જો A=left[begin{array}{lll}1 & 1 & -2 2 & 1 & -3 5 & 4 & -9end{array}rig

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3 and 4 .Determinants and Matrices

easy

જો $A=\left[\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right]$ હોય, તો $|A|$ શોધો.

$1$

$2$

$0$

$3$

Solution

Let $A=\left[\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right]$

By expanding along the first row, we have:

$A=\left[\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right]$

$|A|=1\left|\begin{array}{cc}1 & -3 \\ 4 & -9\end{array}\right|-1\left|\begin{array}{cc}2 & -3 \\ 5 & -9\end{array}\right|-2\left|\begin{array}{cc}2 & 1 \\ 5 & 4\end{array}\right|$

$=1(-9+12)-1(-18+15)-2(8-5)$

$=1(3)-1(-3)-2(3)$

$=3+3-6$

$=6-6$

$=0$

Standard 12

Mathematics

જો સુરેખ સમીકરણ સંહતિ $x + ky + 3z = 0;3x + ky – 2z = 0$ ; $2x + 4y – 3z = 0$ ને શૂન્યતેર ઉકેલ $\left( {x,y,z} \right)$ હોય ,તો $\frac{{xz}}{{{y^2}}} = $. . . . .

hard

(JEE MAIN-2018)

$\left| {\,\begin{array}{*{20}{c}}{1/a}&{{a^2}}&{bc}\\{1/b}&{{b^2}}&{ca}\\{1/c}&{{c^2}}&{ab}\end{array}\,} \right| = $

easy

$m$ ની કેટલી કિમંતો માટે રેખાઓ $x + y – 1 = 0$, $(m – 1) x + (m^2 – 7) y – 5 = 0 \,\,\&\,\, (m – 2) x + (2m – 5) y = 0$ ઓ સંગામી થાય.

normal

જો $\left| {\,\begin{array}{*{20}{c}}{ – {a^2}}&{ab}&{ac}\\{ab}&{ – {b^2}}&{bc}\\{ac}&{bc}&{ – {c^2}}\end{array}\,} \right| = K{a^2}{b^2}{c^2} $ તો $K = $

medium

જો $\left| {\,\begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x – 2}\\{2{x^2} + 3x – 1}&{3x}&{3x – 3}\\{{x^2} + 2x + 3}&{2x – 1}&{2x – 1}\end{array}\,} \right| = Ax – 12$, તો $A$ મેળવો.