NCERT Class 12 Mathematics — Determinants, Exercise 4.4. All 18 questions solved.

Exercise 4.4 turns the cofactors of Exercise 4.3 into an inverse. Two definitions carry the whole exercise.

The adjoint of $\mathrm{A}$ is the transpose of its matrix of cofactors:

$$\mathrm{adj}\,\mathrm{A} = \big[\mathrm{A}_{ij}\big]^{T}$$

and from it the inverse, which exists exactly when $|\mathrm{A}| \neq 0$:

$$\mathrm{A}^{-1} = \frac{1}{|\mathrm{A}|}\,\mathrm{adj}\,\mathrm{A}$$

For a $2\times2$ matrix there is a shortcut worth memorising — swap the entries on the main diagonal, negate the other two:

$$\mathrm{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \quad\Rightarrow\quad \mathrm{adj}\,\mathrm{A} = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$

Key insight. The transpose in the definition of the adjoint is where most marks are lost. Compute all nine cofactors, write them into a grid in their natural positions, and then transpose — cofactor $\mathrm{A}_{12}$ ends up in row 2, column 1 of the adjoint. Skipping that step gives a matrix that is wrong in exactly the four off-diagonal places.

There is also a second route to an inverse, used in questions 13, 15 and 16: if a matrix satisfies a polynomial equation, multiplying through by $\mathrm{A}^{-1}$ rearranges it into a formula for the inverse, with no cofactors at all.

Questions 1 and 2 — find the adjoint.

Question 1

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

Solution. Using the $2\times2$ rule: swap $1$ and $4$, negate $2$ and $3$.

$$\mathrm{adj}\,\mathrm{A} = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$$

Question 2

$\begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}$

Solution. Nine cofactors, then transpose. Taking them row by row:

$$\mathrm{A}_{11} = +\begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = 3, \quad \mathrm{A}_{12} = -\begin{vmatrix} 2 & 5 \\ -2 & 1 \end{vmatrix} = -12, \quad \mathrm{A}_{13} = +\begin{vmatrix} 2 & 3 \\ -2 & 0 \end{vmatrix} = 6$$

$$\mathrm{A}_{21} = -\begin{vmatrix} -1 & 2 \\ 0 & 1 \end{vmatrix} = 1, \quad \mathrm{A}_{22} = +\begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix} = 5, \quad \mathrm{A}_{23} = -\begin{vmatrix} 1 & -1 \\ -2 & 0 \end{vmatrix} = 2$$

$$\mathrm{A}_{31} = +\begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} = -11, \quad \mathrm{A}_{32} = -\begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} = -1, \quad \mathrm{A}_{33} = +\begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = 5$$

Writing these in a grid and transposing:

$$\mathrm{adj}\,\mathrm{A} = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}$$

Questions 3 and 4 — verify $\mathrm{A}(\mathrm{adj}\,\mathrm{A}) = (\mathrm{adj}\,\mathrm{A})\mathrm{A} = |\mathrm{A}|\,\mathrm{I}$.

Question 3

$\begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$

Solution.

$$|\mathrm{A}| = (2)(-6) – (3)(-4) = -12 + 12 = 0$$

$$\mathrm{adj}\,\mathrm{A} = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix}$$

$$\mathrm{A}(\mathrm{adj}\,\mathrm{A}) = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}\begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

Multiplying in the other order gives the zero matrix too, and $|\mathrm{A}|\,\mathrm{I} = 0\cdot\mathrm{I}$ is also zero, so the identity holds. Note what this matrix is telling you: $|\mathrm{A}| = 0$, so it is singular and has no inverse — the adjoint exists but cannot be divided by zero.

Question 4

$\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$

Solution. Expanding along the second column, which has two zeros:

$$|\mathrm{A}| = -(-1)\begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix} = 1(9 + 2) = 11$$

Computing the nine cofactors and transposing gives

$$\mathrm{adj}\,\mathrm{A} = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}$$

$$\mathrm{A}(\mathrm{adj}\,\mathrm{A}) = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} = 11\,\mathrm{I} = |\mathrm{A}|\,\mathrm{I}$$

and the product in the reverse order is the same. The off-diagonal zeros are exactly the “elements times cofactors of a different row” sums from Exercise 4.3 — which is why they vanish.

Questions 5 to 11 — find the inverse, if it exists.

Question 5

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

Solution. $|\mathrm{A}| = 6 + 8 = 14 \neq 0$, so the inverse exists.

$$\mathrm{A}^{-1} = \frac{1}{14}\begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$$

Question 6

$\begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}$

Solution. $|\mathrm{A}| = -2 + 15 = 13 \neq 0$.

$$\mathrm{A}^{-1} = \frac{1}{13}\begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}$$

Question 7

$\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}$

Solution. The matrix is upper triangular, so its determinant is the product of the diagonal: $|\mathrm{A}| = 1 \times 2 \times 5 = 10$.

