Download App

STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
MCQ
An ordered pair $(\alpha , \beta )$ for which the system of linear equations

$\left( {1 + \alpha } \right)x + \beta y + z = 2$ ; $\alpha x + \left( {1 + \beta } \right)y + z = 3$ ; $\alpha x  + \beta y + 2z = 2$ has a unique solution, is

  • ✓
    $(2, 4)$
  • B
    $(-3, 1)$
  • C
    $(-4, 2)$
  • D
    $(1, -3)$
✓

Answer

Correct option: A.
$(2, 4)$
a
$\left( {1 + \alpha } \right)x + \beta y + z = 0$

$\alpha x + \left( {1 + \beta } \right)y + z = 0$

$\alpha x + \beta y + 2z = 0$

$D = \left| {\begin{array}{*{20}{c}}
{1 + \alpha }&\beta &1\\
\alpha &{1 + \beta }&1\\
\alpha &\beta &2
\end{array}} \right|$

${C_1} \to {C_1} + {C_2} + {C_3}$

$D = \left( {\alpha  + \beta  + 2} \right)\left| {\begin{array}{*{20}{c}}
1&\beta &1\\
1&{1 + \beta }&1\\
1&\beta &2
\end{array}} \right|$

${R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}$

$D = \left( {\alpha  + \beta  + 2} \right)\left| {\begin{array}{*{20}{c}}
1&\beta &1\\
0&1&0\\
0&0&1
\end{array}} \right| = \alpha  + \beta  + 2$

For unique solution $\alpha  + \beta  + 2 \ne 0 \Rightarrow \alpha  + \beta  \ne  - 2$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 3 and 4 . determinant and metrices→STD 12 - 3 and 4 . determinant and metrices — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

If $\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 3}\\{x + 2}&{x + 3}&{x + 4}\\{x + a}&{x + b}&{x + c}\end{array}\,} \right| = 0$, then $a,b,c$ are in
→
Consider a square of side $2\ cm$ . It is cut from one of its corner as shown in adjacent figure. Maximum value of sum of the perimeters of two plane figures thus formed is
→
A box contains 6 nails and 10 nuts. Half of the nails and half of the nuts are rusted. If one iten is chosen ar random, the probability that it is rusted or is nail is
→
The area of the region bounded by the curve $x = 2y + 3$ and lines $y = 1$ and $y = –1$ is:
→
$\left| {\,\begin{array}{*{20}{c}}{bc}&{bc' + b'c}&{b'c'}\\{ca}&{ca' + c'a}&{c'a'}\\{ab}&{ab' + a'b}&{a'b'}\end{array}\,} \right|$ is equal to
→
The angle between the straight lines $\frac{\text{x}+1}{2}=\frac{\text{y}-2}{5}=\frac{\text{z}+3}{4}$ and $\frac{\text{x}-1}{1}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{-3}$ is:
→
$\int {\frac{{\left( {x - 3} \right){e^x}}}{{{{\left( {x - 1} \right)}^3}}}\,dx}$ is equal to
→
If $\tan^{-1}\frac{\text{x}+1}{\text{x}-1}+\tan^{-1}\frac{\text{x}-1}{\text{x}}=\tan^{-1}(-7),$ then the value of x is:
→
The corner points of the feasible region determined by the system of linear constraints are $(0,10),(5,5),(15,15),(0,20) .$ Let $z=p x+q y,$ where $p, q\,>\,0 .$ Condition on $p$ and $q$ so that the maximum of $z$ occurs at both the pooints $(15,15)$ and $(0,20)$ is $\ldots \ldots$
→
The vector $\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{k}}$ is to be written as the sum of a vector $\vec{\alpha}$ parallel to $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}$ and $a$ vector $\vec{\beta}$ perpendicular to $\vec{\text{a}}.$ Then $\vec{\alpha}=$
→