Home/Class 12/Maths/

Question and Answer

Question
MathsClass 12

Evaluate \(\left| {\begin{array}{*{20}{c}} 3&{ - 1}&{ - 2} \\ 0&0&{ - 1} \\ 3&{ - 5}&0 \end{array}} \right|\)

Answer

\(-12\)
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

Solution

Given: \(\left| {\begin{array}{*{20}{c}} 3&{ - 1}&{ - 2} \\ 0&0&{ - 1} \\ 3&{ - 5}&0 \end{array}} \right|\)
Expanding along first row, \(3\left| {\begin{array}{*{20}{c}} 0&{ - 1} \\ { - 5}&0 \end{array}} \right| - \left( { - 1} \right)\left| {\begin{array}{*{20}{c}} 0&{ - 1} \\ 3&0 \end{array}} \right| + \left( { - 2} \right)\left| {\begin{array}{*{20}{c}} 0&0 \\ 3&{ - 5} \end{array}} \right|\)
= 3(0 - 5) + 1(0 - (-3)) - 2(0 - 0)
= -15 + 3 - 0 = -12
To Keep Reading This Solution, Download the APP
4.6
star pngstar pngstar pngstar pngstar png
Review from Google Play
To Keep Reading This Solution, Download the APP
4.6
star pngstar pngstar pngstar pngstar png
Review from Google Play
Correct2
Incorrect0
Still Have Question?
Ask Your QuestionFind More Answer
Watch More Related Solutions
Using the property of determinant and without expanding prove that \(\left| {\begin{array}{*{20}{c}} 2&7&{65} \\ 3&8&{75} \\ 5&9&{86} \end{array}} \right| = 0\)
Using the property of determinant and without expanding prove that \(\left| {\begin{array}{*{20}{c}} 2&7&{65} \\ 3&8&{75} \\ 5&9&{86} \end{array}} \right| = 0\)
Math|Class 12
Evaluate \(\left| {\begin{array}{*{20}{c}} 2&4 \\ { - 5}&{ - 1} \end{array}} \right|\)
Evaluate \(\left| {\begin{array}{*{20}{c}} 2&4 \\ { - 5}&{ - 1} \end{array}} \right|\)
Math|Class 12
Evaluate the determinant \(\left|\begin{array}{rrr} {2} & {-1} & {-2} \\ {0} & {2} & {-1} \\ {3} & {-5} & {0} \end{array}\right|\) 
Evaluate the determinant \(\left|\begin{array}{rrr} {2} & {-1} & {-2} \\ {0} & {2} & {-1} \\ {3} & {-5} & {0} \end{array}\right|\) 
Math|Class 12
Find the minor of element 6 in the determinant \(\Delta=\left|\begin{array}{lll} {1} & {2} & {3} \\ {4} & {5} & {6} \\ {7} & {8} & {9} \end{array}\right|\)
Find the minor of element 6 in the determinant \(\Delta=\left|\begin{array}{lll} {1} & {2} & {3} \\ {4} & {5} & {6} \\ {7} & {8} & {9} \end{array}\right|\)
Math|Class 12
If area of triangle is 35 sq units with vertices (2, -6), (5, 4) and (k, 4). Then k is
  • -12, -2
  • 12
  • 12, -2
  • -2
If area of triangle is 35 sq units with vertices (2, -6), (5, 4) and (k, 4). Then k is
  • -12, -2
  • 12
  • 12, -2
  • -2
Math|Class 12
Use product  \(\left[\begin{array}{ccc} {1} & {1} & {2} \\ {0} & {2} & {3} \\ {3} & {2} & {4} \end{array}\right]\left[\begin{array}{ccc} {2} & {0} & {1} \\ {9} & {2} & {3} \\ {6} & {1} & {2} \end{array}\right]\)to solve the system of equation
 x – y + 2z = 1, 2y – 3z = 1, 3x – 2y + 4z = 2
Use product  \(\left[\begin{array}{ccc} {1} & {1} & {2} \\ {0} & {2} & {3} \\ {3} & {2} & {4} \end{array}\right]\left[\begin{array}{ccc} {2} & {0} & {1} \\ {9} & {2} & {3} \\ {6} & {1} & {2} \end{array}\right]\)to solve the system of equation
 x – y + 2z = 1, 2y – 3z = 1, 3x – 2y + 4z = 2
Math|Class 12
Let A be a square matrix of order 3 \(\times\) 3, then | kA| is equal to
  • k2 | A |
  • k |A |
  • k3 | A |
  • 3k | A |
Let A be a square matrix of order 3 \(\times\) 3, then | kA| is equal to
  • k2 | A |
  • k |A |
  • k3 | A |
  • 3k | A |
Math|Class 12
Let A = \(\left[\begin{array}{ccc} {1} & {2} & {1} \\ {2} & {3} & {1} \\ {1} & {1} & {5} \end{array}\right]\). verify that \([adj A] ^{–1} = adj (A ^{–1})\)
Let A = \(\left[\begin{array}{ccc} {1} & {2} & {1} \\ {2} & {3} & {1} \\ {1} & {1} & {5} \end{array}\right]\). verify that \([adj A] ^{–1} = adj (A ^{–1})\)
Math|Class 12
If a, b, c, are in A.P, then the determinant
\(\left|\begin{array}{ccc} {x+2} & {x+3} & {x+2 a} \\ {x+3} & {x+4} & {x+2 b} \\ {x+4} & {x+5} & {x+2 c} \end{array}\right|\) is
  • 1
  • 2x
  • x
  • 0
If a, b, c, are in A.P, then the determinant
\(\left|\begin{array}{ccc} {x+2} & {x+3} & {x+2 a} \\ {x+3} & {x+4} & {x+2 b} \\ {x+4} & {x+5} & {x+2 c} \end{array}\right|\) is
  • 1
  • 2x
  • x
  • 0
Math|Class 12
By using properties of determinants, show that \(\left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac} \\ {ab}&{{b^2} + 1}&{bc} \\ {ca}&{cb}&{{c^2} + 1} \end{array}} \right| = 1 + {a^2} + {b^2} + {c^2}\)
By using properties of determinants, show that \(\left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac} \\ {ab}&{{b^2} + 1}&{bc} \\ {ca}&{cb}&{{c^2} + 1} \end{array}} \right| = 1 + {a^2} + {b^2} + {c^2}\)
Math|Class 12

Load More

More Solution Recommended For You
NCERT Maths for Class 6
NCERT Physics for Class 6
NCERT Chemistry for Class 6
NCERT Biology for Class 6
NCERT Maths for Class 7
NCERT Physics for Class 7
NCERT Chemistry for Class 7
NCERT Biology for Class 7
NCERT Maths for Class 8
NCERT Physics for Class 8
NCERT Chemistry for Class 8
NCERT Biology for Class 8
Previous
Next