Difficult
Suppose the vectors $x_{1}, x_{2}$ and $x_{3}$ are the solutions of the system of linear equations $Ax = b$ when the vector $b$ on the right side is equal to $b_{1}, b_{2}$ and $b_{3}$ respectively. If $x_{1} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, x_{2} = \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix}, x_{3} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, b_{1} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, b_{2} = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix}$ and $b_{3} = \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix}$,then the determinant of $A$ is equal to
- A$1/2$
- B$4$
- C$3/2$
- D$2$
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Let $d \in \mathbb{R}$,and $A = \begin{bmatrix} -2 & 4+d & \sin \theta - 2 \\ 1 & \sin \theta + 2 & d \\ 5 & 2\sin \theta - d & -\sin \theta + 2 + 2d \end{bmatrix}$,where $\theta \in [0, 2\pi]$. If the minimum value of $\det(A)$ is $8$,then a value of $d$ is
DifficultJEE MAIN 2019
View SolutionLet $A = \begin{bmatrix} \frac{1}{6} & \frac{-1}{3} & \frac{-1}{6} \\ \frac{-1}{3} & \frac{2}{3} & \frac{1}{3} \\ \frac{-1}{6} & \frac{1}{3} & \frac{1}{6} \end{bmatrix}$. If $A^{2016l} + A^{2017m} + A^{2018n} = \frac{1}{\alpha} A$, for every $l, m, n \in N$, then the value of $\alpha$ is
If $A = \begin{bmatrix} 1 & a & 3 \\ b & 2 & c \\ 3 & d & 4 \end{bmatrix}$ is a symmetric matrix and $B = \begin{bmatrix} 0 & 5 & b \\ -5 & 0 & -7 \\ 6 & c & 0 \end{bmatrix}$ is a skew-symmetric matrix, then $AB = $
Give the correct order of initials $T$ or $F$ for following statements. Use $T$ if statement is true and $F$ if it is false.
Statement $-1$ : If $A$ is an invertible $3 \times 3$ matrix and $B$ is a $3 \times 4$ matrix,then $A^{-1}B$ is defined.
Statement $-2$ : It is never true that $A + B, A - B$,and $AB$ are all defined.
Statement $-3$ : Every matrix none of whose entries are zero is invertible.
Statement $-4$ : Every invertible matrix is square and has no two rows the same.
Statement $-1$ : If $A$ is an invertible $3 \times 3$ matrix and $B$ is a $3 \times 4$ matrix,then $A^{-1}B$ is defined.
Statement $-2$ : It is never true that $A + B, A - B$,and $AB$ are all defined.
Statement $-3$ : Every matrix none of whose entries are zero is invertible.
Statement $-4$ : Every invertible matrix is square and has no two rows the same.
The equation $\left| \begin{array}{ccc} (1+x)^2 & (1-x)^2 & -(2+x^2) \\ 2x+1 & 3x & 1-5x \\ x+1 & 2x & 2-3x \end{array} \right| + \left| \begin{array}{ccc} (1+x)^2 & 2x+1 & x+1 \\ (1-x)^2 & 3x & 2x \\ 1-2x & 3x-2 & 2x-3 \end{array} \right| = 0$
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