GATE CSE 1996 | Question: 10

Question

Let $A = \begin{bmatrix} a_{11} && a_{12} \\ a_{21} && a_{22} \end{bmatrix} \text { and } B = \begin{bmatrix} b_{11} && b_{12} \\ b_{21} && b_{22} \end{bmatrix}$ be two matrices such that $AB=I$. Let $C = A \begin{bmatrix} 1 && 0 \\ 1 && 1 \end{bmatrix}$ and $CD =I$. Express the elements of $D$ in terms of the elements of $B$.

Comments

caluclating we get the inversse of X  = (1\(det X)) * adjoint of X, and then multiply by B to get the desired matrix

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Note to self:

In this question i computed $D$ as $A^{-1}[MATRIX]^{-1}$ which gave me wrong matrix.

i should have applied the correct property initially.

since $D=C^{-1}$ and $C=A[MATRIX]$ inverse both sides. we get $C^{-1}=[MATRIX]^{-1}A^{-1}$ i wrote it wrong.

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Very Easy Explanation : https://gateoverflow.in/2762/gate-cse-1996-question-10?show=532327#a532327

Answer 1

$AB = I$, B is equal to the inverse of $A$ and vice versa.

So, $B= A^{-1}$

Now $CD =I$, $C$ is equal to the inverse of $D$ and vice versa.

So, $D =C^{-1}$ ​​​​​​

​$=\left(A.\begin{bmatrix} 1& 0 \\ 1&1 \end{bmatrix}\right)^{-1}$

Remark: $(AB)^{-1} = B^{-1}A^{-1} $

​$=\begin{bmatrix} 1& 0 \\ 1&1 \end{bmatrix}^{-1}.A^{-1}$

​$=\begin{bmatrix} 1& 0 \\ {-1}&1 \end{bmatrix}.B$

 ​$=\begin{bmatrix} b_{11}& b_{12} \\ b_{21}-b_{11}&b_{22}-b_{12} \end{bmatrix}$

Comments

$(1 - \lambda)^{2} = 0\\ 2\lambda - \lambda ^{2} = 1\\ 2 - \lambda = \lambda^{-1}\\ X^{-1} = 2I - X$

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If $A/B = A * inverse(B)$

I did the same and:

$inverse(C) = inverse(A) * inverse($\begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix}$)$

 

What am I doing wrong?

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thats the matrix itself after multiplying it with B

Answer 2

Suppose D = $\begin{bmatrix} d_{11} & d_{12} \\ d_{21} & d_{22} \end{bmatrix}$

Now, CD = AB  where C= A$\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$

so,  A$\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ = AB  [here A will be cancel out because on bothside A is a matrix]

$\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ . $\begin{bmatrix} d_{11} & d_{12} \\ d_{21} & d_{22} \end{bmatrix}$ = A$\begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}$

$\begin{bmatrix} d_{11} & d_{12} \\ d_{11} + d_{21} & d_{12} + d_{22} \end{bmatrix}$ = $\begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}$

two matrix equally same means they are element wise equal.by this we can easily calculate, D = $\begin{bmatrix} b_{11} & b_{12} \\ b_{21} - b_{11} & b_{22} - b_{12} \end{bmatrix}$

Comments

in 3rd line why have you not written D after substituting value of C ?

Answer 3

AB=CD

let x=[ 1  0 ]    

         [ 1  1]

AB=AXD

D=inv(X) *B

ANS IS          [b11                b12  ]

                     [ b21-b11   b22-b12  ]

Answer 4

Important concept Here : The inverse of a product is the product of the inverses in reverse order: $(AM)^{-1} = M^{-1}A^{-1}$.

• We are given $AB = I$, which implies $A^{-1} = B$.

• We are given $C = A \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$. Let $M = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$. 

•  $C = AM$.

• We are given $CD = I$, which implies $D = C^{-1}$.

2. Deriving $D$ in terms of $B$

Substitute the expression for $C$ into the equation for $D$:

$$D = (AM)^{-1}$$

By the property of inverse matrices, $(AM)^{-1} = M^{-1} A^{-1}$:

$$D = M^{-1} A^{-1}$$

Since $A^{-1} = B$, we substitute $B$:

$$D = M^{-1} B$$

3. Calculating $M^{-1}$

For $M = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$, the inverse $M^{-1}$ is calculated as:

$$M^{-1} = \frac{1}{(1)(1) - (0)(1)} \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}$$

4. Calculation

Now, compute $D = M^{-1} B$:

$$D = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}$$

$$D = \begin{bmatrix} (1)b_{11} + (0)b_{21} & (1)b_{12} + (0)b_{22} \\ (-1)b_{11} + (1)b_{21} & (-1)b_{12} + (1)b_{22} \end{bmatrix}$$

The elements of matrix $D$ are:

$$D = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} - b_{11} & b_{22} - b_{12} \end{bmatrix}$$

 

Answer 5

Let |1 0| is X.

      |1 1|

C = AX

AB = I, A = 1/B

C = (1/B)*X

CD = (1/B)*XD = 1*I XD = B. // left multiplication by B

D = (1/X)*B. // left multiplication by 1/X

(1/X)*B = |1 -1| * |b11 b12|

                |0 1|     |b21 b22|

            = |(b11-b21) b12-b22|

               | b21 b22 |

            = D

Comments

answer is wrong