Determinants Te...

$$\left| \begin{matrix} a & b-c & c+b \\ a+c & b & c-a \\ a-b & b+a & c \end{matrix} \right| $$ =

To solve  $$x + y = 3 : 3 x - 2 y - 4 = 0$$  by determinant method find  $$D.$$

If $$(k,2-2k),(-k+1,2k),(-4-k,6-2k)$$ are collinear, then $$k=$$

If $$A = \left[ \begin{array} { l l } { 1 } & { 2 } \\ { 2 } & { 1 } \end{array} \right]$$ then  $$adj (A) =?$$

If a,b,c are distinct and $$\left| \begin{matrix} a & { a }^{ 2 } & { a }^{ 3 }-1 \\ b & { b }^{ 2 } & { b }^{ 3 }-1 \\ c & { c }^{ 2 } & { c }^{ 3 }-1 \end{matrix} \right| =0$$ then

If f(x), g(x), h(x) are polynomials in x of degree 2 and F(x)=$$\left| \begin{matrix} f & g & h \\ { f' } & g' & h' \\ f" & g" & h" \end{matrix} \right| ,$$ , then F(x) is equal to

If the point $$\left(\lambda+1,1\right),\left(2\lambda+1,3\right)$$ and $$\left(2\lambda+2,2\lambda\right)$$ are collinear then the possible value of $$\lambda$$ is

Solve $$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{vmatrix}$$

If $${ D }_{ P }=\left| \begin{matrix} P & 15 & 8 \\ { P }^{ 2 } & 35 & 9 \\ { P }^{ 3 } & 25 & 10 \end{matrix} \right| ,$$ then $${ D }_{ 1 }+{ D }_{ 2 }+{ D }_{ 3 }+{ D }_{ 4 }+{ D }_{ 5 }$$ is equal to -

The value of determinant $$\left| \begin{matrix} { bc-a }^{ 2 } & { ac-b }^{ 2 } & ab-c^{ 2 } \\ { ac-b }^{ 2 } & { ab-c }^{ 2 } & { bc-a }^{ 2 } \\ { ab-c }^{ 2 } & { bc-a }^{ 2 } & ac-b^{ 2 } \end{matrix} \right| $$ is