Skip to content
Question

Q.(a) If A=[10−17]A = \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix}, find the value of k such that A2−8A+kI=0A^2 - 8A + kI = 0.

(OR)
(b) If [x−y2x+z2x−y3z+w]=[−15013]\begin{bmatrix} x-y & 2x+z \\ 2x-y & 3z+w \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}, find the values of x, y, z and w.
CBSECBSE Class XII Board 2024Subjective· 2mImportance★★★★★
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Equate corresponding entries of the two sides. (a) gives k=7k=7;

(b) gives x=1,y=2,z=3,w=4x=1,y=2,z=3,w=4.

Two matrices are equal iff they have the same order and every corresponding entry is equal. Here I=[1001]I=\begin{bmatrix}1&0\\0&1\end{bmatrix} and A2=A⋅AA^2=A\cdot A.

Part (a): Given A=[10−17]A=\begin{bmatrix}1&0\\-1&7\end{bmatrix}, find kk with A2−8A+kI=0A^2-8A+kI=0.

  1. Compute A2A^2: A2=[10−17][10−17]=[10−849]A^2=\begin{bmatrix}1&0\\-1&7\end{bmatrix}\begin{bmatrix}1&0\\-1&7\end{bmatrix}=\begin{bmatrix}1&0\\-8&49\end{bmatrix}.
  2. Form A2−8A+kIA^2-8A+kI: [10−849]−8[10−17]+k[1001]=[1−8+k0−8+849−56+k]\begin{bmatrix}1&0\\-8&49\end{bmatrix}-8\begin{bmatrix}1&0\\-1&7\end{bmatrix}+k\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}1-8+k&0\\-8+8&49-56+k\end{bmatrix}.
  3. Set equal to the zero matrix: the (1,1)(1,1) entry gives 1−8+k=0⇒k=71-8+k=0\Rightarrow k=7, and the (2,2)(2,2) entry gives 49−56+k=0⇒k=749-56+k=0\Rightarrow k=7 (consistent). …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.