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Exercise 7.1 · Q14

Q.Find the matrix AA which satisfies the matrix relation
[!FORMULA] A(123456)=(−7−8−9246)A\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6\end{pmatrix} = \begin{pmatrix} -7 & -8 & -9 \\ 2 & 4 & 6\end{pmatrix}

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Since AA multiplies a 2×32\times3 matrix to give a 2×32\times3 result, AA itself must be 2×22\times2; we set A=(abcd)A=\begin{pmatrix}a&b\\c&d\end{pmatrix}, multiply out, and solve.

Step 1. Fix the order of AA. M=(123456)M=\begin{pmatrix}1&2&3\\4&5&6\end{pmatrix} is 2×32\times3 and AMAM is 2×32\times3, so AA must be 2×22\times2. Let A=(abcd)A=\begin{pmatrix}a&b\\c&d\end{pmatrix}.

Step 2. Multiply AMAM symbolically.

AM=(a+4b2a+5b3a+6bc+4d2c+5d3c+6d)AM=\begin{pmatrix}a+4b & 2a+5b & 3a+6b\\ c+4d & 2c+5d & 3c+6d\end{pmatrix}

Step 3. Equate the top row to (−7,−8,−9)(-7,-8,-9).

a+4b=−7a+4b=-7 … (I),\quad 2a+5b=−82a+5b=-8 … (II),\quad 3a+6b=−9⇒a+2b=−33a+6b=-9\Rightarrow a+2b=-3 … (III)

Subtract (III) from (I): (a+4b)−(a+2b)=2b=−7−(−3)=−4⇒b=−2(a+4b)-(a+2b)=2b=-7-(-3)=-4\Rightarrow b=-2. Then from (III), a=−3−2(−2)=1a=-3-2(-2)=1.

Check with (II): 2(1)+5(−2)=2−10=−82(1)+5(-2)=2-10=-8 ✓.

Step 4. Equate the bottom row to (2,4,6)(2,4,6). …

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