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.

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: $$ A^{-1} =\left[\begin{array}{ccc} \dfrac{1}{3} & \dfrac{1}{3} & 0 \\ -\dfrac{1}{9} & \dfrac{2}{9} & -\dfrac{1}{3} \\ \dfrac{5}{9} & -\dfrac{1}{9} & -\dfrac{1}{3} \end{array}\right]$$

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.

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*}

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.

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