Question

Find the inverse of the matrix (if it exists)
A = \left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 0&2&4 \\\ 0&0&5 \end{array}} \right]

Answer 1

In this particular question use the concept that inverse of a matrix is calculated as, {A^{ - 1}} = \dfrac{1}{{\left| A \right|}}{\left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]^T}, where T is the symbol of transpose i.e. all rows changed into columns or all columns changed into rows, and ${a_{ij}}$ is the determinant of the elements except column i and column j and that minor determinants is multiplied by the ${\left( { - 1} \right)^n}$, where n = 1, 2, 3..... 9, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
A = \left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 0&2&4 \\\ 0&0&5 \end{array}} \right]
Now we have to find out the inverse of this matrix.
Now as we know that inverse of the matrix is calculated as,
${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}\left( {adjA} \right)$................ (1)
Where adj (A) = {\left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]^T}
Now if |A|$\ne$0 then inverse of the matrix exists otherwise not.
So first find out the determinant of the matrix i.e. |A|.
\Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}} 1&2&3 \\\ 0&2&4 \\\ 0&0&5 \end{array}} \right|
Now expand this matrix we have,
\Rightarrow \left| A \right| = 1\left| {\begin{array}{*{20}{c}} 2&4 \\\ 0&5 \end{array}} \right| - 2\left| {\begin{array}{*{20}{c}} 0&4 \\\ 0&5 \end{array}} \right| + 3\left| {\begin{array}{*{20}{c}} 0&2 \\\ 0&0 \end{array}} \right|
Now expand the mini determinant we have,
$\Rightarrow \left| A \right| = 1\left( {2\left( 5 \right) - 0\left( 4 \right)} \right) - 2\left( {0\left( 5 \right) - 0\left( 4 \right)} \right) + 3\left( {0 - 0\left( 2 \right)} \right)$
$\Rightarrow \left| A \right| = 1\left( {10} \right) - 2\left( 0 \right) + 3\left( 0 \right)$
$\Rightarrow \left| A \right| = 10 \ne 0$
So the inverse of the matrix exists.
Now find out the adj (A)
adj (A) = {\left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]^T}
Now first find out the internal elements of the above matrix, where ${a_{ij}} = {\left( { - 1} \right)^n}{a_{ij}}$
\Rightarrow {a_{11}} = {\left( { - 1} \right)^1}\left| {\begin{array}{*{20}{c}} 2&4 \\\ 0&5 \end{array}} \right| = - \left( {10 - 0} \right) = - 10
\Rightarrow {a_{12}} = {\left( { - 1} \right)^2}\left| {\begin{array}{*{20}{c}} 0&4 \\\ 0&5 \end{array}} \right| = + \left( {0 - 0} \right) = 0
\Rightarrow {a_{13}} = {\left( { - 1} \right)^3}\left| {\begin{array}{*{20}{c}} 0&2 \\\ 0&0 \end{array}} \right| = - \left( {0 - 0} \right) = 0
\Rightarrow {a_{21}} = {\left( { - 1} \right)^4}\left| {\begin{array}{*{20}{c}} 2&3 \\\ 0&5 \end{array}} \right| = + \left( {10 - 0} \right) = 10
\Rightarrow {a_{22}} = {\left( { - 1} \right)^5}\left| {\begin{array}{*{20}{c}} 1&3 \\\ 0&5 \end{array}} \right| = - \left( {5 - 0} \right) = - 5
\Rightarrow {a_{23}} = {\left( { - 1} \right)^6}\left| {\begin{array}{*{20}{c}} 1&2 \\\ 0&0 \end{array}} \right| = + \left( {0 - 0} \right) = 0
\Rightarrow {a_{31}} = {\left( { - 1} \right)^7}\left| {\begin{array}{*{20}{c}} 2&3 \\\ 2&4 \end{array}} \right| = - \left( {8 - 6} \right) = - 2
\Rightarrow {a_{32}} = {\left( { - 1} \right)^8}\left| {\begin{array}{*{20}{c}} 1&3 \\\ 0&4 \end{array}} \right| = + \left( {4 - 0} \right) = 4

1&2 \\\ 0&2 \end{array}} \right| = - \left( {2 - 0} \right) = - 2$$ So the adj (A) is, $ \Rightarrow $ adj (A) = ${\left[ {\begin{array}{*{20}{c}} { - 10}&0&0 \\\ {10}&{ - 5}&0 \\\ { - 2}&4&{ - 2} \end{array}} \right]^T}$ Now apply the transpose i.e. all rows changed into columns or all columns changed into rows. I.e. first row changed into first column and so on. $ \Rightarrow $ adj (A) = $\left[ {\begin{array}{*{20}{c}} { - 10}&{10}&{ - 2} \\\ 0&{ - 5}&4 \\\ 0&0&{ - 2} \end{array}} \right]$ Now from equation (1) we have, $ \Rightarrow {A^{ - 1}} = \dfrac{1}{{10}}\left[ {\begin{array}{*{20}{c}} { - 10}&{10}&{ - 2} \\\ 0&{ - 5}&4 \\\ 0&0&{ - 2} \end{array}} \right]$ $ \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {\dfrac{{ - 10}}{{10}}}&{\dfrac{{10}}{{10}}}&{\dfrac{{ - 2}}{{10}}} \\\ {\dfrac{0}{{10}}}&{\dfrac{{ - 5}}{{10}}}&{\dfrac{4}{{10}}} \\\ {\dfrac{0}{{10}}}&{\dfrac{0}{{10}}}&{\dfrac{{ - 2}}{{10}}} \end{array}} \right]$ $ \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} { - 1}&1&{ - 0.2} \\\ 0&{ - 0.5}&{0.4} \\\ 0&0&{ - 0.2} \end{array}} \right]$ So this is the required inverse of A. **Note:** Whenever we face such types of questions the key concept we have to remember is that always recall how to calculate the inverse of the matrix whose formula is stated above, in this formula denominator is determinant of A, i.e. |A|, so if this value is zero than inverse does not exist otherwise exist, so first find out the |A| as above, then check whether it is zero or not if not then calculate the adj (A) as above, and simplify as above we will get the required answer.