Construct a 3 × 4 matrix, whose elements are given by a

Hindi

3 and 4 .Determinants and Matrices

easy

Construct a $3 \times 4$ matrix, whose elements are given by $a_{i j}=\frac{1}{2}|-3 i+j|$.

$A=\left[\begin{array}{cccc}1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2}\end{array}\right]$

$A=\left[\begin{array}{cccc}1 & \frac{3}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{3}{2}\end{array}\right]$

$A=\left[\begin{array}{cccc}1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{7}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{-7}{2} & 3 & \frac{5}{2}\end{array}\right]$

$A=\left[\begin{array}{cccc}1 & \frac{-1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & -3 & \frac{5}{2}\end{array}\right]$

Solution

In general, a $3 \times 4$ matrix is given by $A=\left[\begin{array}{llll}a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34}\end{array}\right]$

Given : $a_{ij}=\frac{1}{2}|-3 i+j|$, $i=1,\,2,\,3$ and $j=1,\,2,\,3,\,4$

Thus, we have

${a_{11}} = \frac{1}{2}| – 3 \times 1 + 1| = $ $\frac{1}{2}| – 3 + 1| = \frac{1}{2}| – 2| = \frac{2}{2} = 1$

${a_{21}} = \frac{1}{2}| – 3 \times 2 + 1| = $ $\frac{1}{2}| – 6 + 1| = \frac{1}{2}| – 5| = \frac{5}{2}$

${a_{31}} = \frac{1}{2}| – 3 \times 3 + 1| = $ $\frac{1}{2}| – 9 + 1| = \frac{1}{2}| – 8| = 4$

${a_{12}} = \frac{1}{2}| – 3 \times 1 + 2| = $ $\frac{1}{2}| – 3 + 2| = \frac{1}{2}| – 1| = \frac{1}{2}$

${a_{22}} = \frac{1}{2}| – 3 \times 2 + 2| = $ $\frac{1}{2}| – 6 + 2| = \frac{1}{2}| – 4| = \frac{4}{2} = 2$

${a_{32}} = \frac{1}{2}| – 3 \times 3 + 2| = $ $\frac{1}{2}| – 9 + 2| = \frac{1}{2}| – 7| = \frac{7}{2}$

$a_{13}=\frac{1}{2}|-3 \times 1+3|=\frac{1}{2}|-3+3|=0$

${a_{23}} = \frac{1}{2}| – 3 \times 2 + 3| = $ $\frac{1}{2}| – 6 + 3| = \frac{1}{2}| – 3| = \frac{3}{2}$

${a_{33}} = \frac{1}{2}| – 3 \times 3 + 3| = $ $\frac{1}{2}| – 9 + 3| = \frac{1}{2}| – 6| = \frac{6}{2} = 3$

${a_{14}} = \frac{1}{2}| – 3 \times 1 + 4| = $ $\frac{1}{2}| – 3 + 4| = \frac{1}{2}|1| = \frac{1}{2}$

${a_{24}} = \frac{1}{2}| – 3 \times 2 + 4| = $ $\frac{1}{2}| – 6 + 4| = \frac{1}{2}| – 2| = \frac{2}{2} = 1$

${a_{34}} = \frac{1}{2}| – 3 \times 3 + 4| = $ $\frac{1}{2}| – 9 + 4| = \frac{1}{2}| – 5| = \frac{5}{2}$

Therefore, the required matrix is $A=\left[\begin{array}{cccc}1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2}\end{array}\right]$

Standard 12

Mathematics

Let $A=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right], B=\left[B_1, B_2, B_3\right]$, where $B_1$, $\mathrm{B}_2, \mathrm{~B}_3$ are column matrices, and $\mathrm{AB}_1=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$, $\mathrm{AB}_2=\left[\begin{array}{l}2 \\ 3 \\ 0\end{array}\right], \mathrm{AB}_3=\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right]$ If $\alpha=|B|$ and $\beta$ is the sum of all the diagonal elements of $B$, then $\alpha^3+\beta^3$ is equal to

hard

(JEE MAIN-2024)

The solution of the equation $\left[ {\begin{array}{*{20}{c}}1&0&1\\{ – 1}&1&0\\0&{ – 1}&1\end{array}} \right]\,\left[ \begin{array}{l}x\\y\\z\end{array} \right] = \left[ \begin{array}{l}1\\1\\2\end{array} \right]$ is $(x,y,z)$=

easy

Let $\mathrm{p}$ be an odd prime number and $\mathrm{T}_{\mathrm{p}}$ be the following set of $2 \times 2$ matrices :

$T_p=\left\{A=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{a}\end{array}\right]: \mathrm{a}, \mathrm{b}, \mathrm{c} \in\{0,1, \ldots ., \mathrm{p}-1\}\right\}$

$1.$ The number of $A$ in $T_p$ such that $A$ is either symmetric or skew-symmetric or both, and $\operatorname{det}(\mathrm{A})$ divisible by $\mathrm{p}$ is

$(A)$ $(\mathrm{p}-1)^2$ $(B)$ $2(\mathrm{p}-1)$

$(C)$ $(\mathrm{p}-1)^2+1$ $(D)$ $2 \mathrm{p}-1$

$2.$ The number of $A$ in $T_p$ such that the trace of $A$ is not divisible by $p$ but det $(A)$ is divisible by $p$ is [Note: The trace of a matrix is the sum of its diagonal entries.]

$(A)$ $(\mathrm{p}-1)\left(\mathrm{p}^2-\mathrm{p}+1\right)$ $(B)$ $\mathrm{p}^3-(\mathrm{p}-1)^2$

$(C)$ $(\mathrm{p}-1)^2$ $(D)$ $(p-1)\left(p^2-2\right)$

$3.$ The number of $A$ in $T_p$ such that det $(A)$ is not divisible by $p$ is

$(A)$ $2 \mathrm{p}^2$ $(B)$ $p^3-5 p$ $(C)$ $p^3-3 p$ $(D)$ $p^3-p^2$

Give the answer question $1,2$ and $3.$

normal

(IIT-2010)

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\end{array}} \right]$ $ + \left[ {\begin{array}{{20}{c}}
{2ab}&{2bc} \\
{ – 2ac}&{ – 2ab}
\end{array}} \right]$

easy

If $A,B,C$ are three square matrices such that $AB = AC$ implies $B = C$, then the matrix $A$ is always a/an

normal

Construct a $3 \times 4$ matrix, whose elements are given by $a_{i j}=\frac{1}{2}|-3 i+j|$.

Solution

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