====== Question 5, Exercise 2.3 ======
Solutions of Question 5 of Exercise 2.3 of Unit 02: Matrices and Determinants. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.
=====Question 5(i)=====
Find the multiplicative inverse of the following matrices if it exists by adjoint method $\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & -1\end{array}\right]$.
** Solution. **
Given
\begin{align*}
A &= \left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -1 \\
1 & -2 & -1
\end{array}\right]\\
|A|&= 1 [-1 - 2] + 1 [-2 + 1] + 1 [-4 - 1] \\
&= 1 \cdot (-3) + 1 \cdot (-1) + 1 \cdot (-5) \\
&= -3 - 1 - 5 \\
&= -9
\end{align*}
Thus, $|A| = -9 \neq 0$, so $A$ is non-singular.\\
Let find the cofactor matrix for $A$.\\
\begin{align*}
A_{11} &= (-1)^{1+1} \left|\begin{array}{cc}
1 & -1 \\
-2 & -1
\end{array}\right|\\
&= (1) [(1)(-1) - (-1)(-2)] \\
&= -1 - 2 = -3 \\
A_{12} &= (-1)^{1+2} \left|\begin{array}{cc}
2 & -1 \\
1 & -1
\end{array}\right|\\
&= (-1) [(2)(-1) - (-1)(1)] \\
&= -1 [-2 + 1] = 1 \\
A_{13} &= (-1)^{1+3} \left|\begin{array}{cc}
2 & 1 \\
1 & -2
\end{array}\right|\\
&= (1) [(2)(-2) - (1)(1)]\\
&= -4 - 1 = -5 \\
A_{21} &= (-1)^{2+1} \left|\begin{array}{cc}
-1 & 1 \\
-2 & -1
\end{array}\right| \\
&= (-1) [(-1)(-1) - (1)(-2)]\\
&= -1 [1 + 2] = -3 \\
A_{22} &= (-1)^{2+2} \left|\begin{array}{cc}
1 & 1 \\
1 & -1
\end{array}\right|\\
&= (1) [(1)(-1) - (1)(1)]\\
&= -1 - 1 = -2 \\
A_{23} &= (-1)^{2+3} \left|\begin{array}{cc}
1 & -1 \\
1 & -2
\end{array}\right|\\
&= (-1) [(1)(-2) - (-1)(1)]\\
&= -1 [-2 + 1] = 1 \\
A_{31} &= (-1)^{3+1} \left|\begin{array}{cc}
-1 & 1 \\
1 & -1
\end{array}\right| \\
&= (1) [(-1)(-1) - (1)(1)]\\
&= 1 - 1 = 0 \\
A_{32} &= (-1)^{3+2} \left|\begin{array}{cc}
1 & 1 \\
2 & -1
\end{array}\right| \\
&= (-1) [(1)(-1) - (1)(2)] \\
&= -1 [-1 - 2] = 3 \\
A_{33} &= (-1)^{3+3} \left|\begin{array}{cc}
1 & -1 \\
2 & 1
\end{array}\right|\\
&= (1) [(1)(1) - (-1)(2)]\\
&= 1 + 2 = 3
\end{align*}
\begin{align*}
adj(A) &= \left[\begin{array}{ccc}
-3 & 1 & -5 \\
-3 & -2 & 1 \\
0 & 3 & 3
\end{array}\right]^t\\
&= \left[\begin{array}{ccc}
-3 & -3 & 0 \\
1 & -2 & 3 \\
-5 & 1 & 3
\end{array}\right]
\end{align*}
\begin{align*}
A^{-1} &= \dfrac{1}{-9} \left[\begin{array}{ccc}
-3 & -3 & 0 \\
1 & -2 & 3 \\
-5 & 1 & 3
\end{array}\right]\\
& = \left[\begin{array}{ccc}
\frac{1}{3} & \frac{1}{3} & 0 \\
-\frac{1}{9} & \frac{2}{9} & -\frac{1}{3} \\
\frac{5}{9} & -\frac{1}{9} & -\frac{1}{3}
\end{array}\right]
\end{align*}
Thus, the inverse of $A$ is:
$$
A^{-1} =\left[\begin{array}{ccc}
\dfrac{1}{3} & \dfrac{1}{3} & 0 \\
-\dfrac{1}{9} & \dfrac{2}{9} & -\dfrac{1}{3} \\
\dfrac{5}{9} & -\dfrac{1}{9} & -\dfrac{1}{3}
\end{array}\right]$$
=====Question 5(ii)=====
Find the multiplicative inverse of the following matrices if it exists by adjoint method $\left[\begin{array}{ccc}3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{array}\right]$.
** Solution. **
Do yourself.
=====Question 5(iii)=====
Find the multiplicative inverse of the following matrices if it exists by adjoint method $\left[\begin{array}{ccc}i & 0 & 1 \\ 2 i & -1 & -i \\ 1 & 0 & 4 i\end{array}\right]$.
