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

Exercise 3.2 · Q5: Compute 3A−5B3A-5B

Exercise 3.2Scalar multiplication and difference of matrices · Section 3.4.2
The question

If

A=[2315313234373223]A=\begin{bmatrix}\frac23&1&\frac53\\[1mm]\frac13&\frac23&\frac43\\[1mm]\frac73&2&\frac23\end{bmatrix} andB=[25351152545756525],\text{and}\quad B=\begin{bmatrix}\frac25&\frac35&1\\[1mm]\frac15&\frac25&\frac45\\[1mm]\frac75&\frac65&\frac25\end{bmatrix},

then compute 3A−5B3A-5B⁠.

Exam answer 3A=[235124762]3A = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix} 5B=[235124762]5B = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix} 3A−5B=[2−23−35−51−12−24−47−76−62−2]=[000000000]\begin{aligned} &3A - 5B \\ &= \begin{bmatrix} 2-2 & 3-3 & 5-5 \\ 1-1 & 2-2 & 4-4 \\ 7-7 & 6-6 & 2-2 \end{bmatrix} \\ &= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \end{aligned}

∴ 3A−5B=O\therefore\ 3A - 5B = O⁠, the zero matrix of order 3×33\times3⁠.

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

The one idea this problem is built on

Every entry of AA is a whole number of thirds, so multiplying by 3 clears all the denominators of AA⁠. Every entry of BB is a whole number of fifths, so multiplying by 5 clears all the denominators of BB⁠. Once cleared, 3A3A and 5B5B turn out to be the same matrix of whole numbers, so their difference is the zero matrix OO⁠.

Given, to find, and the plan
Given
Two 3×33\times3 matrices AA and BB with fractional entries.
To find
The matrix 3A−5B3A-5B⁠.
Plan
(1) Check orders. (2) Compute 3A3A entry by entry. (3) Compute 5B5B entry by entry. (4) Subtract. (5) Explain the result.

Basics you need

1A number times a matrix

kA=[k aij]kA=[k\,a_{ij}]⁠: every entry is multiplied by kk⁠. The result has the same order as AA⁠. Basics: adding and scaling matrices

2Difference of matrices

For matrices of the same order, P−Q=[pij−qij]P-Q=[p_{ij}-q_{ij}]⁠.

3Clearing a denominator

3⋅m3=m3\cdot\frac{m}{3}=m and 5⋅m5=m5\cdot\frac{m}{5}=m⁠: multiplying a fraction by its own denominator leaves the numerator.

4The zero matrix

A matrix whose entries are all 0 is the zero matrix OO (of whatever order the context needs). Basics: rows, columns and entries

Solution

  1. Orders

    AA and BB each have 3 rows and 3 columns: both are 3×33\times3⁠. So 3A3A and 5B5B are 3×33\times3⁠, and 3A−5B3A-5B is defined and is 3×33\times3⁠.

    Multiplying by a number never changes the order, so the subtraction at the end is between two 3×33\times3 matrices, which is allowed.

  2. Compute 3A3A

    Multiply each entry of AA by 3:

    3A=[3⋅233⋅13⋅533⋅133⋅233⋅433⋅733⋅23⋅23]=[235124762]\begin{aligned}3A&=\begin{bmatrix}3\cdot\frac23&3\cdot1&3\cdot\frac53\\[1mm]3\cdot\frac13&3\cdot\frac23&3\cdot\frac43\\[1mm]3\cdot\frac73&3\cdot2&3\cdot\frac23\end{bmatrix}\\&=\begin{bmatrix}2&3&5\\1&2&4\\7&6&2\end{bmatrix}\end{aligned}
    Each fraction m3\frac{m}{3} becomes mm (Basic 3). The two whole-number entries also fit the pattern: 1=331=\frac33 becomes 3, and 2=632=\frac63 becomes 6.
    kAkA multiplies every entry by kk⁠. Lesson: scalar multiplication

