Solved Examples

Important Theorems on Determinants and Matrices

Before solving a large variety of problems, let us recall the fundamental results used throughout this chapter.

• Theorem 1: If $A$ is any square matrix of order $n$, then $$A(\operatorname{adj} A) = (\operatorname{adj} A)A = |A|I,$$ where $I$ is the identity matrix of order $n$. This is one of the most important identities in matrix algebra and directly leads to the inverse formula.

• Theorem 2: If $A$ and $B$ are non-singular matrices of the same order, then $AB$ and $BA$ are also non-singular matrices. This follows because $$|AB| = |A||B|,$$ and the product of two non-zero numbers is non-zero.

• Theorem 3: If $A$ and $B$ are square matrices of the same order, then $$|AB| = |A||B|.$$ As an important corollary, for a non-singular matrix of order $n$, $$|\operatorname{adj} A| = |A|^{n-1}.$$

• Theorem 4: A square matrix $A$ is invertible if and only if it is non-singular, that is, $$|A| \neq 0.$$ In that case, $$A^{-1} = \frac{1}{|A|}\operatorname{adj} A.$$

These results will be used repeatedly in the examples below.

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Example 1: Evaluation of a $2 \times 2$ Trigonometric Determinant

Evaluate: $$\Delta = \begin{vmatrix} \sin 30^\circ & \cos 30^\circ \\ -\sin 60^\circ & \cos 60^\circ \end{vmatrix}$$

Solution: Substitute the standard trigonometric values: $$\sin 30^\circ = \frac{1}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{1}{2}.$$ So, $$\Delta = \begin{vmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{vmatrix}.$$ Now use the rule for a $2 \times 2$ determinant: $$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc.$$ Therefore, $$\Delta = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) - \left(\frac{\sqrt{3}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right) = \frac{1}{4} - \left(-\frac{3}{4}\right) = \frac{1}{4} + \frac{3}{4} = 1.$$

Answer: $1$

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Example 2: Equating Determinants

Find the value(s) of $x$ if $$\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}.$$

Solution: First evaluate the determinant on the left-hand side: $$\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = (2)(1) - (4)(5) = 2 - 20 = -18.$$ Now evaluate the determinant on the right-hand side: $$\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix} = (2x)(x) - (4)(6) = 2x^2 - 24.$$ Equating both sides, $$-18 = 2x^2 - 24.$$ So, $$2x^2 = 6 \Rightarrow x^2 = 3.$$ Hence, $$x = \pm \sqrt{3}.$$

Answer: $x = \sqrt{3}, -\sqrt{3}$

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Example 3: Evaluation of a $3 \times 3$ Determinant

Evaluate: $$\Delta = \begin{vmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{vmatrix}$$

Solution: Expand along the first row: $$\Delta = 1 \begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix} - 1 \begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix} + (-2) \begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix}.$$ Now evaluate each minor: $$\begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix} = (1)(-9) - (-3)(4) = -9 + 12 = 3,$$ $$\begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix} = (2)(-9) - (-3)(5) = -18 + 15 = -3,$$ $$\begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix} = (2)(4) - (1)(5) = 8 - 5 = 3.$$ Substitute back: $$\Delta = 1(3) - 1(-3) + (-2)(3) = 3 + 3 - 6 = 0.$$

Answer: $0$

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Example 4: Algebraic $3 \times 3$ Determinant

Evaluate: $$\Delta = \begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix}$$

Solution: Expand along the first row: $$\Delta = x \begin{vmatrix} x+y & x \\ x & y \end{vmatrix} - y \begin{vmatrix} y & x \\ x+y & y \end{vmatrix} + (x+y) \begin{vmatrix} y & x+y \\ x+y & x \end{vmatrix}.$$ Now compute each $2 \times 2$ determinant.

