If [2−1]=[P−1], then P =

# If $\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]\left[\begin{array}{c}2\\ -1\end{array}\right]=\left[\begin{array}{c}P\\ -1\end{array}\right]$, then P =

1. A

0

2. B

–1

3. C

2

4. D

1

Fill Out the Form for Expert Academic Guidance!l

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

$\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]\left[\begin{array}{c}2\\ -1\end{array}\right]=\left[\begin{array}{c}P\\ -1\end{array}\right]$

$⇒\left[\begin{array}{l}1\left(2\right)+3\left(-1\right)\\ 0\left(2\right)+1\left(-1\right)\end{array}\right]=\left[\begin{array}{l}P\\ -1\end{array}\right]⇒\left[\begin{array}{l}-1\\ -1\end{array}\right]=\left[\begin{array}{l}P\\ -1\end{array}\right]⇒P=-1$

## Related content

 Area of Square Area of Isosceles Triangle Pythagoras Theorem Triangle Formulae Perimeter of Triangle Formula Area Formulae Volume of Cone Formula Matrices and Determinants_mathematics Critical Points Solved Examples Type of relations_mathematics  +91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)