  3. Compute 5B5B

    5B=[5⋅255⋅355⋅15⋅155⋅255⋅455⋅755⋅655⋅25]=[235124762]\begin{aligned}5B&=\begin{bmatrix}5\cdot\frac25&5\cdot\frac35&5\cdot1\\[1mm]5\cdot\frac15&5\cdot\frac25&5\cdot\frac45\\[1mm]5\cdot\frac75&5\cdot\frac65&5\cdot\frac25\end{bmatrix}\\&=\begin{bmatrix}2&3&5\\1&2&4\\7&6&2\end{bmatrix}\end{aligned}
    The same idea with fifths: 5⋅m5=m5\cdot\frac{m}{5}=m⁠, and the single whole number 1=551=\frac55 becomes 5.

  4. Subtract

    3A−5B=[2−23−35−51−12−24−47−76−62−2]=[000000000]=O\begin{aligned}3A-5B&=\begin{bmatrix}2-2&3-3&5-5\\1-1&2-2&4-4\\7-7&6-6&2-2\end{bmatrix}\\&=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O\end{aligned}
    Steps 2 and 3 produced identical matrices, so every position of the difference is a number minus itself, which is 0. The answer is the 3×33\times3 zero matrix.

  5. Why it came out to zero

    Compare the numerators: each entry of AA has the same numerator as the entry of BB in the same position, with denominator 3 instead of 5. So every entry of AA is 53\frac53 times the matching entry of BB⁠; for example 23=53⋅25\frac23=\frac53\cdot\frac25⁠. That is, A=53BA=\frac53B⁠, and multiplying by 3 gives 3A=5B3A=5B⁠, so 3A−5B=O3A-5B=O⁠.

    Spotting this relation first would give the answer without any arithmetic, but the full entry-by-entry computation above is the reliable proof; the relation explains why the answer is so simple.
Result 3A−5B=[000000000]=O,3A-5B=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O,

the 3×33\times3 zero matrix. (Textbook answer: the 3×33\times3 zero matrix.)

Check it with numbers

In decimals, the (1,1)(1,1) entries: 3×0.6667≈2.0003\times0.6667\approx2.000 and 5×0.4=25\times0.4=2⁠, difference 0. The (2,3)(2,3) entries: 3×1.3333≈4.0003\times1.3333\approx4.000 and 5×0.8=45\times0.8=4⁠, difference 0. ✓ The relation A=53BA=\frac53B at (3,1)(3,1)⁠: 53⋅75=73\frac53\cdot\frac75=\frac{7}{3}⁠, which is a31a_{31}⁠. ✓

Common mistakes
  1. Multiplying only the fractions. The scalar multiplies every entry, including the whole numbers: 3⋅1=33\cdot1=3 and 3⋅2=63\cdot2=6 in 3A3A⁠.
  2. Writing the answer as the number 0. The difference of two 3×33\times3 matrices is a 3×33\times3 matrix; here it is the zero matrix OO⁠, not the number 0.
  3. Multiplying the numerator and denominator by 3. 3⋅233\cdot\frac23 is 63=2\frac63=2⁠, not 69\frac69⁠.
What to take away
  1. A scalar multiplies every entry; choosing it to match the denominators clears fractions in one go.
  2. The difference of two equal matrices is the zero matrix of the same order.
  3. When an answer is surprisingly simple, look for the relation (here A=53BA=\frac53B⁠) that explains it.

Every fact used, and where it comes from

FactUsed inSource
Scalar multiple kA=[k aij]kA=[k\,a_{ij}]⁠, same order as AASteps 1 to 3, 5Section 3.4.2 (Lesson: scalar multiplication)
Difference P−Q=[pij−qij]P-Q=[p_{ij}-q_{ij}]Step 4Section 3.4.2
Zero matrix OOStep 4Section 3.3 (vii) (Lesson: types of matrices)
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 →