First term: $$\begin{vmatrix} x+y & x \\ x & y \end{vmatrix} = (x+y)y - x^2 = xy + y^2 - x^2.$$ So the first contribution is $$x(xy + y^2 - x^2) = x^2y + xy^2 - x^3.$$

Second term: $$\begin{vmatrix} y & x \\ x+y & y \end{vmatrix} = y^2 - x(x+y) = y^2 - x^2 - xy.$$ So the second contribution is $$-y(y^2 - x^2 - xy) = -y^3 + x^2y + xy^2.$$

Third term: $$\begin{vmatrix} y & x+y \\ x+y & x \end{vmatrix} = xy - (x+y)^2 = xy - (x^2 + 2xy + y^2) = -x^2 - xy - y^2.$$ Hence the third contribution is $$(x+y)(-x^2 - xy - y^2).$$ Now use the identity $$(x+y)(x^2 - xy + y^2) = x^3 + y^3.$$ So, $$(x+y)(x^2 + xy + y^2) = x^3 + 2x^2y + 2xy^2 + y^3,$$ and therefore the third contribution is $$-x^3 - 2x^2y - 2xy^2 - y^3.$$ Now add all three contributions: $$\Delta = (x^2y + xy^2 - x^3) + (-y^3 + x^2y + xy^2) + (-x^3 - 2x^2y - 2xy^2 - y^3).$$ All the mixed terms cancel: $$\Delta = -2x^3 - 2y^3.$$ Now factor using $$x^3 + y^3 = (x+y)(x^2 - xy + y^2).$$ Hence, $$\Delta = -2(x^3 + y^3) = -2(x+y)(x^2 - xy + y^2).$$

Answer: $$-2(x^3 + y^3) = -2(x+y)(x^2 - xy + y^2).$$

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Example 5: Area of a Triangle

Find the area of the triangle with vertices $A(5, 4)$, $B(-2, 4)$, and $C(2, -6)$.

Solution: Use the determinant formula for area: $$\text{Area} = \frac{1}{2} \left| \begin{vmatrix} 5 & 4 & 1 \\ -2 & 4 & 1 \\ 2 & -6 & 1 \end{vmatrix} \right|.$$ Expand along the third column: $$\begin{vmatrix} 5 & 4 & 1 \\ -2 & 4 & 1 \\ 2 & -6 & 1 \end{vmatrix} = 1\begin{vmatrix} -2 & 4 \\ 2 & -6 \end{vmatrix} - 1\begin{vmatrix} 5 & 4 \\ 2 & -6 \end{vmatrix} + 1\begin{vmatrix} 5 & 4 \\ -2 & 4 \end{vmatrix}.$$ Now, $$\begin{vmatrix} -2 & 4 \\ 2 & -6 \end{vmatrix} = 12 - 8 = 4,$$ $$\begin{vmatrix} 5 & 4 \\ 2 & -6 \end{vmatrix} = -30 - 8 = -38,$$ $$\begin{vmatrix} 5 & 4 \\ -2 & 4 \end{vmatrix} = 20 + 8 = 28.$$ So the determinant equals $$4 - (-38) + 28 = 70.$$ Therefore, $$\text{Area} = \frac{1}{2}|70| = 35.$$

Answer: $35$ square units.

Example 6: Equation of a Line

Find the equation of the line joining $A(1, 3)$ and $B(0, 0)$ using determinants.

Solution: Let $P(x,y)$ be any point on the required line. Then the three points $P(x,y)$, $A(1,3)$, and $B(0,0)$ are collinear. Therefore, the area of the triangle formed by them is zero: $$\begin{vmatrix} x & y & 1 \\ 1 & 3 & 1 \\ 0 & 0 & 1 \end{vmatrix} = 0.$$ Expand along the third row: $$0 - 0 + 1\begin{vmatrix} x & y \\ 1 & 3 \end{vmatrix} = 0.$$ So, $$3x - y = 0.$$ Hence, $$y = 3x.$$

Answer: $y = 3x$

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Example 7: Checking Collinearity

Using determinants, show that the points $A(a, b+c)$, $B(b, c+a)$, and $C(c, a+b)$ are collinear.

Solution: The three points are collinear if the determinant used in the area formula is zero: $$\Delta = \frac{1}{2}\begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix}.$$ Apply the column operation $$C_2 \to C_2 + C_1.$$ Then, $$\Delta = \frac{1}{2}\begin{vmatrix} a & a+b+c & 1 \\ b & a+b+c & 1 \\ c & a+b+c & 1 \end{vmatrix}.$$ Now take $(a+b+c)$ common from the second column: $$\Delta = \frac{1}{2}(a+b+c)\begin{vmatrix} a & 1 & 1 \\ b & 1 & 1 \\ c & 1 & 1 \end{vmatrix}.$$ But the second and third columns are identical, so the determinant is zero. Therefore, $$\Delta = 0.$$ Hence the points are collinear.

