CBSE Sample Paper 2020-21 · Part B: Section III · 2 marks

Q20: $|A|$ when $A^2 = 2A$ ⁠. Or: show $A^2 - 5A + 7I = O$ and find $A^{-1}$

Sample 2020-21Ch 4 · Determinants Ch 3 · Matrices

The question

(a) If A is a square matrix of order 3 such that $A^2 = 2A$ ⁠, then find the value of $|A|$ ⁠.

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

Exam answer

(a)

\begin{aligned} A^2 &= 2A \\ \Rightarrow |AA| &= |2A| \\ \Rightarrow |A||A| &= 2^3|A| \end{aligned}

$\because |AB| = |A||B|$ and $|kA| = k^3|A|$ for order 3.

Why?

Equal matrices have equal determinants, so we take the determinant of both sides.

$kA$ multiplies every entry by $k$ ⁠. Each of the 3 rows then has a factor $k$ ⁠, and each row's factor comes out of the determinant, so $|2A| = 2^3|A| = 8|A|$ ⁠.

Watch out: $|2A| = 2|A|$ is a common slip.

\begin{aligned} \Rightarrow |A|(|A| - 8) &= 0 \end{aligned}

Why?

Bring everything to one side: $|A|^2 - 8|A| = 0$ ⁠, then take out the common factor $|A|$ ⁠.

\begin{aligned} |A||A| &= 2^3|A| = 8|A| \\ |A|^2 - 8|A| &= 0 \\ |A|(|A| - 8) &= 0 \end{aligned}

A product is 0 when one of its factors is 0:

\begin{aligned} |A| = 0 \quad &\text{or} \quad |A| - 8 = 0 \\ &\text{so} \quad |A| = 8 \end{aligned}

Do not divide by $|A|$ ⁠: it may be 0. Both values happen: $A = O$ gives $|A| = 0$ ⁠, and $A = 2I$ gives $|A| = 8$ ⁠.

$\therefore\ |A| = 0$ or $8$ ⁠.

(b)

\begin{aligned} 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} \end{aligned}

Why?

Each entry is a row of the first matrix times a column of the second:

\begin{aligned} 3(3) + 1(-1) &= 8 \\ 3(1) + 1(2) &= 5 \\ -1(3) + 2(-1) &= -5 \\ -1(1) + 2(2) &= 3 \end{aligned}

5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix},\ 7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}

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

Why?

$5A$ multiplies every entry of $A$ by 5:

\begin{aligned} 5(3) = 15,&\quad 5(1) = 5 \\ 5(-1) = -5,&\quad 5(2) = 10 \end{aligned}

$7I$ has 7 on the diagonal and 0 elsewhere. Now work out each place, two numbers at a time:

\begin{aligned} (1, 1)&: 8 - 15 = -7,\ -7 + 7 = 0 \\ (1, 2)&: 5 - 5 = 0,\ 0 + 0 = 0 \\ (2, 1)&: -5 + 5 = 0,\ 0 + 0 = 0 \\ (2, 2)&: 3 - 10 = -7,\ -7 + 7 = 0 \end{aligned}

Watch out: in the $(2, 1)$ place, $-5A$ gives $-(-5) = +5$ ⁠.

Hence proved. Also $|A| = 6 + 1 = 7 \ne 0$ ⁠, so $A^{-1}$ exists. Multiplying by $A^{-1}$ ⁠,

Why?

$|A| = 3(2) - 1(-1) = 6 + 1 = 7$ ⁠. We may multiply by $A^{-1}$ only after checking that it exists, that is $|A| \ne 0$ ⁠.

\begin{aligned} A^{-1}(A^2 - 5A + 7I) &= A^{-1}O \\ \Rightarrow A - 5I + 7A^{-1} &= O \\ \Rightarrow 7A^{-1} &= 5I - A \end{aligned}

Why?

Multiply each term on the left by $A^{-1}$ ⁠, using $A^{-1}A = I$ ⁠:

\begin{aligned} A^{-1}A^2 &= A, \quad A^{-1}A = I \\ A^{-1}I &= A^{-1}, \quad A^{-1}O = O \end{aligned}

Then move $A - 5I$ to the right side.

\begin{aligned} A - 5I + 7A^{-1} &= O \\ 7A^{-1} &= -A + 5I \\ &= 5I - A \\ A^{-1} &= \frac{1}{7}(5I - A) \end{aligned}

Here $5I = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$ ⁠.

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

Why?

Subtract entry by entry: $5 - 3 = 2$ ⁠, $0 - 1 = -1$ ⁠, $0 - (-1) = 1$ ⁠, $5 - 2 = 3$ ⁠.

Check: $A \times \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$ ⁠, which is $7I$ ⁠.

\begin{aligned} (1, 1)&: 3(2) + 1(1) = 6 + 1 = 7 \\ (1, 2)&: 3(-1) + 1(3) = -3 + 3 = 0 \\ (2, 1)&: -1(2) + 2(1) = -2 + 2 = 0 \\ (2, 2)&: -1(-1) + 2(3) = 1 + 6 = 7 \end{aligned}

$\therefore\ A^{-1} = \dfrac{1}{7}\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$

Where this comes from

Chapter 4, Determinants: |kA| and |AB|; inverse from a matrix equation.

Question type: (a) One-mark questions on $|kA|$ ⁠, $|\operatorname{adj}A|$ and $|A^{-1}|$ ; (b) A matrix equation, then "hence find $A^{-1}$ "

How to spot it: click to see

(a) Type 10. One-mark questions on $|kA|$ ⁠, $|\operatorname{adj}A|$ and $|A^{-1}|$

Spot it: an MCQ about a matrix you are not given, only its order.

Here: Only the order 3 and $A^2 = 2A$ are given: take determinants, $|A||A| = |2A| = 2^3|A|$ ⁠, so $|A|(|A| - 8) = 0$ and $|A| = 0$ or 8.

(b) Type 7. A matrix equation, then "hence find $A^{-1}$ "

Spot it: "show that $A^2 - 5A + 7I = O$ " (or a similar equation), then "hence find $A^{-1}$ ".

Here: Show $A^2 - 5A + 7I = O$ entry by entry. Then, as $|A| = 7 \ne 0$ ⁠, multiply by $A^{-1}$ ⁠: $A - 5I + 7A^{-1} = O$ ⁠, so $A^{-1} = \frac17(5I - A)$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 3, Matrices.

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Determinants in past papers: question 33 of 34 · see them all

Q20: ∣A∣|A|∣A∣ when A2=2AA^2 = 2AA2=2A⁠. Or: show A2−5A+7I=OA^2 - 5A + 7I = OA2−5A+7I=O and find A−1A^{-1}A−1

Q20: $|A|$ when $A^2 = 2A$ ⁠. Or: show $A^2 - 5A + 7I = O$ and find $A^{-1}$