Class 12 Maths, Worked in Full Past papersGet updates
Chapter 3 · Matrices

Exercise 3.2 · Q8: Find XX from 2X+Y2X+Y

Exercise 3.2Matrix equations with scalar multiples · Sections 3.4.2 to 3.4.4
The question

Find XX⁠, if

Y=[3214]Y=\begin{bmatrix}3&2\\1&4\end{bmatrix} and2X+Y=[10−32].\text{and}\quad 2X+Y=\begin{bmatrix}1&0\\-3&2\end{bmatrix}.
Exam answer

From the given equation, 2X=(2X+Y)−Y2X = (2X + Y) - Y⁠.

2X=[10−32]−[3214]=[1−30−2−3−12−4]=[−2−2−4−2]\begin{aligned} 2X &= \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} \\ &= \begin{bmatrix} 1-3 & 0-2 \\ -3-1 & 2-4 \end{bmatrix} \\ &= \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix} \end{aligned}

∴ X=[−1−1−2−1]\therefore\ X = \begin{bmatrix} -1 & -1 \\ -2 & -1 \end{bmatrix}

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Solve 2X+Y=M2X+Y=M exactly as you would solve 2x+y=m2x+y=m for a number xx⁠: subtract YY from both sides to get 2X=M−Y2X=M-Y⁠, then multiply by 12\frac12⁠. Each move is allowed because matrix addition has an additive inverse (−Y-Y⁠) and an additive identity (OO⁠), and scalar multiplication behaves like ordinary multiplication.

Given, to find, and the plan
Given
The 2×22\times2 matrix YY⁠, and the value of 2X+Y2X+Y⁠, also 2×22\times2⁠.
To find
The matrix XX⁠.
Plan
(1) Find the order of XX⁠. (2) Isolate 2X2X by subtracting YY⁠. (3) Compute M−YM-Y entry by entry. (4) Halve each entry. (5) Check.

Basics you need

1Subtracting matrices

Same order; subtract the entries in matching positions. Basics: adding like with like

2Additive inverse and identity

Y+(−Y)=OY+(-Y)=O and Z+O=ZZ+O=Z for matrices of the same order. These let us "move YY to the other side".

3Scalar multiples

kAkA multiplies every entry by kk⁠, and 12(2X)=X\frac12(2X)=X⁠.

4Solving a linear equation

Undo the operations in reverse order: first undo "+Y+Y", then undo "×2\times2". Basics: solving equations

Solution

  1. The order of XX

    Write MM for the given value of 2X+Y2X+Y⁠:

    M=[10−32].M=\begin{bmatrix}1&0\\-3&2\end{bmatrix}.

    The sum 2X+Y2X+Y is defined only if 2X2X has the same order as YY⁠, which is 2×22\times2⁠. Multiplying by 2 does not change the order, so XX is 2×22\times2⁠.

    This tells us what we are looking for (a 2×22\times2 matrix) and that the subtraction below is between matrices of the same order.

  2. Isolate 2X2X

    Add −Y-Y to both sides of 2X+Y=M2X+Y=M⁠:

    (2X+Y)+(−Y)=M+(−Y)2X+(Y+(−Y))=M−Y2X+O=M−Y2X=M−Y\begin{aligned} (2X+Y)+(-Y)&=M+(-Y)\\ 2X+(Y+(-Y))&=M-Y\\ 2X+O&=M-Y\\ 2X&=M-Y \end{aligned}
    Line 2 uses the associative law; line 3 uses Y+(−Y)=OY+(-Y)=O (additive inverse); line 4 uses 2X+O=2X2X+O=2X (additive identity). This is the same reasoning the textbook spells out in Example 8, and it is what "subtract YY from both sides" means for matrices.
    Lesson: properties of addition

  3. Compute M−YM-Y

    2X=[1−30−2−3−12−4]=[−2−2−4−2]2X=\begin{bmatrix}1-3&0-2\\-3-1&2-4\end{bmatrix}=\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}
    Subtract in each position. Every entry of YY is larger than the matching entry of MM⁠, so every entry of the difference is negative; −3−1=−4-3-1=-4 is the one to do carefully.
    Basics: subtracting matrices

  4. Multiply by 12\frac12

    X=12[−2−2−4−2]=[−1−1−2−1]X=\frac12\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}=\begin{bmatrix}-1&-1\\-2&-1\end{bmatrix}
    12(2X)=(12⋅2)X=1⋅X=X\frac12(2X)=(\frac12\cdot2)X=1\cdot X=X⁠, so multiplying both sides by 12\frac12 leaves XX on the left. On the right, each entry is halved.

  5. Check

    Compute 2X2X⁠, then add YY⁠:

    2X+Y=[−2+3−2+2−4+1−2+4]=[10−32].\begin{aligned}2X+Y&=\begin{bmatrix}-2+3&-2+2\\-4+1&-2+4\end{bmatrix}\\&=\begin{bmatrix}1&0\\-3&2\end{bmatrix}.\end{aligned}

    This is MM⁠, the given matrix. ✓

    Substituting the answer back into the original equation is the surest test that no sign was lost.
Result X=[−1−1−2−1]X=\begin{bmatrix}-1&-1\\-2&-1\end{bmatrix}

(Textbook answer: the same matrix.)

Check it with numbers

Entry by entry, the equation 2xij+yij=mij2x_{ij}+y_{ij}=m_{ij} gives xij=mij−yij2x_{ij}=\frac{m_{ij}-y_{ij}}{2}⁠: 1−32=−1\frac{1-3}{2}=-1⁠, 0−22=−1\frac{0-2}{2}=-1⁠, −3−12=−2\frac{-3-1}{2}=-2⁠, 2−42=−1\frac{2-4}{2}=-1⁠. These are the four entries found. ✓

Common mistakes
  1. Halving before subtracting. 12M−Y\frac12M-Y is not XX⁠; undo "+Y+Y" first, then the "×2\times2".
  2. Adding YY instead of subtracting it. Moving YY across the equals sign changes its sign.
  3. −3−1-3-1 written as −2-2⁠. Two negatives in a subtraction like this add up: −3−1=−4-3-1=-4⁠.
What to take away
  1. Linear matrix equations with sums and scalar multiples are solved like number equations.
  2. Each step is backed by a property: additive inverse, additive identity, scalar rules.
  3. Finish by substituting back.

Every fact used, and where it comes from

FactUsed inSource
Sum defined only for equal orders; scalar multiple keeps the orderStep 1Sections 3.4.1, 3.4.2
Associative law; additive identity OO⁠; additive inverse −Y-YStep 2Section 3.4.3 (Lesson: properties)
Difference M−Y=[mij−yij]M-Y=[m_{ij}-y_{ij}]Step 3Section 3.4.2 (Lesson: addition and scalar multiplication)
Scalar multiplication, 12(2X)=X\frac12(2X)=XStep 4Sections 3.4.2, 3.4.4
The same method for 2A+3X=5B2A+3X=5BStep 2Example 8
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →