Inverse of thefollowing matrix using elementary Row transformations would be - Wegin vmatrix 183 - 11108181 1386 - 1 e n d v m a t r i x A- beginbmatrix -5 -38211382 -111-38-3 - 1 e n d b m a t r i x B- beginbmatrix -5 -382113 -2 -1 - l38-3 - 1endbmatrix C- Wbeginbmatrix -5 -3821138-28-1 1-3 -381 e n d b m a t r i x D- beginbmatrix -5 -3 - 2 3828-111-3 -3 - 1 e n d b m a t r i x

The correct option is C [ 5 amp;+3 amp;2; 3 amp; 2 amp; 1; 3 amp;+3 amp;1 ] We know, for matrix A, A = IA I = PA → P=A^ 1 = 1 amp;3 amp;1 ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Elementary Transformations

Inverse of th...

Question

Inverse of thefollowing matrix using elementary Row transformations would be

$= ∣ ∣ ∣ ∣ \begin{matrix} 1 & 3 & 1 0 & 1 & 1 3 & 6 & 1 \end{matrix} ∣ ∣ ∣ ∣$

$⎡ ⎢ ⎣ \begin{matrix} - 5 & + 3 & 2 3 & - 2 & - 1 3 & + 3 & 1 \end{matrix} ⎤ ⎥ ⎦$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$⎡ ⎢ ⎣ \begin{matrix} - 5 & + 3 & 2 3 & 2 & - 1 - 3 & + 3 & 1 \end{matrix} ⎤ ⎥ ⎦$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$⎡ ⎢ ⎣ \begin{matrix} - 5 & + 3 & 2 3 & - 2 & - 1 - 3 & + 3 & 1 \end{matrix} ⎤ ⎥ ⎦$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$⎡ ⎢ ⎣ \begin{matrix} - 5 & + 3 & 2 3 & 2 & - 1 - 3 & + 3 & 1 \end{matrix} ⎤ ⎥ ⎦$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$⎡ ⎢ ⎣ \begin{matrix} - 5 & + 3 & 2 3 & - 2 & - 1 - 3 & + 3 & 1 \end{matrix} ⎤ ⎥ ⎦$

We know, for matrix A,

A = IA

I = PA

$\to P = A^{- 1}$

$= ∣ ∣ ∣ ∣ \begin{matrix} 1 & 3 & 1 0 & 1 & 1 3 & 6 & 1 \end{matrix} ∣ ∣ ∣ ∣$

A = IA

Lets convert the LHS into a Identity Matrix and apply the same transformations on the identity matrix on the right. The final matrix obtained in the right will give the inverse.

$∣ ∣ ∣ ∣ \begin{matrix} 1 & 3 & 1 0 & 1 & 1 3 & 6 & 1 \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣ A .$

$R_{3} \to R_{3} - 3 R_{1} ∣ ∣ ∣ ∣ \begin{matrix} 1 & 3 & 1 0 & 1 & 1 0 & - 3 & - 2 \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 - 3 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣ A .$

$R_{3} \to R_{3} + 3 R_{2} t h e n R_{1} \to R_{1} - 3 R_{2} ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & - 2 0 & 1 & 1 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & - 3 & 0 0 & 1 & 0 - 3 & + 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ A$

$R_{2} \to R_{2} - R_{3} t h e n R_{1} \to R_{1} + 2 R_{3} ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} - 5 & + 3 & 2 3 & - 2 & - 1 - 3 & + 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ A$

$I = A^{- 1} A A^{- 1} = ⎡ ⎢ ⎣ \begin{matrix} - 5 & + 3 & 2 3 & - 2 & - 1 - 3 & + 3 & 1 \end{matrix} ⎤ ⎥ ⎦$

Suggest Corrections

Similar questions

Which of the following would be the Inverse of the matrix A found using Elementary Row Transformations where

A = $∣ ∣ ∣ ∣ \begin{matrix} 1 & 3 & 1 0 & 1 & 1 3 & 6 & 1 \end{matrix} ∣ ∣ ∣ ∣$

Find a matrix X such that A-X=B, given

$A = ⎛ ⎜ ⎝ \begin{matrix} 3 & 7 & 2 - 4 & - 3 & 0 5 & 1 & - 2 \end{matrix} ⎞ ⎟ ⎠ a n d B = ⎛ ⎜ ⎝ \begin{matrix} 8 & - 6 & 1 2 & 0 & - 3 3 & 4 & - 1 \end{matrix} ⎞ ⎟ ⎠$

Q. Let

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 0 & 1 & 1 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

. Then for positive integer

n, A^{n}

Q. Let

[\begin{matrix} a_{n + 1} b_{n + 1} \end{matrix}] = [\begin{matrix} \sqrt{3} + 1 & 1 - \sqrt{3} \sqrt{3} - 1 & 1 + \sqrt{3} \end{matrix}] [\begin{matrix} a_{n} b_{n} \end{matrix}],

n \in N .

a_{1} = b_{1} = 1

and

a_{22} = 2^{n} .

Then the value of

n

If $A = [\begin{matrix} 1 & 2 - 3 & 1 \end{matrix}] and B = [\begin{matrix} 11 & - 1 4 & 3 \end{matrix}]$

Find Matrix X, given that
X-5A=3B

Join BYJU'S Learning Program

Inverse of thefollowing matrix using elementary Row transformations would be =∣∣ ∣∣131011361∣∣ ∣∣

Inverse of thefollowing matrix using elementary Row transformations would be

$= ∣ ∣ ∣ ∣ \begin{matrix} 1 & 3 & 1 0 & 1 & 1 3 & 6 & 1 \end{matrix} ∣ ∣ ∣ ∣$