Question

A=[a 0 0 2] B=[0 -b 1 0] and M=[1 -1 1 1] and BA=M² , find the values of a and b .​

Answer 1

$\:\begin{gathered}\begin{gathered}\bf\: Given-\begin{cases} &\sf{A = \: \begin{bmatrix} a & 0\\ 0 & 2\end{bmatrix}} \\ \\ &\sf{B = \: \begin{bmatrix} 0 & - b\\ 1 & 0\end{bmatrix}} \\ \\ &\sf{M = \: \begin{bmatrix} 1 & - 1\\ 1 & 1\end{bmatrix}} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{values \: of \: a \: and \: b}\end{cases}\end{gathered}\end{gathered}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:A = \: \begin{bmatrix} a & 0\\ 0 & 2\end{bmatrix}$

$\rm :\longmapsto\:B = \: \begin{bmatrix} 0 & -b\\ 1 & 0\end{bmatrix}$

$\rm :\longmapsto\:M = \: \begin{bmatrix} 1 & - 1\\ 1 & 1\end{bmatrix}$

According to statement,

$\rm :\longmapsto\:BA = {M}^{2}$

$\rm :\implies\:BA = M \times M$

$\rm :\longmapsto\: \: \begin{bmatrix} 0 & -b\\ 1 & 0\end{bmatrix} \times \: \begin{bmatrix} a & 0\\ 0 & 2\end{bmatrix} = \: \begin{bmatrix} 1 & - 1\\ 1 & 1\end{bmatrix} \times \: \begin{bmatrix} 1 & - 1\\ 1 & 1\end{bmatrix}$

$\rm\: \begin{bmatrix} 0 \times a - b \times 0 & 0 \times 0-2 \times b\\ a \times 1 + 0 \times 0& 1 \times 0 + 0 \times 2\end{bmatrix} = \: \begin{bmatrix} 1 \times 1 - 1 \times 1 & - 1 \times 1 - 1 \times 1\\ 1 \times 1 + 1 \times 1 & - 1 \times 1 + 1 \times 1\end{bmatrix}$

$\rm :\longmapsto\: \: \begin{bmatrix} 0 & -2b\\ a & 0\end{bmatrix} = \: \begin{bmatrix} 0 & - 2\\ 2 & 0\end{bmatrix}$

So, on comparing we get

$\rm :\longmapsto\: \boxed{ \bf \: b = 1} \: \: \: and \: \: \: \boxed{ \bf \: a = 2}$

Additional Information :-

1. Matrix multiplication is not Commutative always. i.e AB or BA is not equal always.

2. Matrix multiplication is associative. i.e A(BC) = (AB)C

3. Matrix multiplication is Distributive A(B + C) = AB + AC

4. Let A and B are two matrices of order m × n and p × q, then matrix multiplication AB is possible only when number of columns = number of rows, i.e. n = p.