Answer: Proved.

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Example 8: Finding Coordinate Given Area

If the area of a triangle is $4$ square units and its vertices are $(k, 0)$, $(4, 0)$, and $(0, 2)$, find the value(s) of $k$.

Solution: Use the area formula: $$\frac{1}{2}\left|\begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix}\right| = 4.$$ So, $$\begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = \pm 8.$$ Expand along the second column, since it has two zeros. Only the entry 2 contributes: $$-2\begin{vmatrix} k & 1 \\ 4 & 1 \end{vmatrix} = \pm 8.$$ Thus, $$-2(k-4) = \pm 8.$$ Case 1: $$-2(k-4)=8 \Rightarrow k-4=-4 \Rightarrow k=0.$$ Case 2: $$-2(k-4)=-8 \Rightarrow k-4=4 \Rightarrow k=8.$$

Answer: $k = 0$ or $k = 8$

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Example 9: Collinearity Condition

If the points $(a, 0)$, $(0, b)$, and $(1, 1)$ are collinear, prove that $\frac{1}{a} + \frac{1}{b} = 1$.

Solution: Since the points are collinear, $$\begin{vmatrix} a & 0 & 1 \\ 0 & b & 1 \\ 1 & 1 & 1 \end{vmatrix} = 0.$$ Expand along the first row: $$a\begin{vmatrix} b & 1 \\ 1 & 1 \end{vmatrix} - 0 + 1\begin{vmatrix} 0 & b \\ 1 & 1 \end{vmatrix} = 0.$$ So, $$a(b-1) + (0-b) = 0,$$ which gives $$ab - a - b = 0.$$ Thus, $$ab = a+b.$$ Dividing both sides by $ab$, $$1 = \frac{1}{b} + \frac{1}{a}.$$ Hence, $$\frac{1}{a} + \frac{1}{b} = 1.$$

Answer: Proved.

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Example 10: Minors and Cofactors of a $2 \times 2$ Matrix

Write the minors and cofactors of the elements of the determinant $$\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}.$$

Solution: The elements are $$a_{11}=2, \quad a_{12}=-4, \quad a_{21}=0, \quad a_{22}=3.$$ Now find the minors: $$M_{11} = |3| = 3,$$ $$M_{12} = |0| = 0,$$ $$M_{21} = |-4| = -4,$$ $$M_{22} = |2| = 2.$$ Now use the cofactor rule $A_{ij}=(-1)^{i+j}M_{ij}$: $$A_{11} = (+1)(3)=3,$$ $$A_{12} = (-1)(0)=0,$$ $$A_{21} = (-1)(-4)=4,$$ $$A_{22} = (+1)(2)=2.$$

Answer: Minors: $M_{11}=3,\; M_{12}=0,\; M_{21}=-4,\; M_{22}=2$. Cofactors: $A_{11}=3,\; A_{12}=0,\; A_{21}=4,\; A_{22}=2$.

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Example 11: Expanding using Cofactors

Evaluate $$\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$$ using the cofactors of the elements of the second row.

Solution: Expansion along the second row is $$\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}.$$ Here, $$a_{21}=2, \quad a_{22}=0, \quad a_{23}=1.$$ Now compute the needed cofactors: $$A_{21} = (-1)^{2+1}\begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = - (9-16)=7,$$ $$A_{23} = (-1)^{2+3}\begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = -(10-3)=-7.$$ Then, $$\Delta = 2(7) + 0 + 1(-7) = 14-7 = 7.$$

Answer: $7$

Example 12: Zero Property of Cofactors

For $$\Delta = \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix},$$ verify that $$a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0.$$

Solution: The elements of the first row are $$a_{11}=2, \quad a_{12}=-3, \quad a_{13}=5.$$ Now compute cofactors of the second row.