** Solution. **
Given
\begin{align*}
B &= \left[\begin{array}{ccc}
i & 0 & 1 \\
2i & -1 & -i \\
1 & 0 & 4i
\end{array}\right]\\
|B| &= i \left[(-1)(4i) - (-i)(0)\right] + 1 \left[(2i)(0) - (-1)(1)]\right] \\
&= i \left[-4i\right] + 1 [0 + 1] \\
&= -4i^2 + 1 \\
&= 4 + 1 \\
&= 5
\end{align*}
Thus, $|B| = 5 \neq 0$, so $B$ is non-singular.\\
\begin{align*}
B_{11} &= (-1)^{1+1} \left|\begin{array}{cc}
-1 & -i \\
0 & 4i
\end{array}\right|\\
&= (1) [(-1)(4i) - (-i)(0)] = -4i \\
B_{12} &= (-1)^{1+2} \left|\begin{array}{cc}
2i & -i \\
1 & 4i
\end{array}\right|\\
&= (-1) [(2i)(4i) - (-i)(1)] \\
&= -[8i^2 + i] \\
&= -[8(-1) + i] = 8 - i \\
B_{13} &= (-1)^{1+3} \left|\begin{array}{cc}
2i & -1 \\
1 & 0
\end{array}\right| \\
&= (1) [(2i)(0) - (-1)(1)] = 1 \\
B_{21} &= (-1)^{2+1} \left|\begin{array}{cc}
0 & 1 \\
0 & 4i
\end{array}\right|\\
&= (-1) [(0)(4i) - (1)(0)] = 0 \\
B_{22} &= (-1)^{2+2} \left|\begin{array}{cc}
i & 1 \\
1 & 4i
\end{array}\right|\\
&= (1) [(i)(4i) - (1)(1)] \\
&= 4i^2 - 1 = 4(-1) - 1 \\
&= -4 - 1 = -5 \\
B_{23} &= (-1)^{2+3} \left|\begin{array}{cc}
i & 0 \\
1 & 0
\end{array}\right|\\
&= (-1) [(i)(0) - (0)(1)] = 0 \\
B_{31} &= (-1)^{3+1} \left|\begin{array}{cc}
0 & 1 \\
-1 & -i
\end{array}\right| \\
&= (1) [(0)(-i) - (1)(-1)] = 1 \\
B_{32} &= (-1)^{3+2} \left|\begin{array}{cc}
i & 1 \\
2i & -i
\end{array}\right| \\
&= (-1) [(i)(-i) - (1)(2i)]\\
&= -[-1 - 2i] = 1 + 2i \\
B_{33} &= (-1)^{3+3} \left|\begin{array}{cc}
i & 0 \\
2i & -1
\end{array}\right|\\
&= (1) [(i)(-1) - (0)(2i)] = -i
\end{align*}
\begin{align*}
adj(B) &= \left[\begin{array}{ccc}
-4i & 8 - i & 1 \\
0 & -5 & 0 \\
1 & 1 + 2i & -i
\end{array}\right]^t \\
&= \left[\begin{array}{ccc}
-4i & 0 & 1 \\
8 - i & -5 & 1 + 2i \\
1 & 0 & -i
\end{array}\right]\end{align*}
\begin{align*}
B^{-1} &= \dfrac{1}{|B|} adj(B)\\
& = \dfrac{1}{5} \left[\begin{array}{ccc}
-4i & 0 & 1 \\
8 - i & -5 & 1 + 2i \\
1 & 0 & -i
\end{array}\right]\\
& = \left[\begin{array}{ccc}
-\dfrac{4i}{5} & 0 & \dfrac{1}{5} \\
\dfrac{8 - i}{5} & -1 & \dfrac{1 + 2i}{5} \\
\dfrac{1}{5} & 0 & -\dfrac{i}{5}
\end{array}\right]
\end{align*}
Thus, the inverse of $B$ is:
\begin{align*}
B^{-1} &= \left[\begin{array}{ccc}
-\dfrac{4i}{5} & 0 & \dfrac{1}{5} \\
\dfrac{8 - i}{5} & -1 & \dfrac{1 + 2i}{5} \\
\dfrac{1}{5} & 0 & -\dfrac{i}{5}
\end{array}\right]
\end{align*}
=====Question 5(iv)=====
Find the multiplicative inverse of the matrix if it exists by adjoint method $\left[\begin{array}{ccc}3 & -i & i \\ 2 & 1 & -3 i \\ 4 i & 2 & 6\end{array}\right]$.
** Solution. **
Do yourself.
====Go to ====
[[math-11-nbf:sol:unit02:ex2-3-p4|< Question 4]]
[[math-11-nbf:sol:unit02:ex2-3-p6|Question 6 >]]