}\\1&{ – 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 4}\\1&{ – 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 4}\\0&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 1}&2\end{array}} \right]A $$

Applying $R_1 \to R_1 + 2R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 1}&3\\{ – 1}&2\end{array}} \right]A $$

Applying $R_2 \to \frac{1}{2}R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ – 1}&3\\{\frac{{ – 1}}{2}}&1\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}{ – 1}&3\\{\frac{{ – 1}}{2}}&1\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}6&{ – 3}\\{ – 2}&1\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}6&{ – 3}\\{ – 2}&1\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}6&{ – 3}\\{ – 2}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 + 3R_2$

$$ \left[ {\begin{array}{*{20}{c}}0&0\\{ – 2}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&3\\0&1\end{array}} \right]A $$

We have all zeroes in the first row of the left hand side matrix of the above equation.

Therefore, $A^{-1}$ does not exist.
$$ \left[ {\begin{array}{*{20}{c}}2&{ – 3}\\{ – 1}&2\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}2&{ – 3}\\{ – 1}&2\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}2&{ – 3}\\{ – 1}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 + R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 1}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 + R_1$

$$ \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1\\1&2\end{array}} \right]A $$

Applying $R_1 \to R_1 + R_2$

$$ \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&3\\1&2\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{c}}2&3\\1&2\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{c}}2&1\\4&2\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{c}}2&1\\4&2\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{c}}2&1\\4&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_1$

$$ \left[ {\begin{array}{*{20}{c}}2&1\\0&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{ – 2}&1\end{array}} \right]A $$

We have all zeroes in the second row of the left hand side matrix of the above equation. Therefore, $A^{-1}$ does not exist.
$$ \left[ {\begin{array}{*{20}{r}}2&{ – 3}&3\\2&2&3\\3&{ – 2}&2\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{r}}2&{ – 3}&3\\2&2&3\\3&{ – 2}&2\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{r}}2&{ – 3}&3\\2&2&3\\3&{ – 2}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{r}}0&{ – 5}&0\\2&2&3\\3&{ – 2}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&{ – 1}&0\\0&1&0\\0&0&1\end{array}} \right]A $$

Applying $R_3 \to R_3 – R_2$

$$ \left[ {\begin{array}{*{20}{r}}0&{ – 5}&0\\2&2&3\\1&{ – 4}&{ – 1}\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&{ – 1}&0\\0&1&0\\0&{ – 1}&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_3$

$$ \left[ {\begin{array}{*{20}{r}}0&{ – 5}&0\\0&{10}&5\\1&{ – 4}&{ – 1}\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&{ – 1}&0\\0&3&{ – 2}\\0&{ – 1}&1\end{array}} \right]A $$

Applying $R_1 \leftrightarrow R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&{ – 4}&{ – 1}\\0&{10}&5\\0&{ – 5}&0\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&{ – 1}&1\\0&3&{ – 2}\\1&{ – 1}&0\end{array}} \right]A $$

Applying $R_3 \leftrightarrow R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&{ – 4}&{ – 1}\\0&{ – 5}&0\\0&{10}&5\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&{ – 1}&1\\1&{ – 1}&0\\0&3&{ – 2}\end{array}} \right]A $$

Applying $R_3 \to R_3 + 2R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&{ – 4}&{ – 1}\\0&{ – 5}&0\\0&0&5\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&{ – 1}&1\\1&{ – 1}&0\\2&1&{ – 2}\end{array}} \right]A $$

Applying $R_2 \to -R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&{ – 4}&{ – 1}\\0&5&0\\0&0&5\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&{ – 1}&1\\{ – 1}&1&0\\2&1&{ – 2}\end{array}} \right]A $$

Applying $R_2 \to \frac{1}{5}R_2$ and $R_3 \to \frac{1}{5}R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&{ – 4}&{ – 1}\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&{ – 1}&1\\{\frac{{ – 1}}{5}}&{\frac{1}{5}}&0\\{\frac{2}{5}}&{\frac{1}{5}}&{\frac{{ – 2}}{5}}\end{array}} \right]A $$

Applying $R_1 \to R_1 + 4R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&0&{ – 1}\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – \frac{4}{5}}&{ – \frac{1}{5}}&1\\{\frac{{ – 1}}{5}}&{\frac{1}{5}}&0\\{\frac{2}{5}}&{\frac{1}{5}}&{\frac{{ – 2}}{5}}\end{array}} \right]A $$

Applying $R_1 \to R_1 + R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{\frac{{ – 2}}{5}}&0&{\frac{3}{5}}\\{\frac{{ – 1}}{5}}&{\frac{1}{5}}&0\\{\frac{2}{5}}&{\frac{1}{5}}&{\frac{{ – 2}}{5}}\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{r}}{\frac{{ – 2}}{5}}&0&{\frac{3}{5}}\\{\frac{{ – 1}}{5}}&{\frac{1}{5}}&0\\{\frac{2}{5}}&{\frac{1}{5}}&{\frac{{ – 2}}{5}}\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{r}}1&3&{ – 2}\\{ – 3}&0&{ – 5}\\2&5&0\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{r}}1&3&{ – 2}\\{ – 3}&0&{ – 5}\\2&5&0\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{r}}1&3&{ – 2}\\{ – 3}&0&{ – 5}\\2&5&0\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 + 3R_1$

$$ \left[ {\begin{array}{*{20}{r}}1&3&{ – 2}\\0&9&{ – 11}\\2&5&0\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&0&0\\3&1&0\\0&0&1\end{array}} \right]A $$