For $A_{21}$: $$A_{21} = (-1)^{2+1}\begin{vmatrix} -3 & 5 \\ 5 & -7 \end{vmatrix} = -\big(21-25\big)=4.$$ For $A_{22}$: $$A_{22} = (-1)^{2+2}\begin{vmatrix} 2 & 5 \\ 1 & -7 \end{vmatrix} = -14-5=-19.$$ For $A_{23}$: $$A_{23} = (-1)^{2+3}\begin{vmatrix} 2 & -3 \\ 1 & 5 \end{vmatrix} = -(10+3)=-13.$$ Now evaluate: $$(2)(4) + (-3)(-19) + (5)(-13) = 8 + 57 - 65 = 0.$$ Thus the expression is verified.

Answer: Verified.

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Example 13: Adjoint of a $2 \times 2$ Matrix

Find the adjoint of the matrix $$A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}.$$

Solution: For a $2 \times 2$ matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix},$$ its adjoint is $$\operatorname{adj} A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.$$ So here, $$\operatorname{adj} A = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}.$$

Answer: $$\begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$$

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Example 14: Inverse of a $2 \times 2$ Matrix

Find the inverse of $$A = \begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}.$$

Solution: First compute the determinant: $$|A| = (2)(3) - (-2)(4) = 6+8=14.$$ Since $|A| \neq 0$, the inverse exists. Now, $$\operatorname{adj} A = \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}.$$ Thus, $$A^{-1} = \frac{1}{14}\begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 3/14 & 1/7 \\ -2/7 & 1/7 \end{bmatrix}.$$

Answer: $$\begin{bmatrix} 3/14 & 1/7 \\ -2/7 & 1/7 \end{bmatrix}$$

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Example 15: Property $A(\operatorname{adj} A) = |A|I$

Verify $A(\operatorname{adj} A) = |A|I$ for the matrix $$A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}.$$

Solution: First compute the determinant: $$|A| = (2)(-6) - (3)(-4) = -12 + 12 = 0.$$ So, $$|A|I = 0\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$$ Now find the adjoint: $$\operatorname{adj} A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix}.$$ Then, $$A(\operatorname{adj} A) = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}\begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} -12+12 & -6+6 \\ 24-24 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$$ Thus, $$A(\operatorname{adj} A)=|A|I.$$

Answer: Verified.

Example 16: Inverse of a $3 \times 3$ Matrix

Find the inverse of $$A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}.$$

Solution: First find the determinant. Expanding along the first column: $$|A| = 1\begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} + 0 + 3\begin{vmatrix} -1 & 2 \\ 2 & -3 \end{vmatrix}.$$ So, $$|A| = 1(8-6) + 3(3-4) = 2 - 3 = -1.$$ Hence the inverse exists.

Now compute the cofactors: $$A_{11}=2, \quad A_{12}=-9, \quad A_{13}=-6,$$ $$A_{21}=0, \quad A_{22}=-2, \quad A_{23}=-1,$$ $$A_{31}=-1, \quad A_{32}=3, \quad A_{33}=2.$$ Thus the cofactor matrix is $$\begin{bmatrix} 2 & -9 & -6 \\ 0 & -2 & -1 \\ -1 & 3 & 2 \end{bmatrix},$$ and therefore $$\operatorname{adj} A = \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix}.$$ Now, $$A^{-1} = \frac{1}{-1}\operatorname{adj} A = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}.$$

Answer: $$\begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$$

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Example 17: Matrix Polynomial to find Inverse

If $$A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix},$$ show that $A^2 - 5A + 7I = O$. Hence find $A^{-1}$.

Solution: First compute $$A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}.$$ Now evaluate $$A^2 - 5A + 7I:$$ $$\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$$ So the relation is verified.

Now post-multiply by $A^{-1}$: $$(A^2 - 5A + 7I)A^{-1} = O.$$ Using associativity, $$A - 5I + 7A^{-1} = O.$$ Thus, $$7A^{-1} = 5I - A.$$ So, $$A^{-1} = \frac{1}{7}(5I - A) = \frac{1}{7}\left(\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\right) = \frac{1}{7}\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}.$$

Answer: $$A^{-1} = \frac{1}{7}\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$$

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Example 18: Verifying Reversal Law $(AB)^{-1} = B^{-1}A^{-1}$

Verify $(AB)^{-1} = B^{-1}A^{-1}$ for $$A = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 4 & 0 \\ 2 & 5 \end{bmatrix}.$$

