Board 2023 Class 12 Maths Q14 Solution: x when (3I + 4A)(3I − 4A) = x²I

The question

If $A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ and $(3I + 4A)(3I - 4A) = x^2I$ ⁠, then the value(s) x is/are :

(a) $\pm\sqrt{7}$

(b) $0$

(c) $\pm 5$

(d) $25$

Exam answer

\begin{aligned} A^2 &= \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I \end{aligned}

Why?

Multiply row by column. For example, row 1 times column 1:

0(0) + 1(-1) = -1

The other entries are $0(1) + 1(0) = 0$ ⁠, $-1(0) + 0(-1) = 0$ and $-1(1) + 0(0) = -1$ ⁠.

Since $AI = IA = A$ ⁠,

\begin{aligned} &(3I + 4A)(3I - 4A) \\ &= 9I - 12A + 12A - 16A^2 \\ &= 9I + 16I = 25I \end{aligned}

Why?

Expand term by term: $3I \cdot 3I = 9I$ ⁠, $3I(-4A) = -12A$ ⁠, $4A(3I) = 12A$ and $4A(-4A) = -16A^2$ ⁠.

The middle terms cancel only because $I$ and $A$ commute ( $AI = IA$ ⁠). For two general matrices, $(P + Q)(P - Q)$ is not $P^2 - Q^2$ ⁠.

Then $A^2 = -I$ turns $-16A^2$ into $+16I$ ⁠.

\begin{aligned} &9I - 12A + 12A - 16A^2 \\ &= 9I - 16A^2 \\ &= 9I - 16(-I) \\ &= 9I + 16I \\ &= (9 + 16)I = 25I \end{aligned}

$\therefore\ x^2I = 25I$ ⁠, so $x^2 = 25 \Rightarrow x = \pm 5$

Why?

Compare the numbers in front of $I$ ⁠. Both $5^2$ and $(-5)^2$ equal 25.

\begin{aligned} x^2 &= 25 \\ x &= \pm\sqrt{25} \\ x &= \pm 5 \end{aligned}

Watch out: option (d), 25, is the value of $x^2$ ⁠, not of $x$ ⁠.

Answer: (c) $\pm 5$

Where this comes from

Chapter 3, Matrices: Multiplication of matrices.

Question type: Polynomials in $A$

How to spot it: click to see

Type 7. Polynomials in $A$ ⁠: show that it gives $O$

Spot it: $A^2$ or $A^3$ mixed with numbers and $I$ ⁠, as in "show that $A^2 - 5A + 7I = O$ " or "find $k$ ".

Here: Expanding $(3I + 4A)(3I - 4A)$ leaves $9I - 16A^2$ ⁠, so find $A^2 = AA$ row by column: it is $-I$ ⁠. Then $25I = x^2I$ gives $x = \pm 5$ ⁠.

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Q14: x when (3I+4A)(3I−4A)=x2I(3I + 4A)(3I - 4A) = x^2I(3I+4A)(3I−4A)=x2I

Q14: x when $(3I + 4A)(3I - 4A) = x^2I$