The inverse of [ * 20 c 1&2&3 0&1&2 0&0&1 ] is - Vidyadip

MCQ

The inverse of $\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&0&1\end{array}} \right]$ is

A
$\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\0&1&{ - 2}\\0&0&0\end{array}} \right]$
✓
$\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\0&1&{ - 2}\\0&0&1\end{array}} \right]$
C
$\left[ {\begin{array}{*{20}{c}}1&2&1\\0&1&2\\0&0&1\end{array}} \right]$
D
None of these

✓

Answer

Correct option: B.

$\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\0&1&{ - 2}\\0&0&1\end{array}} \right]$

b
(b) Let $A = \left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&0&1\end{array}} \right]$$ \Rightarrow $ $|A| = 1\,(1 + 0) = 1$

$Adj\,(A) = \left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{21}}}&{{A_{31}}}\\{{A_{12}}}&{{A_{22}}}&{{A_{32}}}\\{{A_{13}}}&{{A_{23}}}&{{A_{33}}}\end{array}} \right]$

$Adj\,(A) = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\0&1&{ - 2}\\0&0&1\end{array}} \right]$

$ \Rightarrow $${A^{ - 1}} = \frac{{Adj\,(A)}}{{|A|}} = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\0&1&{ - 2}\\0&0&1\end{array}} \right]$.

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

Let $\vec{a}=2 \hat{i}-\hat{j}+5 \hat{k}$ and $\vec{b}=\alpha \hat{i}+\beta \hat{j}+2 \hat{k}$. If $((\vec{a} \times \vec{b}) \times \hat{i}) \cdot \hat{k}=\frac{23}{2}$, then $|\vec{b} \times 2 \hat{j}|$ is equal to.

Let $A$ and $B$ be two symmetric matrices of order $3.$

Statement $-1$: $A(BA)$ and $(AB)A$ are symmetric matrices.

Statement $-2:$ $AB$ is symmetric matrix if matrix multiplication of $A$ with $B$ is commutative.

Choose the correct answer The slope of the normal to the curve $y=2 x^2+3 \sin x$ at $x=0$ is:

In linear programming, objective function and objective constraints are:

In Graphical solution the redundant constraint is:

The derivative of $F[f\{ \phi (x)\} ]$ is

A unit vector perpendicular to the plane containing the vectors $i - j + k$ and $ - i + j + k$ is

Let $a_1=1, a_2, a_3, a_4, \ldots$. be consecutive natural numbers. Then $\tan ^{-1}\left(\frac{1}{1+ a _1 a _2}\right)+\tan ^{-1}\left(\frac{1}{1+ a _2 a _3}\right)$ $+\ldots . .+\tan ^{-1}\left(\frac{1}{1+ a _{2021} a _{2022}}\right)$ is equal to

It is given that the $A $ and $B$ are such that $P\left( A \right) = \frac{1}{4}\;,P\left( {A{\rm{|}}B} \right) = \frac{1}{2}$ and $P\left( {B{\rm{|}}A} \right) = \frac{2}{3}$, then $P(B) $= . .. . .

If for non-zero $x,$ $af(x) + bf\left( {\frac{1}{x}} \right) = \frac{1}{x} - 5,$ where $a \ne b,$ then $\int_1^2 {f(x)\,dx = } $