Find the matrix A such that
Question:

Find the matrix A such that

(i) $\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] A=\left[\begin{array}{lll}3 & 3 & 5 \\ 1 & 0 & 1\end{array}\right]$

(ii) $A\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{rrr}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]$

(iii) $\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right] A=\left[\begin{array}{lll}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right]$

(iv) $\left[\begin{array}{lll}2 & 1 & 3\end{array}\right]\left[\begin{array}{ccc}-1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]=A$

(v) $\left[\begin{array}{rc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right] \mathrm{A}=\left[\begin{array}{rcc}-1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15\end{array}\right]$

(vi) $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{rrr}-7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9\end{array}\right]$

Solution:

(i)

Let $A=\left[\begin{array}{lll}x & y & z \\ a & b & c\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{lll}x & y & z \\ a & b & c\end{array}\right]=\left[\begin{array}{lll}3 & 3 & 5 \\ 1 & 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}x+a & y+b & z+c \\ 0+a & 0+b & 0+c\end{array}\right]=\left[\begin{array}{lll}3 & 3 & 5 \\ 1 & 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}x+a & y+b & z+c \\ a & b & c\end{array}\right]=\left[\begin{array}{lll}3 & 3 & 5 \\ 1 & 0 & 1\end{array}\right]$

The corresponding elements of two equal matrices are equal.

$\Rightarrow x+a=3$       …(1)

$y+b=3$                          …(2)

$z+c=5$                         …(3)

$\Rightarrow a=1, b=0$ and $c=1$

Putting the value of $a$ in eq. (1), we get

$x+1=3$

$\Rightarrow x=3-1$

$\therefore x=2$

Putting the value of $b$ in eq. (2), we get

$y+b=3$

$\Rightarrow y+0=3$

$\therefore y=3$

Putting the value of $c$ in eq. (3), we get

$z+1=5$

$\Rightarrow z=5-1$

$\therefore z=4$

$\therefore A=\left[\begin{array}{lll}2 & 3 & 4 \\ 1 & 0 & 1\end{array}\right]$

(ii) Let $A=\left[\begin{array}{ll}w & x \\ y & z\end{array}\right]$

$\Rightarrow A\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}w & x \\ y & z\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}w+4 x & 2 w+5 x & 3 w+6 x \\ y+4 z & 2 y+5 z & 3 y+6 z\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]$

The correspnding elements of two equal matrices are equal.

$\therefore 3 w+6 x=-9$           …(1)

$y+4 z=2$

$y=2-4 z$                    …(2)

$w+4 x=-7$

$\Rightarrow w=-7-4 x$    …(3)

$2 y+5 z=4$       ….(4)

Putting the value of $w$ in eq. $(1)$, we get

$3(-7-4 x)+6 x=-9$

$\Rightarrow-21-12 x+6 x=-9$

$\Rightarrow-6 x=12$

$\Rightarrow x=-2$’

Putting the value of $x$ in eq. $(3)$, we get

$w=-7-4(-2)$

$\Rightarrow w=-7+8$

$\Rightarrow w=1$

Putting the value of $y$ in eq. $(4)$, we get

$y=2-4(0)$

$\Rightarrow y=2$

$\therefore A=\left[\begin{array}{cc}1 & -2 \\ 2 & 0\end{array}\right]$

(iii)

Let $A=\left[\begin{array}{lll}x & y & z\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]\left[\begin{array}{lll}x & y & z\end{array}\right]=\left[\begin{array}{lll}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}4 x & 4 y & 4 z \\ x & y & z \\ 3 x & 3 y & 3 z\end{array}\right]=\left[\begin{array}{lll}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right]$

The corresponding elements of two equal matrices are equal.

$\Rightarrow 4 x=-4$      …(1)

$4 y=8$                          …(2)

$4 z=4$                          …(3)

$\Rightarrow x=-1, y=2$ and $z=1$

$\therefore A=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]$

(iv) Let $A=[x]$

$\Rightarrow\left[\begin{array}{lll}2 & 1 & 3\end{array}\right]\left[\begin{array}{ccc}-1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]=A$

$\Rightarrow\left[\begin{array}{lll}2 & 1 & 3\end{array}\right]\left[\begin{array}{ccc}-1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]=[x]$

$\Rightarrow\left[\begin{array}{lll}-2-1+0 & 0+1+3 & -2+0+3\end{array}\right]\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]=[x]$

$\Rightarrow\left[\begin{array}{lll}-3 & 4 & 1\end{array}\right]\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]=[x]$

$\Rightarrow[-3+0-1]=[x]$

$\Rightarrow[-4]=[x]$

The corresponding elements of two equal matrices are equal.

$\therefore x=-4$

$\therefore A=[-4]$

(v) $\left[\begin{array}{rc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right] A=\left[\begin{array}{rcc}-1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15\end{array}\right]$

Let $A=\left[\begin{array}{lll}x & y & z \\ a & b & c\end{array}\right]$

$\Rightarrow\left[\begin{array}{rc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right]\left[\begin{array}{lll}x & y & z \\ a & b & c\end{array}\right]=\left[\begin{array}{rcc}-1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}2 x-a & 2 y-b & 2 z-c \\ x & y & z \\ -3 x+4 a & -3 y+4 b & -3 z+4 c\end{array}\right]=\left[\begin{array}{rcc}-1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15\end{array}\right]$

By comparing the elements of second row, we get

$x=1, y=-2, z=-5$

By comparing the elements of first row, we get

$2 x-a=-1$

$\Rightarrow 2-a=-1$

$\Rightarrow a=3$

Also,

$2 y-b=-8$

$\Rightarrow-4-b=-8$

$\Rightarrow b=4$

And,

$2 z-c=-10$

$\Rightarrow-10-c=-10$

$\Rightarrow c=0$

$\therefore A=\left[\begin{array}{ccc}1 & -2 & -5 \\ 3 & 4 & 0\end{array}\right]$

(vi) $A\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{rrr}-7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9\end{array}\right]$

Let $A=\left[\begin{array}{ll}x & a \\ y & b \\ z & c\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}x & a \\ y & b \\ z & c\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{rrr}-7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9\end{array}\right]$

$\Rightarrow\left[\begin{array}{lll}x+4 a & 2 x+5 a & 3 x+6 a \\ y+4 b & 2 y+5 b & 3 y+6 b \\ z+4 c & 2 z+5 c & 3 z+6 c\end{array}\right]=\left[\begin{array}{rrr}-7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9\end{array}\right]$

By comparing the corresponding elements, we get

$x+4 a=-7$ and $2 x+5 a=-8$

$\Rightarrow a=-2$ and $x=1$

Also,

$y+4 b=2$ and $2 y+5 b=4$

$\Rightarrow b=0$ and $y=2$

And,

$z+4 c=11$ and $2 z+5 c=10$

$\Rightarrow c=4$ and $z=-5$

$\therefore A=\left[\begin{array}{cc}1 & -2 \\ 2 & 0 \\ -5 & 4\end{array}\right]$

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