Solution: First compute $$AB = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 4 & 0 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} 10 & 5 \\ 6 & 5 \end{bmatrix}.$$ Now, $$|AB| = 10\cdot 5 - 5\cdot 6 = 50-30 = 20.$$ So, $$(AB)^{-1} = \frac{1}{20}\begin{bmatrix} 5 & -5 \\ -6 & 10 \end{bmatrix} = \begin{bmatrix} 1/4 & -1/4 \\ -3/10 & 1/2 \end{bmatrix}.$$ Next, $$|A| = 2\cdot 1 - 1\cdot 1 = 1,$$ so $$A^{-1} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}.$$ Also, $$|B| = 4\cdot 5 - 0\cdot 2 = 20,$$ so $$B^{-1} = \frac{1}{20}\begin{bmatrix} 5 & 0 \\ -2 & 4 \end{bmatrix}.$$ Now multiply: $$B^{-1}A^{-1} = \frac{1}{20}\begin{bmatrix} 5 & 0 \\ -2 & 4 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} = \frac{1}{20}\begin{bmatrix} 5 & -5 \\ -6 & 10 \end{bmatrix}.$$ Thus, $$B^{-1}A^{-1} = (AB)^{-1}.$$

Answer: Verified.

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Example 19: Checking Consistency of a System

Examine the consistency of the system: $$x + 3y = 5$$ $$2x + 6y = 8$$

Solution: Write the system in matrix form: $$A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}, \qquad B = \begin{bmatrix} 5 \\ 8 \end{bmatrix}.$$ Now, $$|A| = 6-6=0.$$ So the matrix is singular. Next, $$\operatorname{adj} A = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}.$$ Then, $$(\operatorname{adj} A)B = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\begin{bmatrix} 5 \\ 8 \end{bmatrix} = \begin{bmatrix} 30-24 \\ -10+8 \end{bmatrix} = \begin{bmatrix} 6 \\ -2 \end{bmatrix} \neq O.$$ Hence the system is inconsistent.

Answer: Inconsistent.

Example 20: Solving a $2 \times 2$ System

Solve the system using matrix method: $$5x + 2y = 4$$ $$7x + 3y = 5$$

Solution: Write $$A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}, \qquad X = \begin{bmatrix} x \\ y \end{bmatrix}, \qquad B = \begin{bmatrix} 4 \\ 5 \end{bmatrix}.$$ Now, $$|A| = 5\cdot 3 - 7\cdot 2 = 15-14=1 \neq 0.$$ So, $$A^{-1} = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}.$$ Hence, $$X = A^{-1}B = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4 \\ 5 \end{bmatrix} = \begin{bmatrix} 12-10 \\ -28+25 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}.$$ Therefore, $$x=2, \qquad y=-3.$$

Answer: $x = 2, y = -3$

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Example 21: Solving a $3 \times 3$ System

Solve by matrix method: $$x - y + z = 4$$ $$2x + y - 3z = 0$$ $$x + y + z = 2$$

Solution: Here, $$A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}.$$ First compute the determinant: $$|A| = 1\begin{vmatrix} 1 & -3 \\ 1 & 1 \end{vmatrix} - (-1)\begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} + 1\begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix}.$$ So, $$|A| = 1(1+3) + 1(2+3) + 1(2-1) = 4+5+1=10.$$ Thus a unique solution exists.

The cofactors are: $$A_{11}=4, \quad A_{12}=-5, \quad A_{13}=1,$$ $$A_{21}=2, \quad A_{22}=0, \quad A_{23}=-2,$$ $$A_{31}=2, \quad A_{32}=5, \quad A_{33}=3.$$ Thus, $$\operatorname{adj} A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}.$$ Now, $$X = \frac{1}{10}\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}\begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}.$$ So, $$x=2, \qquad y=-1, \qquad z=1.$$

Answer: $x = 2, y = -1, z = 1$

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Example 22: Word Problem to Linear System

The sum of three numbers is 6. If we multiply the third number by 3 and add the second number to it, we get 11. By adding first and third numbers, we get double of the second number. Find the numbers.

