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$\frac{1}{2}\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$
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$\frac{1}{2}\begin{bmatrix} -2 & 4 \\ 3 & 5 \end{bmatrix}$
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$\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$
- None of these.
135 questions across 4 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
$\frac{1}{2}\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$
$\frac{1}{2}\begin{bmatrix} -2 & 4 \\ 3 & 5 \end{bmatrix}$
$\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$
$19$
$\frac{1}{19}$
$-19$
$-\frac{1}{19}$
$\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$
$\text{xyz}\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$
$\frac{1}{\text{xyz}}\begin{bmatrix}\text{x} & 0 & 0\\ 0 & \text{y} & 0 \\ 0 & 0 & \text{z}\end{bmatrix}$
$\frac{1}{\text{xyz}}\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$|\text{adj A}|=|\text{A}|^2$
$(\text{A}^{-1})^{-1}=\text{A}$
$(\text{AB})^{-1}=\text{B}^{-1}\text{A}^{-1},$ where $\text{B}\neq\big[\text{b}_{\text{ij}}\big]_{3\times3}\text{ and |B|}\neq0$
$\begin{bmatrix}1 & \frac{\tan\alpha}{2} \\ -\frac{\tan\alpha}{2} & 1 \end{bmatrix}$
Verify that (adjoint A) A = |A|I = A (adjoint A) for the above matrices.
$\begin{bmatrix}2 & 5 \\ -3 & 1 \end{bmatrix}$
$\text{G}(\beta)=\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}$
Show that$\big[\text{G}(\beta)\big]^{-1}=\text{G}(-\beta)$
$\text{G}(\beta)=\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}$
Show that$\big[\text{F}(\alpha)\big]^{-1}=\text{F}(-\alpha)$
$\begin{bmatrix}0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix}$
$\begin{bmatrix}3 & 10 \\ 2 & 7 \end{bmatrix}$
Pick question groups from the list above, set marks and difficulty, and export a branded PDF with step-by-step answer keys. First 3 chapters free — no signup.