3 Marks Question

Question 13 Marks

If $\Delta=\left|\begin{array}{ccc} A x & x^2 & 1 \\ B y & y^2 & 1 \\ C z & z^2 & 1\end{array}\right|$ and $\Delta_1=\left|\begin{array}{ccc} A & B & C \\ x & y & z \\ z y & z x & x y\end{array}\right|$ then prove that $\Delta-\Delta_1=0$

Answer

$
\begin{aligned}
\Delta_1 & =\left|\begin{array}{ccc}
A & B & C \\
x & y & z \\
z y & z x & x y
\end{array}\right|=\left|\begin{array}{lll}
A & x & y z \\
B & y & z x \\
C & z & x y
\end{array}\right| \\
& =\frac{1}{x y z}\left|\begin{array}{lll}
A x & x^2 & x y z \\
B y & y^2 & x y z \\
C z & z^2 & x y z
\end{array}\right| \\
& =\frac{x y z}{x y z}\left|\begin{array}{lll}
A x & x^2 & 1 \\
B y & y^2 & 1 \\
C z & z^2 & 1
\end{array}\right|=\Delta
\end{aligned}
$
$\Delta-\Delta_1=0$ Hence proved.

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Question 23 Marks

If $A$ and $B$ has same order invertible square matrix, then prove that :$
(AB)^{-1}=B^{-1} \cdot A^{-1}
$

Answer

If A has invertible matrix, then according to definition this matrix will be square. Hence A and B has same order square matrix. From this AB multiplication will be possible.
$\begin{array}{ll}\text { Because } & | A | \neq 0,|B| \neq 0 \\ \therefore & | AB |=| A || B | \neq 0\end{array}$
Hence matrix AB also has invertible.
Now taking a matrix $C$ such that :
$\begin{aligned} C & = B ^{-1} A^{-1} \\ ( AB ) C & =( AB )\left( B ^{-1} A^{-1}\right) \\ & = A \cdot\left( BB ^{-1}\right) A ^{-1} \quad(\text { Associativity }) \\ & = A \cdot IA ^{-1} \quad\left[\because BB ^{-1}= I \right] \\ & = AA ^{-1}= I \end{aligned}$

Thus$
\begin{aligned}
C(AB) & =\left(B^{-1} A^{-1}\right) AB \\
& =B^{-1}\left(A^{-1} A\right) B \\
& =B^{-1}(I) B \\
& =B^{-1} B \\
& =I
\end{aligned}
$
$
\begin{aligned}
& =B^{-1}(I) B \quad\left[\because A^{-1} A=I\right] \\
& =B^{-1} B \\
& =I \\
\therefore \quad(AB) C & =I=C(AB)
\end{aligned}
$
Hence $A B$ matrix has only inverse matrix $C$.$
\therefore \quad(AB)^{-1}=C=B^{-1} A^{-1} \text { Hence proved. }
$

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