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Given \( A=\left[\begin{array}{ll}2 & 1 \\ 2 & 1\end{array}\right] ; B=\left[\begin{array}{ll}9 & 3 \\ 3 & 1\end{array}\right] \). I is a unit matrix of order Find all possible matrix \( X \) in the following cases.(i) \( A X=A \) (ii) \( XA =1 \) (iii) \( X B=O \) but \( B X \neq O \). |
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Answer» \(A = \begin{bmatrix}2&1\\2&1\end{bmatrix} ; B = \begin{bmatrix}9&3\\3&1\end{bmatrix}\) (i) \(AX = A\) Let \(X = \begin{bmatrix}a&b\\c&d\end{bmatrix}\) \(\begin{bmatrix}2&1\\2&1\end{bmatrix} \begin{bmatrix}a&b\\c&d\end{bmatrix}= \begin{bmatrix}2&1\\2&1\end{bmatrix} \) ⇒ \(\begin{bmatrix}2a +c&2b + d\\2a +c&2b+d\end{bmatrix} = \begin{bmatrix}2&1\\2&1\end{bmatrix} \) \(\therefore 2a + c = 2\) ⇒ \(c = 2 - 2a = 2(1 - a) ; a\in R\) \(\therefore 2b + d = 1\) ⇒ \(d = 1 - 2b; b \in R\) So, there is infinite many such matrices X exist for which AX = A as |A| = 0. X is of the type \(\begin{bmatrix}a&b\\2(1 -a)&1-2b\end{bmatrix}\). (ii) \(XA = I\) \(\begin{bmatrix}a&b\\c&d\end{bmatrix} \begin{bmatrix}2&1\\2&1\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix} \) \(\begin{bmatrix}2(a +b)&a + b\\2(c + d)&c+d\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix} \) \(\therefore 2(a +b) = 1\) ⇒ \(a + b = \frac 12 \) (By comparing a11 elements of both matrices) and \(a +b = 0\) (By comparing a12 elements of both matrices) (Not possible at same time) Hence, no such matrix X exists for Which XA = I. (iii) \(XB = 0 \) but \(BX \ne 0\) \(\begin{bmatrix}a&b\\c&d\end{bmatrix} \begin{bmatrix}9&3\\3&1\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix} \) ⇒ \(\begin{bmatrix}3(3a +b)&3a + b\\3(3c + d)&3c+d\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix} \) ⇒ \(3a + b = 0\) and \(3c + d = 0\) ⇒ \(b = -3a \) and \(d= -3c; a,c \in R\) But \(BX \ne 0\) \(\begin{bmatrix}9&3\\3&1\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix} \ne \begin{bmatrix}0&0\\0&0\end{bmatrix} \) ⇒ \(\begin{bmatrix}3(3a +c)&3(3b + d)\\3a+c&3b+d\end{bmatrix} \ne\begin{bmatrix}0&0\\0&0\end{bmatrix} \) ⇒ \(3a + c \ne 0\) & \(3b +d \ne 0\) ⇒ \(3b - 3 c \ne 0\) ⇒ \(b - c \ne 0\) ⇒ \(b \ne c\) ⇒ \(-3a \ne c\) Hence, \(X = \begin{bmatrix}a&b\\c&d\end{bmatrix} = \begin{bmatrix}a&-3a\\c&-3c\end{bmatrix} ; c\ne -3a\) Hence, there is infinite many such X matrices exist for which \(XB = 0\) but \(BX \ne 0\) and of the type \(X =\begin{bmatrix}a&-3a\\c&-3c\end{bmatrix} ; c\ne -3a\). |
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