Applying $R_3 \to R_3 – 2R_1$

$$ \left[ {\begin{array}{*{20}{r}}1&3&{ – 2}\\0&9&{ – 11}\\0&{ – 1}&4\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&0&0\\3&1&0\\{ – 2}&0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 + 3R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&0&{10}\\0&9&{ – 11}\\0&{ – 1}&4\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 5}&0&3\\3&1&0\\{ – 2}&0&1\end{array}} \right]A $$

Interchanging $R_2$ and $R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&0&{10}\\0&{ – 1}&4\\0&9&{ – 11}\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 5}&0&3\\{ – 2}&0&1\\3&1&0\end{array}} \right]A $$

Applying $R_3 \to R_3 + 9R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&0&{10}\\0&{ – 1}&4\\0&0&{25}\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 5}&0&3\\{ – 2}&0&1\\{ – 15}&1&9\end{array}} \right]A $$

Applying $R_2 \to -R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&0&{10}\\0&1&{ – 4}\\0&0&{25}\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 5}&0&3\\2&0&{ – 1}\\{ – 15}&1&9\end{array}} \right]A $$

Applying $R_3 \to \frac{1}{25}R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&0&{10}\\0&1&{ – 4}\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 5}&0&3\\2&0&{ – 1}\\{\frac{{ – 15}}{25}}&{\frac{1}{25}}&{\frac{9}{25}}\end{array}} \right]A $$

Applying $R_1 \to R_1 – 10R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&{ – 4}\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&{\frac{{ – 2}}{5}}&{\frac{{ – 3}}{5}}\\2&0&{ – 1}\\{\frac{{ – 3}}{5}}&{\frac{1}{25}}&{\frac{9}{25}}\end{array}} \right]A $$

Applying $R_2 \to R_2 + 4R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&{\frac{{ – 2}}{5}}&{\frac{{ – 3}}{5}}\\{\frac{{ – 2}}{5}}&{\frac{4}{25}}&{\frac{11}{25}}\\{\frac{{ – 3}}{5}}&{\frac{1}{25}}&{\frac{9}{25}}\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{r}}1&{\frac{{ – 2}}{5}}&{\frac{{ – 3}}{5}}\\{\frac{{ – 2}}{5}}&{\frac{4}{25}}&{\frac{11}{25}}\\{\frac{{ – 3}}{5}}&{\frac{1}{25}}&{\frac{9}{25}}\end{array}} \right]$
$$ \left[ {\begin{array}{*{20}{r}}2&0&{ – 1}\\5&1&0\\0&1&3\end{array}} \right] $$

Solution:

Let us take $A = \left[ {\begin{array}{*{20}{r}}2&0&{ – 1}\\5&1&0\\0&1&3\end{array}} \right]$

We know that, A = IA

$$ \left[ {\begin{array}{*{20}{r}}2&0&{ – 1}\\5&1&0\\0&1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]A $$

Interchanging $R_1$ and $R_2$

$$ \left[ {\begin{array}{*{20}{r}}5&1&0\\2&0&{ – 1}\\0&1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}0&1&0\\1&0&0\\0&0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – 2R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&1&2\\2&0&{ – 1}\\0&1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 2}&1&0\\1&0&0\\0&0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_1$

$$ \left[ {\begin{array}{*{20}{r}}1&1&2\\0&{ – 2}&{ – 5}\\0&1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 2}&1&0\\5&{ – 2}&0\\0&0&1\end{array}} \right]A $$

Applying $R_2 \to -R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&1&2\\0&2&5\\0&1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 2}&1&0\\{ – 5}&2&0\\0&0&1\end{array}} \right]A $$

Applying $R_2 \to R_2 – R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&1&2\\0&1&2\\0&1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ – 2}&1&0\\{ – 5}&2&{ – 1}\\0&0&1\end{array}} \right]A $$

Applying $R_1 \to R_1 – R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&2\\0&1&3\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}3&{ – 1}&1\\{ – 5}&2&{ – 1}\\0&0&1\end{array}} \right]A $$

Applying $R_3 \to R_3 – R_2$

$$ \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&2\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}3&{ – 1}&1\\{ – 5}&2&{ – 1}\\5&{ – 2}&2\end{array}} \right]A $$

Applying $R_2 \to R_2 – 2R_3$

$$ \left[ {\begin{array}{*{20}{r}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}3&{ – 1}&1\\{ – 15}&6&{ – 5}\\5&{ – 2}&2\end{array}} \right]A $$

Hence, $A^{-1} = \left[ {\begin{array}{*{20}{r}}3&{ – 1}&1\\{ – 15}&6&{ – 5}\\5&{ – 2}&2\end{array}} \right]$
Matrices A and B will be inverse of each other only if
1. AB = BA
2. AB = BA = 0
3. AB = 0, BA = I
4. AB = BA = I
Solution:

(D) Matrices A and B will be inverse of each other only if, AB = BA = I.

Exercise 3.4 class 12 maths solutions Matrices

Using elementary transformations, find the inverse of each of the matrices, if it exists in questions 1 to 17.

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