If system of equations x - ky - z = 0 , kx - y - z = 0 and x + y

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3 and 4 .Determinants and Matrices

medium

If the system of equations $x - ky - z = 0$, $kx - y - z = 0$ and $x + y - z = 0$ has a non zero solution, then the possible value of k are

$-1, 2$

$1, 2$

$0, 1$

$-1, 1$

(IIT-2000)

Solution

(d) For non-trivial solution $\Delta = 0$
==> $\left| {\,\begin{array}{*{20}{c}}1&{ – k}&{ – 1}\\k&{ – 1}&{ – 1}\\1&1&{ – 1}\end{array}\,} \right|\, = \,0$ ==> $k = 1,\, – 1.$

Standard 12

Mathematics

Evaluate $\Delta=\left|\begin{array}{lll}3 & 2 & 3 \\ 2 & 2 & 3 \\ 3 & 2 & 3\end{array}\right|$

easy

Let $S=\left\{A=\left(\begin{array}{lll}0 & 1 & c \\ 1 & a & d \\ 1 & b & e\end{array}\right): a, b, c, d, e \in\{0,1\}\right.$ and $\left.|A| \in\{-1,1\}\right\}$, where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is. . . . .

medium

(IIT-2024)

The area of a triangle is $5$ and two of its vertices are $A(2, 1), B(3, -2)$. The third vertex which lies on line $y = x + 3$ is-

normal

Number of triplets of $a, b \, \& \,c$ for which the system of equations,$ax – by = 2a – b$ and $(c + 1) x + cy = 10 – a + 3 b$ has infinitely many solutions and $x = 1, y = 3$ is one of the solutions, is :