Bright Tutorials Bright Tutorials
ICSE Class 10 Mathematics Question 23 of 28

Short Answer Questions 2 — Question 23

Back to all questions
23
Question

Question 100

If A = [−1320],B=[1−203],C=[1−4] and D=[41]\begin{bmatrix*}[r] -1 & 3 \\ 2 & 0 \end{bmatrix*}, B = \begin{bmatrix*}[r] 1 & -2 \\ 0 & 3 \end{bmatrix*}, C = \begin{bmatrix*}[r] 1 & -4 \end{bmatrix*} \text{ and } D = \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*}

(a) Is the product AC possible? Justify your answer.

(b) Find the matrix X, such that X = AB + B2 − DC

Answer

(a) Order of matrix A = 2 × 2, Order of matrix C = 1 × 2

The product AC is not possible as the no. of columns in A is not equal to the no. of rows in C.

Hence, product AC is not possible.

(b) Given,

⇒X=AB+B2−DC⇒X=[−1320][1−203]+[1−203][1−203]−[41][1−4]⇒X=[−1×1+3×0−1×−2+3×32×1+0×02×−2+0×3]+[1×1+−2×01×−2+−2×30×1+3×00×−2+3×3]−[4×14×−41×11×−4]⇒X=[−1+02+92+0−4+0]+[1+0−2−60+00+9]−[4−161−4]⇒X=[−1112−4]+[1−809]−[4−161−4]⇒X=[−1+1−411+(−8)−(−16)2+0−1(−4)+9−(−4)]⇒X=[−411−8+161−4+9+4]⇒X=[−41919]\Rightarrow X = AB + B^2 - DC \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -1 & 3 \\ 2 & 0 \end{bmatrix*}\begin{bmatrix*}[r] 1 & -2 \\ 0 & 3 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & -2 \\ 0 & 3 \end{bmatrix*}\begin{bmatrix*}[r] 1 & -2 \\ 0 & 3 \end{bmatrix*} - \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & -4 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -1 \times 1 + 3 \times 0 & -1 \times -2 + 3 \times 3 \\ 2 \times 1 + 0 \times 0 & 2 \times -2 + 0 \times 3 \end{bmatrix*} + \begin{bmatrix*}[r] 1 \times 1 + -2 \times 0 & 1 \times -2 + -2 \times 3 \\ 0 \times 1 + 3 \times 0 & 0 \times -2 + 3 \times 3 \end{bmatrix*} - \begin{bmatrix*}[r] 4 \times 1 & 4 \times -4 \\ 1 \times 1 & 1 \times -4 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -1 + 0 & 2 + 9 \\ 2 + 0 & -4 + 0 \end{bmatrix*} + \begin{bmatrix*}[r] 1 + 0 & -2 - 6 \\ 0 + 0 & 0 + 9 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & -16 \\ 1 & -4 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -1 & 11 \\ 2 & -4 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & -8 \\ 0 & 9 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & -16 \\ 1 & -4 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -1 + 1 - 4 & 11 + (-8) - (-16) \\ 2 + 0 - 1 & (-4) + 9 - (-4) \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -4 & 11 - 8 + 16 \\ 1 & -4 + 9 + 4 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -4 & 19 \\ 1 & 9 \end{bmatrix*}

Hence, X = [−41919].\begin{bmatrix*}[r] -4 & 19 \\ 1 & 9 \end{bmatrix*}.