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Karnataka Board PUCPUC Science 2nd PUC Class 12

PUC Science 2nd PUC Class 12 - Karnataka Board PUC Question Bank Solutions for Mathematics

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Mathematics
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Prove that :

\[\begin{vmatrix}a^2 & a^2 - \left( b - c \right)^2 & bc \\ b^2 & b^2 - \left( c - a \right)^2 & ca \\ c^2 & c^2 - \left( a - b \right)^2 & ab\end{vmatrix} = \left( a - b \right) \left( b - c \right) \left( c - a \right) \left( a + b + c \right) \left( a^2 + b^2 + c^2 \right)\]
[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined

If \[I = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}, J = \begin{bmatrix}0 & 1 \\ - 1 & 0\end{bmatrix} and B = \begin{bmatrix}\cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\] then B equals ) 

[3] Matrices
Chapter: [3] Matrices
Concept: undefined >> undefined

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Prove that :

\[\begin{vmatrix}1 & a^2 + bc & a^3 \\ 1 & b^2 + ca & b^3 \\ 1 & c^2 + ab & c^3\end{vmatrix} = - \left( a - b \right) \left( b - c \right) \left( c - a \right) \left( a^2 + b^2 + c^2 \right)\]

 

[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined

Prove that :

\[\begin{vmatrix}a^2 & bc & ac + c^2 \\ a^2 + ab & b^2 & ac \\ ab & b^2 + bc & c^2\end{vmatrix} = 4 a^2 b^2 c^2\]
[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined

Prove that :

\[\begin{vmatrix}x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4\end{vmatrix} = 16 \left( 3x + 4 \right)\]
[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined

Prove that :

\[\begin{vmatrix}1 & 1 + p & 1 + p + q \\ 2 & 3 + 2p & 4 + 3p + 2q \\ 3 & 6 + 3p & 10 + 6p + 3q\end{vmatrix} = 1\]

 

[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined

Prove that :

\[\begin{vmatrix}a & b - c & c - b \\ a - c & b & c - a \\ a - b & b - a & c\end{vmatrix} = \left( a + b - c \right) \left( b + c - a \right) \left( c + a - b \right)\]

 

[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined

Prove that

\[\begin{vmatrix}a^2 & 2ab & b^2 \\ b^2 & a^2 & 2ab \\ 2ab & b^2 & a^2\end{vmatrix} = \left( a^3 + b^3 \right)^2\]
[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined

Prove that

\[\begin{vmatrix}a^2 + 1 & ab & ac \\ ab & b^2 + 1 & bc \\ ca & cb & c^2 + 1\end{vmatrix} = 1 + a^2 + b^2 + c^2\]
[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined
\[\begin{vmatrix}1 & a & a^2 \\ a^2 & 1 & a \\ a & a^2 & 1\end{vmatrix} = \left( a^3 - 1 \right)^2\]
[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined
\[\begin{vmatrix}a + b + c & - c & - b \\ - c & a + b + c & - a \\ - b & - a & a + b + c\end{vmatrix} = 2\left( a + b \right) \left( b + c \right) \left( c + a \right)\]
[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined

2x − y = 17
3x + 5y = 6

[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined

3x + y = 19
3x − y = 23

[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined

2x − y = − 2
3x + 4y = 3

[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined

3x + ay = 4
2x + ay = 2, a ≠ 0

[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined

2x + 3y = 10
x + 6y = 4

[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined

5x + 7y = − 2
4x + 6y = − 3

[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined

9x + 5y = 10
3y − 2x = 8

[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined

Given: x + 2y = 1
            3x + y = 4

[4] Determinants
Chapter: [4] Determinants
Concept: undefined >> undefined

The trace of the matrix \[A = \begin{bmatrix}1 & - 5 & 7 \\ 0 & 7 & 9 \\ 11 & 8 & 9\end{bmatrix}\], is

[3] Matrices
Chapter: [3] Matrices
Concept: undefined >> undefined
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