Solution: Let the numbers be $x, y, z$. Then: $$x+y+z=6,$$ $$y+3z=11,$$ $$x+z=2y \Rightarrow x-2y+z=0.$$ Thus, $$A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix}.$$ As shown earlier, $$|A|=9,$$ and $$\operatorname{adj} A = \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}.$$ Therefore, $$X = \frac{1}{9}\begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}\begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix} = \frac{1}{9}\begin{bmatrix} 9 \\ 18 \\ 27 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}.$$ Hence the numbers are $1, 2, 3$.

Answer: The numbers are $1, 2, 3$.

JEE Main Level

Example 23: Determinant of Adjoint

If $A$ is a $3 \times 3$ non-singular matrix such that $|A| = 2$, then find the value of $|3\operatorname{adj}(2A)|$.

Solution: Let $$X = \operatorname{adj}(2A).$$ Then since $X$ is a $3 \times 3$ matrix, $$|3X| = 3^3|X| = 27|X|.$$ So, $$|3\operatorname{adj}(2A)| = 27|\operatorname{adj}(2A)|.$$ Now use the adjoint property for a matrix of order 3: $$|\operatorname{adj}(2A)| = |2A|^{3-1} = |2A|^2.$$ Also, $$|2A| = 2^3|A| = 8\cdot 2 = 16.$$ Thus, $$|\operatorname{adj}(2A)| = 16^2 = 256.$$ Therefore, $$|3\operatorname{adj}(2A)| = 27 \cdot 256 = 6912.$$

Answer: $6912$

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Example 24: System with Non-Trivial Solutions

The system of linear equations $$x + \lambda y - z = 0,$$ $$\lambda x - y - z = 0,$$ $$x + y - \lambda z = 0$$ has a non-trivial solution for what values of $\lambda$?

Solution: A homogeneous system has a non-trivial solution if and only if the determinant of the coefficient matrix is zero. Thus, $$\begin{vmatrix} 1 & \lambda & -1 \\ \lambda & -1 & -1 \\ 1 & 1 & -\lambda \end{vmatrix} = 0.$$ Expand along the first row: $$1\begin{vmatrix} -1 & -1 \\ 1 & -\lambda \end{vmatrix} - \lambda\begin{vmatrix} \lambda & -1 \\ 1 & -\lambda \end{vmatrix} + (-1)\begin{vmatrix} \lambda & -1 \\ 1 & 1 \end{vmatrix} = 0.$$ Now evaluate: $$1(\lambda+1) - \lambda(-\lambda^2+1) - (\lambda+1)=0.$$ The first and third terms cancel, giving $$-\lambda(-\lambda^2+1)=0 \Rightarrow \lambda(\lambda^2-1)=0.$$ So, $$\lambda(\lambda-1)(\lambda+1)=0.$$ Hence, $$\lambda = 0,\; 1,\; -1.$$

Answer: $\lambda = 0, 1, -1$

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Example 25: Range of Determinant

Let $$A = \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix}, \quad 0 \le \theta \le 2\pi.$$ Find the range of values of $|A|$.

Solution: Expand along the first row: $$|A| = 1\begin{vmatrix} 1 & \sin\theta \\ -\sin\theta & 1 \end{vmatrix} - \sin\theta\begin{vmatrix} -\sin\theta & \sin\theta \\ -1 & 1 \end{vmatrix} + 1\begin{vmatrix} -\sin\theta & 1 \\ -1 & -\sin\theta \end{vmatrix}.$$ Now evaluate each minor: $$\begin{vmatrix} 1 & \sin\theta \\ -\sin\theta & 1 \end{vmatrix} = 1 + \sin^2\theta,$$ $$\begin{vmatrix} -\sin\theta & \sin\theta \\ -1 & 1 \end{vmatrix} = -\sin\theta + \sin\theta = 0,$$ $$\begin{vmatrix} -\sin\theta & 1 \\ -1 & -\sin\theta \end{vmatrix} = \sin^2\theta + 1.$$ So, $$|A| = (1+\sin^2\theta) + 0 + (1+\sin^2\theta) = 2(1+\sin^2\theta).$$ Since $$0 \le \sin^2\theta \le 1,$$ we get $$1 \le 1+\sin^2\theta \le 2.$$ Multiplying by 2, $$2 \le |A| \le 4.$$

Answer: $[2, 4]$

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