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Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMatrices1 Mark
Question
Let A = $\left[\begin{array}{ll} {2} & {4} \\ {3} & {2} \end{array}\right]$, B = $\left[\begin{array}{ll} {1} & {3} \\ {-2} & {5} \end{array}\right]$, C = $\left[\begin{array}{cc} {-2} & {5} \\ {3} & {4} \end{array}\right]$. Find each of the following:
  1. A + B
  2. A - B
  3. 3A - C
  4. AB
  5. BA
✓

Answer

  1. A + B = $\left[\begin{array}{ll} {2} & {4} \\ {3} & {2} \end{array}\right]+\left[\begin{array}{cc} {1} & {3} \\ {-2} & {5} \end{array}\right]$
    = $\left[\begin{array}{cc} {2+1} & {4+3} \\ {3-2} & {2+5} \end{array}\right]$
    = $\left[\begin{array}{ll} {3} & {7} \\ {1} & {7} \end{array}\right]$
  2. A – B = $\left[\begin{array}{ll} {2} & {4} \\ {3} & {2} \end{array}\right]-\left[\begin{array}{cc} {1} & {3} \\ {-2} & {5} \end{array}\right]$
    = $\left[\begin{array}{cc} {2-1} & {4-3} \\ {3-(-2)} & {2-5} \end{array}\right]$
    = $\left[\begin{array}{cc} {1} & {1} \\ {5} & {-3} \end{array}\right]$
  3. We have,
    3A - C = $3\left[\begin{array}{cc} {2} & {4} \\ {3} & {2} \end{array}\right]-\left[\begin{array}{cc} {-2} & {5} \\ {3} & {4} \end{array}\right]$
    = $\left[\begin{array}{ll} {3 \times 2} & {3 \times 4} \\ {3 \times 3} & {3 \times 2} \end{array}\right]-\left[\begin{array}{cc} {-2} & {5} \\ {3} & {4} \end{array}\right]$
    = $\left[\begin{array}{cc} {6} & {12} \\ {9} & {6} \end{array}\right]-\left[\begin{array}{cc} {-2} & {5} \\ {3} & {4} \end{array}\right]$
    = $\left[\begin{array}{cc} {6+2} & {12-5} \\ {9-3} & {6-4} \end{array}\right]$
    = $\left[\begin{array}{ll} {8} & {7} \\ {6} & {2} \end{array}\right]$
  4. We have,
    AB = $\left[\begin{array}{cc} {2} & {4} \\ {3} & {2} \end{array}\right]\left[\begin{array}{cc} {1} & {3} \\ {-2} & {5} \end{array}\right]$
    = $\left[\begin{array}{ll} {2(1)+4(-2)} & {2(3)+4(5)} \\ {3(1)+2(-2)} & {3(3)+2(5)} \end{array}\right]$
    = $\left[\begin{array}{ll} {2-8} & {6+20} \\ {3-4} & {9+10} \end{array}\right]$
    = $\left[\begin{array}{cc} {-6} & {26} \\ {-1} & {19} \end{array}\right]$
  5. We have
    BA = $\left[\begin{array}{cc} {1} & {3} \\ {-2} & {5} \end{array}\right]\left[\begin{array}{ll} {2} & {4} \\ {3} & {2} \end{array}\right]$
    = $\left[\begin{array}{cc} {1(2)+3(3)} & {1(4)+3(2)} \\ {-2(2)+5(3)} & {-2(4)+5(2)} \end{array}\right]$
    = $\left[\begin{array}{rr} {2+9} & {4+6} \\ {-4+15} & {-8+10} \end{array}\right]$
    = $\left[\begin{array}{ll} {11} & {10} \\ {11} & {2} \end{array}\right]$

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