Matrices and Determinants · Mathematics · WB JEE

The values of x for which the given matrix $$\left[ {\matrix{ { - x} & x & 2 \cr 2 & x & { - x} \cr x & { - 2} & { - x} \cr } } \right]$$ will be non-singular are

If the matrix $$\left[ {\matrix{ a & b \cr c & d \cr } } \right]$$ is commutative with the matrix $$\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]$$ then

If A is a square matrix. Then

If A² $$-$$ A + I = 0, then the inverse of the matrix A is

If A and B are square matrices of the same order and AB = 3I, then A^$$-$$1 is equal to

If the matrices $$A = \left[ {\matrix{ 2 & 1 & 3 \cr 4 & 1 & 0 \cr } } \right]$$ and $$B = \left[ {\matrix{ 1 & { - 1} \cr 0 & 2 \cr 5 & 0 \cr } } \right]$$, then AB will be

If $$\omega$$ is an imaginary cube root of unity and $$\left| {\matrix{ {x + {\omega ^2}} & \omega & 1 \cr \omega & {{\omega ^2}} & {1 + x} \cr 1 & {x + \omega } & {{\omega ^2}} \cr } } \right| = 0$$, then one of the values of x is

If $$A = \left[ {\matrix{ 1 & 2 \cr { - 4} & { - 1} \cr } } \right]$$ then A^$$-$$1 is

If A and B are two matrices such that A + B and AB are both defined, then

If $$A = \left( {\matrix{ 3 & {x - 1} \cr {2x + 3} & {x + 2} \cr } } \right)$$ is a symmetric matrix, then the value of x is

If $$z = \left| {\matrix{ 1 & {1 + 2i} & { - 5i} \cr {1 - 2i} & { - 3} & {5 + 3i} \cr {5i} & {5 - 3i} & 7 \cr } } \right|$$, then $$(i = \sqrt { - 1} )$$

If one of the cube roots of 1 be $$\omega$$, then $$\left| {\matrix{ 1 & {1 + {\omega ^2}} & {{\omega ^2}} \cr {1 - i} & { - 1} & {{\omega ^2} - 1} \cr { - i} & { - 1 + \omega } & { - 1} \cr } } \right| = $$

$$\left| {\matrix{ {a - b} & {b - c} & {c - a} \cr {b - c} & {c - a} & {a - b} \cr {c - a} & {a - b} & {b - c} \cr } } \right| = $$

If $$A=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right)$$ and $$\theta=\frac{2 \pi}{7}$$, then $$A^{100}=A \times A \times \ldots .(100$$ times) is equal to

$$ \text { If }\left|\begin{array}{lll} x^k & x^{k+2} & x^{k+3} \\ y^k & y^{k+2} & y^{k+3} \\ z^k & z^{k+2} & z^{k+3} \end{array}\right|=(x-y)(y-z)(z-x)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \text {, then } $$

If $$\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right] \cdot A \cdot\left[\begin{array}{cc}-3 & 2 \\ 5 & -3\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$$, then $$A=$$

Let $$A=\left(\begin{array}{ccc}1 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & 1 & 1\end{array}\right), B=\left(\begin{array}{l}2 \\ 1 \\ 7\end{array}\right)$$

Then for the validity of the result $$\mathrm{AX}=\mathrm{B}, \mathrm{X}$$ is

Let $$A=\left[\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]$$, then

Let A and B are orthogonal matrices and det A + det B = 0. Then

Let $$A = \left( {\matrix{ 2 & 0 & 3 \cr 4 & 7 & {11} \cr 5 & 4 & 8 \cr } } \right)$$. Then

If the matrix M_r is given by $${M_r} = \left( {\matrix{ r & {r - 1} \cr {r - 1} & r \cr } } \right)$$ for r = 1, 2, 3, ... then det (M₁) + det (M₂) + ... + det (M₂₀₀₈) =

Let $$\alpha,\beta$$ be the roots of the equation $$a{x^2} + bx + c = 0,a,b,c$$ real and $${s_n} = {\alpha ^n} + {\beta ^n}$$ and $$\left| {\matrix{ 3 & {1 + {s_1}} & {1 + {s_2}} \cr {1 + {s_1}} & {1 + {s_2}} & {1 + {s_3}} \cr {1 + {s_2}} & {1 + {s_3}} & {1 + {s_4}} \cr } } \right| = k{{{{(a + b + c)}^2}} \over {{a^4}}}$$ then $$k = $$

Let $$A = \left( {\matrix{ 0 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right),B = \left( {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 0 & 0 & 0 \cr } } \right)$$ and $$P\left( {\matrix{ 0 & 1 & 0 \cr x & 0 & 0 \cr 0 & 0 & y \cr } } \right)$$ be an orthogonal matrix such that $$B = PA{P^{ - 1}}$$ holds. Then

Under which of the following condition(s) does(do) the system of equations $$\left( {\matrix{ 1 & 2 & 4 \cr 2 & 1 & 2 \cr 1 & 2 & {(a - 4)} \cr } } \right)\left( {\matrix{ x \cr y \cr z \cr } } \right) = \left( {\matrix{ 6 \cr 4 \cr a \cr } } \right)$$ possesses(possess) unique solution ?

If $$\Delta (x) = \left| {\matrix{ {x - 2} & {{{(x - 1)}^2}} & {{x^3}} \cr {x - 1} & {{x^2}} & {{{(x + 1)}^3}} \cr x & {{{(x + 1)}^2}} & {{{(x + 2)}^3}} \cr } } \right|$$, then coefficient of x in $$\Delta$$x is

If $$p = \left[ {\matrix{ 1 & \alpha & 3 \cr 1 & 3 & 3 \cr 2 & 4 & 4 \cr } } \right]$$ is the adjoint of the $$3 \times 3$$ matrix A and det A = 4, then $$\alpha$$ is equal to

If $$A = \left( {\matrix{ 1 & 1 \cr 0 & i \cr } } \right)$$ and $${A^{2018}} = \left( {\matrix{ a & b \cr c & d \cr } } \right)$$, then $$(a + d)$$ equals

The solution of $$\det (A - \lambda {I_2}) = 0$$ be 4 and 8 and $$A = \left( {\matrix{ 2 & 2 \cr x & y \cr } } \right)$$. Then

(I₂ is identity matrix of order 2)

If A, B are two square matrices such that AB = A and BA = B, then prove that B² = B.

If $$\mathrm{a}_{\mathrm{i}}, \mathrm{b}_{\mathrm{i}}, \mathrm{c}_{\mathrm{i}} \in \mathbb{R}(\mathrm{i}=1,2,3)$$ and $$x \in \mathbb{R}$$ and $$\left|\begin{array}{lll}\mathrm{a}_1+b_1 x & a_1 x+b_1 & c_1 \\ \mathrm{a}_2+b_2 x & a_2 x+b_2 & c_2 \\ \mathrm{a}_3+b_3 x & a_3 x+b_3 & c_3\end{array}\right|=0$$, then

Let $$\Delta = \left| {\matrix{ {\sin \theta \cos \phi } & {\sin \theta \sin \phi } & {\cos \theta } \cr {\cos \theta \cos \phi } & {\cos \theta \sin \phi } & { - \sin \theta } \cr { - \sin \theta \sin \phi } & {\sin \theta \cos \phi } & 0 \cr } } \right|$$. Then