Computing the cofactors and transposing:

$$\mathrm{A}^{-1} = \frac{1}{10}\begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}$$

The inverse of an upper triangular matrix is upper triangular — a useful check.

Question 8

$\begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$

Solution. Lower triangular, so $|\mathrm{A}| = (1)(3)(-1) = -3$.

$$\mathrm{A}^{-1} = \frac{-1}{3}\begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}$$

Question 9

$\begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}$

Solution. Expanding along the first row:

$$|\mathrm{A}| = 2(-1 – 0) – 1(4 – 0) + 3(8 – 7) = -2 – 4 + 3 = -3$$

$$\mathrm{A}^{-1} = \frac{-1}{3}\begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix}$$

Question 10

$\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$

Solution.

$$|\mathrm{A}| = 1(8 – 6) + 1(0 + 9) + 2(0 – 6) = 2 + 9 – 12 = -1$$

Since $|\mathrm{A}| = -1$, dividing by it simply negates the adjoint:

$$\mathrm{A}^{-1} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$$

Question 11

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$

Solution. Expanding along the first row:

$$|\mathrm{A}| = 1\big(-\cos^2\alpha – \sin^2\alpha\big) = -(\cos^2\alpha + \sin^2\alpha) = -1$$

Computing the adjoint and dividing by $-1$ returns the original matrix:

$$\mathrm{A}^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix} = \mathrm{A}$$

A matrix equal to its own inverse is called involutory, and you can confirm it directly: $\mathrm{A}^2 = \mathrm{I}$, using $\cos^2\alpha + \sin^2\alpha = 1$ throughout.

Question 12

Let $\mathrm{A} = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$ and $\mathrm{B} = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$. Verify that $(\mathrm{AB})^{-1} = \mathrm{B}^{-1}\mathrm{A}^{-1}$.

Solution. First the left side.

$$\mathrm{AB} = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}$$

$$|\mathrm{AB}| = (67)(61) – (87)(47) = 4087 – 4089 = -2$$

$$(\mathrm{AB})^{-1} = \frac{1}{-2}\begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} = \frac{1}{2}\begin{bmatrix} -61 & 87 \\ 47 & -67 \end{bmatrix}$$

Now the right side. $|\mathrm{A}| = 15 – 14 = 1$ and $|\mathrm{B}| = 54 – 56 = -2$:

$$\mathrm{A}^{-1} = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}, \qquad \mathrm{B}^{-1} = \frac{1}{-2}\begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}$$

$$\mathrm{B}^{-1}\mathrm{A}^{-1} = \frac{1}{-2}\begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = \frac{1}{-2}\begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}$$

The two agree. The reversal of order is the point: $(\mathrm{AB})^{-1}$ is $\mathrm{B}^{-1}\mathrm{A}^{-1}$, not $\mathrm{A}^{-1}\mathrm{B}^{-1}$, because matrix multiplication does not commute.

Question 13

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

Solution.

$$\mathrm{A}^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$$

$$\mathrm{A}^2 – 5\mathrm{A} + 7\mathrm{I} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} – \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

For the inverse, the word “hence” means: use the equation, not the adjoint. Multiply through by $\mathrm{A}^{-1}$, which exists since $|\mathrm{A}| = 7 \neq 0$:

$$\mathrm{A} – 5\mathrm{I} + 7\mathrm{A}^{-1} = \mathrm{O} \quad\Rightarrow\quad \mathrm{A}^{-1} = \frac{1}{7}\big(5\mathrm{I} – \mathrm{A}\big)$$

$$= \frac{1}{7}\left(\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} – \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\right) = \frac{1}{7}\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$$

Question 14

For the matrix $\mathrm{A} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$, find the numbers $a$ and $b$ such that $\mathrm{A}^2 + a\mathrm{A} + b\mathrm{I} = \mathrm{O}$.

Solution.

$$\mathrm{A}^2 = \begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix}$$

$$\mathrm{A}^2 + a\mathrm{A} + b\mathrm{I} = \begin{bmatrix} 11 + 3a + b & 8 + 2a \\ 4 + a & 3 + a + b \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

Two matrices are equal only if every entry matches, giving four equations. The off-diagonal ones are simplest:

$$8 + 2a = 0 \Rightarrow a = -4, \qquad 4 + a = 0 \Rightarrow a = -4$$

Consistent. Substituting into $3 + a + b = 0$ gives $b = 1$, and the remaining entry checks: $11 + 3(-4) + 1 = 0$.

Question 15

For the matrix $\mathrm{A} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$, show that $\mathrm{A}^3 – 6\mathrm{A}^2 + 5\mathrm{A} + 11\mathrm{I} = \mathrm{O}$. Hence, find $\mathrm{A}^{-1}$.

Solution. Building up the powers:

$$\mathrm{A}^2 = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}, \qquad \mathrm{A}^3 = \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix}$$

Substituting confirms $\mathrm{A}^3 – 6\mathrm{A}^2 + 5\mathrm{A} + 11\mathrm{I} = \mathrm{O}$.

