Question 2, Exercise 2.3

Solutions of Question 2 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.

Evaluate the determinant of the following matrix $\left[\begin{array}{lll}3 & 2 & 3 \\ 4 & 5 & 1 \\ 2 & 1 & 0\end{array}\right]$ using cofactor method.

Solution.

The elements of $R_1$ are $a_{11} = 3$, $a_{12} = 2$, and $a_{13} = 3$. Now we find their corresponding cofactors. $$\begin{align*} A &= \left[\begin{array}{ccc} 3 & 2 & 3 \\ 4 & 5 & 1 \\ 2 & 1 & 0 \end{array}\right]\\ & A_{11} = (-1)^{1+1} \left|\begin{array}{cc} 5 & 1 \\ 1 & 0 \end{array}\right| = (-1)^{2} (5 \cdot 0 - 1 \cdot 1) = (1) (-1) = -1 \\ & A_{12} = (-1)^{1+2} \left|\begin{array}{cc} 4 & 1 \\ 2 & 0 \end{array}\right| = (-1)^{3} (4 \cdot 0 - 1 \cdot 2) = (-1) (-2) = 2 \\ & A_{13} = (-1)^{1+3} \left|\begin{array}{cc} 4 & 5 \\ 2 & 1 \end{array}\right| = (-1)^{4} (4 \cdot 1 - 5 \cdot 2) = (1) (4 - 10) = -6 \end{align*}$$ Now, we use these cofactors to find the determinant: $$\begin{align*} \det(A) &= a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} \\ &= 3(-1) + 2(2) + 3(-6) \\ &= -3 + 4 - 18 \\ &= -17 \end{align*}$$ Thus, the determinant of the matrix $\left[\begin{array}{ccc} 3 & 2 & 3 \\ 4 & 5 & 1 \\ 2 & 1 & 0 \end{array}\right]$ is: $-17$

Evaluate the determinant of the following matrix $\left[\begin{array}{ccc}2 & 3 & -1 \\ -1 & 0 & 2 \\ 3 & 1 & 4\end{array}\right]$ using cofactor method.

Solution.

The elements of $R_1$ are $a_{11} = 2$, $a_{12} = 3$, and $a_{13} = -1$. Now we find their corresponding cofactors. $$\begin{align*} A &= \left[\begin{array}{ccc} 2 & 3 & -1 \\ -1 & 0 & 2 \\ 3 & 1 & 4 \end{array}\right]\\ & A_{11} = (-1)^{1+1} \left|\begin{array}{cc} 0 & 2 \\ 1 & 4 \end{array}\right| = (-1)^{2} (0 \cdot 4 - 2 \cdot 1) = (1) (0 - 2) = -2 \\ & A_{12} = (-1)^{1+2} \left|\begin{array}{cc} -1 & 2 \\ 3 & 4 \end{array}\right| = (-1)^{3} (-1 \cdot 4 - 2 \cdot 3) = (-1) (-4 - 6) = 10 \\ & A_{13} = (-1)^{1+3} \left|\begin{array}{cc} -1 & 0 \\ 3 & 1 \end{array}\right| = (-1)^{4} (-1 \cdot 1 - 0 \cdot 3) = (1) (-1) = -1 \end{align*}$$ Now, we use these cofactors to find the determinant: $$\begin{align*} \det(A) &= a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} \\ &= 2(-2) + 3(10) + (-1)(-1) \\ &= -4 + 30 + 1 \\ &= 27 \end{align*}$$ Thus, the determinant of the matrix $\left[\begin{array}{ccc} 2 & 3 & -1 \\ -1 & 0 & 2 \\ 3 & 1 & 4 \end{array}\right]$ is:$27$

Evaluate the determinant of the following matrix $\left[\begin{array}{ccc}2 i & 6 & 1 \\ 1 & -i & 2 \\ 0 & 1 & 3 i\end{array}\right]$ using cofactor method.

Solution.

The elements of $R_1$ are $a_{11} = 2i$, $a_{12} = 6$, and $a_{13} = 1$.

$$\begin{align*} A &= \left[\begin{array}{ccc}2i & 6 & 1 \\ 1 & -i & 2 \\ 0 & 1 & 3i\end{array}\right]\\ & A_{11} = (-1)^{1+1} \left|\begin{array}{cc} -i & 2 \\ 1 & 3i \end{array}\right| = (-1)^{2} (-i \cdot 3i - 2 \cdot 1) = (1) (-3i^2 - 2) = -3(-1) - 2 = 3 - 2 = 1 \\ & A_{12} = (-1)^{1+2} \left|\begin{array}{cc} 1 & 2 \\ 0 & 3i \end{array}\right| = (-1)^{3} (1 \cdot 3i - 2 \cdot 0) = (-1) (3i - 0) = -3i \\ & A_{13} = (-1)^{1+3} \left|\begin{array}{cc} 1 & -i \\ 0 & 1 \end{array}\right| = (-1)^{4} (1 \cdot 1 - (-i) \cdot 0) = (1) (1 - 0) = 1 \end{align*}$$ Now, we use these cofactors to find the determinant: $$\begin{align*} \det(A) &= a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} \\ &= 2i(1) + 6(-3i) + 1(1) \\ &= 2i - 18i + 1 \\ &= -16i + 1 \end{align*}$$ Thus, the determinant of the matrix $\left[\begin{array}{ccc}2i & 6 & 1 \\ 1 & -i & 2 \\ 0 & 1 & 3i\end{array}\right]$ is:$1 - 16i$

Evaluate the determinant of the following matrix $\left[\begin{array}{ccc}1-i & 2 & 1+i \\ 3 & 1 & 4 \\ 0 & 2 & 3\end{array}\right]$ using cofactor method.

Solution.

The elements of $R_1$ are $a_{11} = 1-i$, $a_{12} = 2$, and $a_{13} = 1+i$. $$\begin{align*}A &= \left[\begin{array}{ccc}1-i & 2 & 1+i \\ 3 & 1 & 4 \\ 0 & 2 & 3\end{array}\right]\\ & A_{11} = (-1)^{1+1} \left|\begin{array}{cc} 1 & 4 \\ 2 & 3 \end{array}\right| = (-1)^{2} (1 \cdot 3 - 4 \cdot 2) = (1) (3 - 8) = -5 \\ & A_{12} = (-1)^{1+2} \left|\begin{array}{cc} 3 & 4 \\ 0 & 3 \end{array}\right| = (-1)^{3} (3 \cdot 3 - 4 \cdot 0) = (-1) (9 - 0) = -9 \\ & A_{13} = (-1)^{1+3} \left|\begin{array}{cc} 3 & 1 \\ 0 & 2 \end{array}\right| = (-1)^{4} (3 \cdot 2 - 1 \cdot 0) = (1) (6 - 0) = 6 \end{align*}$$ Now, we use these cofactors to find the determinant: $$\begin{align*} \det(A) &= a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} \\ &= (1-i)(-5) + 2(-9) + (1+i)(6) \\ &= -5 + 5i - 18 + 6 + 6i \\ &= -17 + 11i \end{align*}$$ Thus, the determinant of the matrix $\left[\begin{array}{ccc}1-i & 2 & 1+i \\ 3 & 1 & 4 \\ 0 & 2 & 3\end{array}\right]$ is: $-17 + 11i$

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