Home/Class 12/Maths/

Question and Answer

Verify A (adj. A) = (adj. A) A = |A|:
\(\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right]\)
loading
settings
Speed
00:00
12:14
fullscreen
Verify A (adj. A) = (adj. A) A = |A|:
\(\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right]\)

Answer

Let \(A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right]\)  
\(\Rightarrow \left| A \right| = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right]\) 
\(\therefore {A_{11}} =+ \left| {\begin{array}{*{20}{c}} 0&{ - 2} \\ 0&3 \end{array}} \right| = + 0 + 0 = 0,\)\({A_{12}} = - \left| {\begin{array}{*{20}{c}} 3&{ - 2} \\ 1&3 \end{array}} \right| = - \left( {9 + 2} \right) = - 11\) 
\({A_{13}} = + \left| {\begin{array}{*{20}{c}} 3&0 \\ 1&0 \end{array}} \right| = + \left( {0 - 0} \right) = 0,\)\({A_{21}} = - \left| {\begin{array}{*{20}{c}} { - 1}&2 \\ 0&3 \end{array}} \right| - \left( { - 3 - 0} \right) = 3\)  
\({A_{22}} = + \left| {\begin{array}{*{20}{c}} 1&2 \\ 1&3 \end{array}} \right| = 3 - 2 = 1,\)\({A_{23}} = - \left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ 1&0 \end{array}} \right| = - \left( {0 + 1} \right) = - 1\) 
\({A_{31}} = + \left| {\begin{array}{*{20}{c}} { - 1}&2 \\ 0&{ - 2} \end{array}} \right| = 2 - 0 = 2,\) \({A_{32}} = - \left| {\begin{array}{*{20}{c}} 1&2 \\ 3&{ - 2} \end{array}} \right| = - \left( { - 2 - 6} \right) = 8\) 
\({A_{33}} = + \left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ 3&0 \end{array}} \right| = 3 + 0 = 3\) 
\(\therefore adj.A = \left| {\begin{array}{*{20}{c}} 0&{ - 11}&0 \\ 3&1&{ - 1} \\ 2&8&3 \end{array}} \right|\) 
\(= \left| {\begin{array}{*{20}{c}} 0&3&2 \\ { - 11}&1&8 \\ 0&{ - 1}&3 \end{array}} \right|\) 
\(\therefore A.\left( {adj.A} \right) = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0&3&2 \\ { - 11}&1&8 \\ 0&{ - 1}&3 \end{array}} \right]\) 
\(\left[ {\begin{array}{*{20}{c}} {0 + 11 + 0}&{3 - 1 - 2}&{2 - 8 + 6} \\ {0 - 0 - 0}&{9 + 0 + 2}&{6 + 0 - 6} \\ {0 + 0 + 0}&{3 + 0 - 3}&{2 + 0 + 9} \end{array}} \right]\) 
\( = \left[ {\begin{array}{*{20}{c}} {11}&0&0 \\ 0&{11}&0 \\ 0&0&{11} \end{array}} \right]\)...(i)
Again  (adj. A). A \(= \left[ {\begin{array}{*{20}{c}} 0&3&2 \\ { - 11}&1&8 \\ 0&{ - 1}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right]\) 
\(= \left[ {\begin{array}{*{20}{c}} {0 + 9 + 2}&{0 + 0 + 0}&{0 - 6 + 6} \\ { - 11 + 3 + 8}&{11 + 0 + 0}&{ - 22 - 2 + 24} \\ {0 - 3 + 3}&{0 - 0 + 0}&{0 + 2 + 9} \end{array}} \right]\) 
\(= \left[ {\begin{array}{*{20}{c}} {11}&0&0 \\ 0&{11}&0 \\ 0&0&{11} \end{array}} \right]\)….(ii)
And \(\left| A \right| = \left| {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right|\) 
\(= 1\left( {0 - 0} \right) - \left( { - 1} \right)\left( {9 + 2} \right) + 2\left( {0 - 0} \right) = 0 + 11 + 0 = 11\) 
Also \(\left| A \right|I = \left| A \right|{I_3} = 11\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {11}&0&0 \\ 0&{11}&0 \\ 0&0&{11} \end{array}} \right]\)...(iii)
\(\therefore\) From eq. (i), (ii) and (iii)  A. (adj. A) = (adj. A). A = |A|
To Keep Reading This Answer, Download the App
4.6
star pngstar pngstar pngstar pngstar png
Review from Google Play
To Keep Reading This Answer, Download the App
4.6
star pngstar pngstar pngstar pngstar png
Review from Google Play
Correct44
Incorrect0
Watch More Related Solutions
Find the inverse of the matrix (if it exists) given \(\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&{\cos \alpha }&{\sin \alpha } \\ 0&{\sin \alpha }&{ - \cos \alpha } \end{array}} \right]\)
Find the equation of the line joining (1, 2) and (3, 6) using determinant.
If  \(A = \left| {\begin{array}{*{20}{c}} 2&{ - 3}&5 \\ 6&0&4 \\ 1&5&{ - 7} \end{array}} \right|,\)Verify that det A = det (A')
Solve the system of linear equation, using matrix method x - y + z = 4; 2x + y - 3z = 0; x + y + z = 2
Solve the system of equations \(\begin{aligned} &2 x+5 y=1\\ &3 x+2 y=7 \end{aligned}\)
By using properties of determinant, show that \(\left| {\begin{array}{*{20}{c}} {x + y + 2z}&x&y \\ z&{y + z + 2x}&y \\ z&x&{z + x + 2y} \end{array}} \right| = 2{\left( {x + y + z} \right)^3}\)
If \(A = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 4&2 \end{array}} \right]\), then show that |2A| = 4|A|
Solve the system of linear equation, using matrix method 2x - y = - 2; 3x + 4y = 3
Using cofactors of elements of third column, evaluate \(\Delta = \left| {\begin{array}{*{20}{c}} 1&x&{yz} \\ 1&y&{zx} \\ 1&z&{xy} \end{array}} \right|\)
Evaluate the determinant \(\Delta=\left|\begin{array}{rrr} {1} & {2} & {4} \\ {-1} & {3} & {0} \\ {4} & {1} & {0} \end{array}\right|\)

Load More