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Mathematics
Prove that
\[\begin{vmatrix}a^2 + 1 & ab & ac \\ ab & b^2 + 1 & bc \\ ca & cb & c^2 + 1\end{vmatrix} = 1 + a^2 + b^2 + c^2\]
Chapter: [4] Determinants
Concept: undefined >> undefined
Concept: undefined >> undefined
\[\begin{vmatrix}1 & a & a^2 \\ a^2 & 1 & a \\ a & a^2 & 1\end{vmatrix} = \left( a^3 - 1 \right)^2\]
Chapter: [4] Determinants
Concept: undefined >> undefined
Concept: undefined >> undefined
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\[\begin{vmatrix}a + b + c & - c & - b \\ - c & a + b + c & - a \\ - b & - a & a + b + c\end{vmatrix} = 2\left( a + b \right) \left( b + c \right) \left( c + a \right)\]
Chapter: [4] Determinants
Concept: undefined >> undefined
Concept: undefined >> undefined
3x + ay = 4
2x + ay = 2, a ≠ 0
Chapter: [4] Determinants
Concept: undefined >> undefined
Concept: undefined >> undefined
The trace of the matrix \[A = \begin{bmatrix}1 & - 5 & 7 \\ 0 & 7 & 9 \\ 11 & 8 & 9\end{bmatrix}\], is
Chapter: [3] Matrices
Concept: undefined >> undefined
Concept: undefined >> undefined
3x + y + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11
Chapter: [4] Determinants
Concept: undefined >> undefined
Concept: undefined >> undefined
x − 4y − z = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1
Chapter: [4] Determinants
Concept: undefined >> undefined
Concept: undefined >> undefined
6x + y − 3z = 5
x + 3y − 2z = 5
2x + y + 4z = 8
Chapter: [4] Determinants
Concept: undefined >> undefined
Concept: undefined >> undefined
2y − 3z = 0
x + 3y = − 4
3x + 4y = 3
Chapter: [4] Determinants
Concept: undefined >> undefined
Concept: undefined >> undefined
5x − 7y + z = 11
6x − 8y − z = 15
3x + 2y − 6z = 7
Chapter: [4] Determinants
Concept: undefined >> undefined
Concept: undefined >> undefined
2x − 3y − 4z = 29
− 2x + 5y − z = − 15
3x − y + 5z = − 11
Chapter: [4] Determinants
Concept: undefined >> undefined
Concept: undefined >> undefined
x + y + z + 1 = 0
ax + by + cz + d = 0
a2x + b2y + x2z + d2 = 0
Chapter: [4] Determinants
Concept: undefined >> undefined
Concept: undefined >> undefined