Now multiply through by $\mathrm{A}^{-1}$ and rearrange:

$$\mathrm{A}^2 – 6\mathrm{A} + 5\mathrm{I} + 11\mathrm{A}^{-1} = \mathrm{O}$$

$$\mathrm{A}^{-1} = -\frac{1}{11}\big(\mathrm{A}^2 – 6\mathrm{A} + 5\mathrm{I}\big)$$

$$= -\frac{1}{11}\begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix}$$

Question 16

If $\mathrm{A} = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$, verify that $\mathrm{A}^3 – 6\mathrm{A}^2 + 9\mathrm{A} – 4\mathrm{I} = \mathrm{O}$ and hence find $\mathrm{A}^{-1}$.

Solution.

$$\mathrm{A}^2 = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix}, \qquad \mathrm{A}^3 = \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix}$$

Substituting gives the zero matrix, as required.

Multiplying the identity by $\mathrm{A}^{-1}$:

$$\mathrm{A}^2 – 6\mathrm{A} + 9\mathrm{I} – 4\mathrm{A}^{-1} = \mathrm{O} \quad\Rightarrow\quad \mathrm{A}^{-1} = \frac{1}{4}\big(\mathrm{A}^2 – 6\mathrm{A} + 9\mathrm{I}\big)$$

$$= \frac{1}{4}\begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}$$

Note the sign difference from question 15: there the constant term was $+11$ and the inverse picked up a minus; here it is $-4$ and it does not. Rearranging carefully beats memorising a pattern.

Question 17

Let A be a nonsingular square matrix of order $3\times3$. Then $|\mathrm{adj}\,\mathrm{A}|$ is equal to

Solution. Start from the identity verified in questions 3 and 4 and take determinants of both sides.

$$\mathrm{A}(\mathrm{adj}\,\mathrm{A}) = |\mathrm{A}|\,\mathrm{I}$$

$$|\mathrm{A}|\,|\mathrm{adj}\,\mathrm{A}| = \big||\mathrm{A}|\,\mathrm{I}\big| = |\mathrm{A}|^3$$

using $|k\mathrm{B}| = k^n|\mathrm{B}|$ with $n = 3$ on the right — the same scaling rule as Exercise 4.1 question 4. Since A is nonsingular, $|\mathrm{A}| \neq 0$ and we may divide:

$$|\mathrm{adj}\,\mathrm{A}| = |\mathrm{A}|^{2}$$

In general $|\mathrm{adj}\,\mathrm{A}| = |\mathrm{A}|^{n-1}$; option (C) is what you get by forgetting to divide.

Question 18

If A is an invertible matrix of order 2, then $\det(\mathrm{A}^{-1})$ is equal to

Solution. Take determinants of the defining property of an inverse, using the fact that the determinant of a product is the product of the determinants.

$$\mathrm{A}\mathrm{A}^{-1} = \mathrm{I} \quad\Rightarrow\quad \det(\mathrm{A})\cdot\det(\mathrm{A}^{-1}) = \det(\mathrm{I}) = 1$$

$$\det(\mathrm{A}^{-1}) = \frac{1}{\det(\mathrm{A})}$$

The order 2 is a red herring — the result holds for any invertible square matrix.

Common mistakes

• Forgetting to transpose the cofactor matrix. The adjoint is $[\mathrm{A}_{ij}]^T$. Omitting the transpose gives the right nine numbers in the wrong four places, and every $3\times3$ inverse in questions 7 to 11 comes out wrong.
• Dividing by the determinant before checking it is non-zero. Question 3 has $|\mathrm{A}| = 0$: the adjoint exists, the inverse does not.
• Reversing the order in question 12. $(\mathrm{AB})^{-1} = \mathrm{B}^{-1}\mathrm{A}^{-1}$. Matrix multiplication does not commute, so $\mathrm{A}^{-1}\mathrm{B}^{-1}$ is a genuinely different matrix.
• Ignoring the word “hence”. Questions 13, 15 and 16 want the inverse derived from the polynomial identity just proved, not recomputed from cofactors. The cofactor route gives the same answer but misses the point of the question.
• Mishandling the sign when rearranging for $\mathrm{A}^{-1}$. In question 15 the constant is $+11\mathrm{I}$ and the inverse carries a minus; in question 16 it is $-4\mathrm{I}$ and it does not. Rearrange each time rather than pattern-matching.
• Applying $|\mathrm{adj}\,\mathrm{A}| = |\mathrm{A}|^3$ for a $3\times3$. The exponent is $n – 1 = 2$. Question 17 offers the cubed version as a distractor.

Practise next

• Exercise 4.5 — solving systems of linear equations with the matrix inverse, which is what all of this machinery was built for.
• Chapter 3, Exercise 3.4 — invertible matrices from the matrices chapter, covering the same ground from the other side.